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(a)​ ​Flower​ ​extracts​ ​can​ ​be​ ​used​ ​as​ ​Acid-base​ ​indicators.​ ​Give​ ​two​ ​limitations​ ​of​ ​such ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​indicators (b)​ ​The​ ​diagram​ ​below​ ​shows​ ​spots​ ​of​ ​pure​ ​substances​ ​W,​ ​X​,​ ​and​ ​Y​ ​on​ ​a chromatography ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​paper.​ ​ ​ ​Spot​ ​Z​ ​is​ ​that​ ​of​ ​a​ ​mixture After​ ​development​ ​W,​ ​X​,​ ​and​ ​Y​ ​were​ ​found​ ​to​ ​have​ ​moved​ ​9cm​3​,​ ​4cm​3​​ ​and​ ​7cm​3 respectively. Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13. A​ ​beekeeper​ ​found​ ​that​ ​when​ ​stung​ ​by​ ​a​ ​bee,​ ​application​ ​of​ ​a​ ​little​ ​solution​ ​of​ ​sodium ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​carbonate​ ​helped​ ​to​ ​relieve​ ​the​ ​irritation​ ​of​ ​the​ ​affected​ ​area.​ ​Explain 14. 10g​ ​of​ ​sodium​ ​hydrogen​ ​carbonate​ ​were​ ​dissolved​ ​in​ ​20cm​3​​ ​of​ ​water​ ​in​ ​a​ ​boiling​ ​tube.
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Z​ ​has​ ​separated​ ​into​ ​two​ ​spots​ ​which​ ​have​ ​moved​ ​7cm​3​​ ​and​ ​9cm​3​:- On​ ​the​ ​diagram:- ​ ​I.​ ​Label​ ​the​ ​baseline​ ​and​ ​solvent​ ​front II.​ ​Show​ ​the​ ​position​ ​of​ ​all​ ​the​ ​spots​ ​after​ ​development III.​ ​Identify​ ​the​ ​substances​ ​present​ ​in​ ​mixture​ ​Z 13. A​ ​beekeeper​ ​found​ ​that​ ​when​ ​stung​ ​by​ ​a​ ​bee,​ ​application​ ​of​ ​a​ ​little​ ​solution​ ​of​ ​sodium ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​hydrogen​ ​carbonate​ ​helped​ ​to​ ​relieve​ ​the​ ​irritation​ ​of​ ​the​ ​affected​ ​area.​ ​Explain 14. 10g​ ​of​ ​sodium​ ​hydrogen​ ​carbonate​ ​were​ ​dissolved​ ​in​ ​20cm​3​​ ​of​ ​water​ ​in​ ​a​ ​boiling​ ​tube. Lemon ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​juice​ ​was​ ​then​ ​added​ ​dropwise​ ​with​ ​shaking​ ​until​ ​there​ ​was​ ​no​ ​further​ ​change. (a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15.
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Lemon ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​juice​ ​was​ ​then​ ​added​ ​dropwise​ ​with​ ​shaking​ ​until​ ​there​ ​was​ ​no​ ​further​ ​change. (a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow.
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(a)​ ​Explain​ ​the​ ​observation​ ​which​ ​was​ ​made​ ​in​ ​the​ ​boiling​ ​tube​ ​when​ ​the​ ​reaction​ ​was​ ​in progress (b)​ ​What​ ​observations​ ​would​ ​be​ ​made​ ​if​ ​the​ ​lemon​ ​juice​ ​had​ ​been​ ​added​ ​to​ ​copper turnings​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​boiling​ ​tube? 15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2.
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15. (a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder.
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(a)​ ​Complete​ ​the​ ​table​ ​below​ ​to​ ​show​ ​the​ ​colour​ ​of​ ​the​ ​given​ ​indicator​ ​in​ ​acidic​ ​and​ ​basic ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​solutions: Indicator Colour​ ​in​ ​acidic​ ​solution Basic​ ​solution Methyl​ ​orange Pink Phenolphthalein Pink 16.​ ​ ​ ​ ​ ​ ​ ​Solutions​ ​can​ ​be​ ​classified​ ​as​ ​acids,​ ​bases​ ​or​ ​neutral.​ ​The​ ​table​ ​below​ ​shows​ ​ ​solutions and​ ​their ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​pH​ ​values:- Solutions PH​ ​VALUES K L M 1.5 7.0 14.0 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Select​ ​any​ ​pair​ ​that​ ​would​ ​react​ ​to​ ​form​ ​a​ ​solution​ ​of​ ​PH​ ​7 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Identify​ ​two​ ​solutions​ ​that​ ​would​ ​react​ ​with​ ​aluminium​ ​hydroxide.​ ​Explain Air​ ​and​ ​combustion 1.​ ​ ​ ​ ​ ​ ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​prepare​ ​a​ ​sample​ ​of​ ​oxygen​ ​gas.​ ​Study​ ​it​ ​and​ ​answer ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​questions​ ​that​ ​follow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer 3.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​Oxygen​ ​can​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​a​ ​chemical​ ​equation​ ​of​ ​the​ ​reaction​ ​to​ ​produce​ ​oxygen 2. Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer 3. (a)​ ​What​ ​is​ ​rust?
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Air​ ​was​ ​passed​ ​through​ ​several​ ​reagents​ ​as​ ​shown​ ​below: (a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​takes​ ​place​ ​in​ ​the​ ​chamber​ ​containing ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Magnesium​ ​powder (b)​ ​Name​ ​one​ ​gas​ ​which​ ​escapes​ ​from​ ​the​ ​chamber​ ​containing​ ​magnesium​ ​powder. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer 3. (a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Give​ ​a​ ​reason​ ​for​ ​your​ ​answer 3. (a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6.
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(a)​ ​What​ ​is​ ​rust? (b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air.
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(b)​ ​Give​ ​two​ ​methods​ ​that​ ​can​ ​be​ ​used​ ​to​ ​prevent​ ​rusting (c)​ ​Name​ ​one​ ​substance​ ​which​ ​speeds​ ​up​ ​the​ ​rusting​ ​process 4.​ ​ ​ ​ ​ ​ ​3.0g​ ​of​ ​ ​clean​ ​ ​magnesium​ ​ ​ribbon​ ​8.0g​ ​ ​of​ ​ ​clean​ ​ ​copper​ ​metal​ ​ ​were​ ​ ​burnt​ ​ ​separately​ ​ ​in ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​equal​ ​volume​ ​ ​of​ ​ ​air​ ​and​ ​ ​both​ ​metals​ ​ ​reacted​ ​completely​ ​ ​with​ ​ ​air; ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​State​ ​and​ ​explain​ ​where​ ​there​ ​was​ ​greater​ ​change​ ​in​ ​volume​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Mg​ ​=24​ ​ ​ ​Cu​ ​=​ ​64 ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​dilute​ ​sulphuric​ ​acid​ ​and​ ​product​ ​of​ ​burnt copper 5.​ ​ ​ ​ ​ ​ ​ ​ ​Oxygen​ ​is​ ​obtained​ ​on​ ​large​ ​scale​ ​by​ ​the​ ​fractional​ ​distillation​ ​of​ ​air​ ​as​ ​shown​ ​on​ ​the​ ​flow ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​chart​ ​bellow. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air. State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment 7.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​a)​ ​Identify​ ​the​ ​substance​ ​that​ ​is​ ​removed​ ​at​ ​the​ ​filtration​ ​stage ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Explain​ ​why​ ​Carbon​ ​(IV)​ ​oxide​ ​and​ ​water​ ​are​ ​removed​ ​before​ ​liquefaction​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​ ​c)​ ​Identify​ ​the​ ​component​ ​that​ ​is​ ​collected​ ​at​ ​-186°C 6. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​study​ ​some​ ​properties​ ​of​ ​air. State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment 7. A​ ​form​ ​two​ ​student​ ​in​ ​an​ ​attempt​ ​to​ ​stop​ ​rusting​ ​put​ ​copper​ ​and​ ​Zinc​ ​ ​in​ ​contact​ ​with​ ​iron ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​as​ ​shown:- ​ ​ ​ ​ ​ ​ ​(a)​ ​State​ ​whether​ ​rusting​ ​occurred​ ​after​ ​one​ ​week​ ​if​ ​the​ ​set-ups​ ​were​ ​left​ ​out ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Explain​ ​your​ ​answer​ ​in​ ​(a)​ ​above 8. In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.
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State​ ​and​ ​explain​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment 7. A​ ​form​ ​two​ ​student​ ​in​ ​an​ ​attempt​ ​to​ ​stop​ ​rusting​ ​put​ ​copper​ ​and​ ​Zinc​ ​ ​in​ ​contact​ ​with​ ​iron ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​as​ ​shown:- ​ ​ ​ ​ ​ ​ ​(a)​ ​State​ ​whether​ ​rusting​ ​occurred​ ​after​ ​one​ ​week​ ​if​ ​the​ ​set-ups​ ​were​ ​left​ ​out ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Explain​ ​your​ ​answer​ ​in​ ​(a)​ ​above 8. In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass?
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In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​clean​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened?
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​After​ ​cooling​ ​the​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days.
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c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment.
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d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 9.​ ​ ​ ​ ​ ​ ​ ​In​ ​an​ ​experiment​ ​a​ ​gas​ ​jar​ ​containing​ ​some​ ​damp​ ​iron​ ​fillings​ ​was​ ​inverted​ ​in​ ​a​ ​water trough ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​containing​ ​some​ ​water​ ​as​ ​shown​ ​in​ ​the​ ​diagram​ ​below.​ ​The​ ​set-up​ ​was​ ​left​ ​un-disturbed for​ ​three ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​days.​ ​Study​ ​it​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow: (a)​ ​Why​ ​were​ ​the​ ​iron​ ​filings​ ​moistened? b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment. d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide (e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10.
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b)​ ​State​ ​and​ ​explain​ ​the​ ​observation​ ​made​ ​after​ ​three​ ​days. (c)​ ​State​ ​two​ ​conclusions​ ​made​ ​from​ ​the​ ​experiment. d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide (e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10. ​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass?
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d)​ ​Draw​ ​a​ ​labelled​ ​set-up​ ​of​ ​apparatus​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​oxygen​ ​using ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Sodium​ ​Peroxide (e)​ ​State​ ​t​wo​ ​uses​ ​of​ ​oxygen 10. ​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11.
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​ ​ ​In​ ​an​ ​experiment,​ ​a​ ​piece​ ​of​ ​magnesium​ ​ribbon​ ​was​ ​cleaned​ ​with​ ​steel​ ​wool.​ ​2.4g​ ​of​ ​the clean ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​magnesium​ ​ribbon​ ​was​ ​placed​ ​in​ ​a​ ​crucible​ ​and​ ​completely​ ​burnt​ ​in​ ​oxygen.​ ​After cooling​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​product​ ​weighed​ ​4.0g a)​ ​Explain​ ​why​ ​it​ ​is​ ​necessary​ ​to​ ​clean​ ​magnesium​ ​ribbon b)​ ​What​ ​observation​ ​was​ ​made​ ​in​ ​the​ ​crucible​ ​after​ ​burning​ ​magnesium​ ​ribbon? c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up?
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c)​ ​Why​ ​was​ ​there​ ​an​ ​increase​ ​in​ ​mass? d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​.
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d)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​major​ ​chemical​ ​reaction​ ​which​ ​took​ ​place​ ​in​ ​the​ ​crucible e)​ ​The​ ​product​ ​in​ ​the​ ​crucible​ ​was​ ​shaken​ ​with​ ​water​ ​and​ ​filtered.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​which​ ​was​ ​made​ ​when​ ​red​ ​and​ ​blue​ ​litmus​ ​paper​ ​were​ ​dropped​ ​into​ ​the filtrate 11. ​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air: (i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous.
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​ ​The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​F​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​sodium peroxide​ ​and water water ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i) Name​ ​gas F​…………………………………………………………………………… (ii)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​round​ ​bottomed​ ​flask​ ​was​ ​found​ ​to be ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​a​ ​strong​ ​base.​ ​Explain​ ​why​ ​this​ ​was​ ​so (iii)​ ​Which​ ​property​ ​of​ ​gas​ ​F​ ​makes​ ​it​ ​be​ ​collected​ ​by​ ​the​ ​method​ ​used​ ​in​ ​the​ ​set-up? (iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air: (i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous. ​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case.
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(iv)​ ​Give​ ​one​ ​industrial​ ​use​ ​of​ ​gas​ ​F 12.​ ​. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​properties​ ​of​ ​the​ ​components​ ​of​ ​air: (i)​ ​State​ ​two​ ​observations​ ​made​ ​during​ ​the​ ​experiment (ii)​ ​Write​ ​two​ ​chemical​ ​equations​ ​for​ ​the​ ​reactions​ ​which​ ​occurred (iii)​ ​The​ ​experiment​ ​was​ ​repeated​ ​using​ ​burning​ ​magnesium​ ​in​ ​place​ ​of​ ​phosphorous. ​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish.
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​ ​ ​ ​ ​ ​ ​There​ ​was​ ​ ​greater​ ​rise​ ​of​ ​water​ ​than​ ​in​ ​the​ ​first​ ​case.​ ​Explain​ ​this​ ​observation (iv)​ ​After​ ​the​ ​two​ ​experiments,​ ​the​ ​water​ ​in​ ​each​ ​trough​ ​was​ ​tested​ ​using​ ​blue​ ​and​ ​red litmus ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​papers.​ ​State​ ​and​ ​explain​ ​the​ ​observations​ ​of​ ​each​ ​case. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish. ​ ​ ​ ​ ​ ​The​ ​ash​ ​was​ ​found​ ​to​ ​comprise​ ​of​ ​magnesium​ ​Oxide​ ​and​ ​Magnesium​ ​nitride ​ ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​leading​ ​to​ ​formation​ ​of​ ​the​ ​magnesium​ ​nitride ​ ​ ​ ​ ​ ​(ii)​ ​A​ ​little​ ​water​ ​was​ ​added​ ​to​ ​the​ ​products​ ​in​ ​the​ ​Petri​ ​dish.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​made.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Phosphorous​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​magnesium​ ​experiment ​ ​ ​ ​ ​ ​(v)​ ​Briefly​ ​explain​ ​how​ ​a​ ​sample​ ​of​ ​nitrogen​ ​gas​ ​can​ ​be​ ​isolated​ ​from​ ​air​ ​in​ ​the​ ​laboratory 13. (a)​ ​A​ ​ ​group​ ​of​ ​students​ ​burnt​ ​a​ ​piece​ ​of​ ​Mg​ ​ribbon​ ​in​ ​air​ ​and​ ​its​ ​ash​ ​collected​ ​in​ ​a​ ​Petri dish. ​ ​ ​ ​ ​ ​The​ ​ash​ ​was​ ​found​ ​to​ ​comprise​ ​of​ ​magnesium​ ​Oxide​ ​and​ ​Magnesium​ ​nitride ​ ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​leading​ ​to​ ​formation​ ​of​ ​the​ ​magnesium​ ​nitride ​ ​ ​ ​ ​ ​(ii)​ ​A​ ​little​ ​water​ ​was​ ​added​ ​to​ ​the​ ​products​ ​in​ ​the​ ​Petri​ ​dish.​ ​State​ ​and​ ​explain​ ​the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation​ ​ ​made. ​ ​ ​ ​ ​(iii)​ ​A​ ​piece​ ​of​ ​blue​ ​litmus​ ​paper​ ​was​ ​dipped​ ​into​ ​the​ ​solution​ ​formed​ ​in​ ​(b)​ ​above. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​State​ ​the​ ​ ​ ​observation​ ​made. 14.
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​ ​ ​ ​ ​(iii)​ ​A​ ​piece​ ​of​ ​blue​ ​litmus​ ​paper​ ​was​ ​dipped​ ​into​ ​the​ ​solution​ ​formed​ ​in​ ​(b)​ ​above. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​State​ ​the​ ​ ​ ​observation​ ​made. 14. A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M..................................................................................
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14. A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................
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A​ ​form​ ​one​ ​class​ ​carried​ ​out​ ​an​ ​experiment​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air.​ ​The diagram ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​below​ ​shows​ ​the​ ​set-up​ ​of​ ​the​ ​experiment​ ​and​ ​also​ ​the​ ​observation​ ​made. ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15.
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​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​At​ ​the​ ​beginning (ii) observation​ ​at​ ​the​ ​end​ ​of​ ​the​ ​experiment ​ ​ ​ ​ ​(a)​ ​(i)​ ​Identify​ ​substance​ ​M.................................................................................. ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16.
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​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​State​ ​two​ ​reasons​ ​for​ ​the​ ​suitability​ ​of​ ​substance​ ​M​ ​for​ ​this​ ​experiment ​ ​ ​ ​ ​(b)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​reaction​ ​of​ ​substance​ ​M​ ​and​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​(c)​ ​(i)​ ​Using​ ​the​ ​letters​ ​Y​ ​and​ ​X​ ​write​ ​an​ ​expression​ ​for​ ​the​ ​percentage​ ​of​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​The​ ​expression​ ​in​ ​(c)(i)​ ​above​ ​gives​ ​lower​ ​value​ ​than​ ​the​ ​expected.​ ​Explain ​ ​ ​ ​ ​ ​(d)​ ​(i)​ ​Explain​ ​the​ ​observation​ ​made​ ​when​ ​litmus​ ​paper​ ​is​ ​dipped​ ​into​ ​the​ ​beaker​ ​at​ ​the​ ​end​ ​of the ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​experiment ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Name​ ​the​ ​active​ ​part​ ​of​ ​air ................................................................................................ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(iii)​ ​Suggest​ ​another​ ​method​ ​that​ ​can​ ​be​ ​used​ ​to​ ​determine​ ​the​ ​active​ ​part​ ​of​ ​air ​ ​ ​15. A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid.
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A​ ​piece​ ​of​ ​phosphorous​ ​was​ ​burnt​ ​in​ ​excess​ ​air.​ ​The​ ​product​ ​obtained​ ​was​ ​shaken​ ​with​ ​a small ​ ​amount​ ​of​ ​hot​ ​water​ ​to​ ​make​ ​a​ ​solution ​ ​ ​ ​ ​i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​burning​ ​of​ ​phosphorus​ ​in​ ​excess​ ​air ​ ​ ​ ​ ​ii)​ ​The​ ​solution​ ​obtained​ ​in​ ​(b)​ ​above​ ​as​ ​found​ ​to​ ​have​ ​pH​ ​of​ ​2.​ ​ ​Give​ ​reasons​ ​for​ ​this ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​observation 16. Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid. 18.Below​ ​is​ ​a​ ​list​ ​of​ ​oxides.
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Study​ ​the​ ​set-up​ ​below​ ​and​ ​answer​ ​the​ ​questions​ ​that​ ​follow:- (a)​ ​State​ ​two​ ​observations​ ​that​ ​would​ ​be​ ​made​ ​after​ ​one​ ​week.​ ​Explain (b)​ ​Write​ ​the​ ​equation​ ​of​ ​the​ ​reaction​ ​taking​ ​place​ ​in​ ​the​ ​test-tube 17. Fe​3​O​4​​ ​and​ ​FeO​ ​are​ ​oxides​ ​of​ ​iron​ ​which​ ​can​ ​be​ ​produced​ ​in​ ​the​ ​laboratory (a)​ ​Write​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​which​ ​can​ ​be​ ​used​ ​to​ ​produce​ ​each​ ​of​ ​the oxides (b)​ ​Wire​ ​an​ ​ionic​ ​equation​ ​for​ ​the​ ​reaction​ ​between​ ​the​ ​oxide,​ ​Fe​3​O​4​​ ​and​ ​a​ ​dilute​ ​acid. 18.Below​ ​is​ ​a​ ​list​ ​of​ ​oxides. MgO,​ ​N​2​O​,​ ​K​2​O,​ ​CaO​ ​ans​ ​Al​2​O​3 Select:- a)​ ​A​ ​neutral​ ​oxide. b)​ ​A​ ​highly​ ​water​ ​soluble​ ​basic​ ​oxide. c)​ ​An​ ​oxide​ ​which​ ​can​ ​react​ ​with​ ​both​ ​sodium​ ​hydroxide​ ​solution​ ​and​ ​dilute​ ​hydrochloric acid. 19.
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b)​ ​A​ ​highly​ ​water​ ​soluble​ ​basic​ ​oxide. c)​ ​An​ ​oxide​ ​which​ ​can​ ​react​ ​with​ ​both​ ​sodium​ ​hydroxide​ ​solution​ ​and​ ​dilute​ ​hydrochloric acid. 19. The​ ​diagram​ ​below​ ​shows​ ​students​ ​set-up​ ​for​ ​the​ ​preparation​ ​and​ ​collection​ ​of​ ​oxygen gas (a)​ ​Name​ ​substance​ ​X​ ​used (b)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​the​ ​reaction​ ​of​ ​sodium​ ​peroxide​ ​with​ ​the​ ​substance​ ​named in​ ​1(a) 5.​ ​Water​ ​and​ ​hydrogen 1. (a)​ ​Hydrogen​ ​can​ ​reduce​ ​coppers​ ​Oxide​ ​but​ ​not​ ​alluminium​ ​oxide.​ ​Explain (b)​ ​When​ ​water​ ​reacts​ ​with​ ​potassium​ ​metal​ ​the​ ​hydrogen​ ​produced​ ​ignites​ ​explosively ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​on​ ​the​ ​surface​ ​of​ ​water. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​What​ ​causes​ ​this​ ​ignition? ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2.
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(a)​ ​Hydrogen​ ​can​ ​reduce​ ​coppers​ ​Oxide​ ​but​ ​not​ ​alluminium​ ​oxide.​ ​Explain (b)​ ​When​ ​water​ ​reacts​ ​with​ ​potassium​ ​metal​ ​the​ ​hydrogen​ ​produced​ ​ignites​ ​explosively ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​on​ ​the​ ​surface​ ​of​ ​water. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(i)​ ​What​ ​causes​ ​this​ ​ignition? ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2. In​ ​an experiment,​ ​dry​ ​hydrogen​ ​gas​ ​was​ ​passed​ ​over​ ​hot​ ​copper​ ​(II)​ ​oxide​ ​in​ ​a​ ​combustion ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​tube​ ​as​ ​ ​shown​ ​in​ ​the​ ​diagram​ ​below:- (a)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​the​ ​other​ ​product,​ ​substance​ ​R​ ​could​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​laboratory. (b)​ ​Describe​ ​how​ ​copper​ ​could​ ​be​ ​obtained​ ​from​ ​the​ ​mixture​ ​containing​ ​copper​ ​(II)​ ​oxide 3.​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​the​ ​reaction​ ​between​ ​metals​ ​and​ ​water.
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​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(ii)​ ​Write​ ​an​ ​equation​ ​to​ ​show​ ​how​ ​this​ ​ignition​ ​occurs 2. In​ ​an experiment,​ ​dry​ ​hydrogen​ ​gas​ ​was​ ​passed​ ​over​ ​hot​ ​copper​ ​(II)​ ​oxide​ ​in​ ​a​ ​combustion ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​tube​ ​as​ ​ ​shown​ ​in​ ​the​ ​diagram​ ​below:- (a)​ ​Complete​ ​the​ ​diagram​ ​to​ ​show​ ​how​ ​the​ ​other​ ​product,​ ​substance​ ​R​ ​could​ ​be​ ​collected ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​laboratory. (b)​ ​Describe​ ​how​ ​copper​ ​could​ ​be​ ​obtained​ ​from​ ​the​ ​mixture​ ​containing​ ​copper​ ​(II)​ ​oxide 3.​ ​ ​ ​ ​ ​The​ ​setup​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​the​ ​reaction​ ​between​ ​metals​ ​and​ ​water. ​ ​ ​ ​ ​(a)​ ​Identify​ ​solid​ ​X​ ​and​ ​state​ ​its​ ​purpose ​ ​ ​ ​ ​ ​ ​ ​ ​Solid​ ​X ………………..……………………………………………………………………….. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​Purpose ……………………………………………………………………………………….. ​ ​ ​ ​ ​(b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​produces​ ​the​ ​flame. 4. Gas​ ​P​ ​was​ ​passed​ ​over​ ​heated​ ​magnesium​ ​ribbon​ ​and​ ​hydrogen​ ​gas​ ​was​ ​collected​ ​as shown in the​ ​diagram​ ​below: (i)​ ​Name​ ​gas​ ​P ...............................................................................................................
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​ ​ ​ ​ ​(b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​produces​ ​the​ ​flame. 4. Gas​ ​P​ ​was​ ​passed​ ​over​ ​heated​ ​magnesium​ ​ribbon​ ​and​ ​hydrogen​ ​gas​ ​was​ ​collected​ ​as shown in the​ ​diagram​ ​below: (i)​ ​Name​ ​gas​ ​P ............................................................................................................... (ii)​ ​Write​ ​an​ ​equation​ ​of​ ​the​ ​reaction​ ​that​ ​takes​ ​place​ ​in​ ​the​ ​combustion​ ​tube (iii)​ ​State​ ​one​ ​precaution​ ​necessary​ ​at​ ​the​ ​end​ ​of​ ​this​ ​experiment 5.​ ​ ​ ​ ​ ​ ​ ​When​ ​hydrogen​ ​is​ ​burnt​ ​and​ ​the​ ​product​ ​cooled,​ ​the​ ​following​ ​results​ ​are​ ​obtained​ ​as shown ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​diagram​ ​below: ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​formation​ ​of​ ​liquid​ ​Y ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Give​ ​a​ ​chemical​ ​test​ ​for​ ​liquid​ ​Y 6. Jane​ ​set-up​ ​the​ ​experiment​ ​as​ ​shown​ ​below​ ​to​ ​collect​ ​a​ ​gas.​ ​The​ ​wet​ ​sand​ ​was​ ​heated before heating​ ​Zinc granules (a)​ ​Complete​ ​the​ ​diagram​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​the​ ​gas (b)​ ​Why​ ​was​ ​it​ ​necessary​ ​to​ ​heat​ ​wet​ ​sand​ ​before​ ​heating​ ​Zinc​ ​granules? 7.
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(ii)​ ​Write​ ​an​ ​equation​ ​of​ ​the​ ​reaction​ ​that​ ​takes​ ​place​ ​in​ ​the​ ​combustion​ ​tube (iii)​ ​State​ ​one​ ​precaution​ ​necessary​ ​at​ ​the​ ​end​ ​of​ ​this​ ​experiment 5.​ ​ ​ ​ ​ ​ ​ ​When​ ​hydrogen​ ​is​ ​burnt​ ​and​ ​the​ ​product​ ​cooled,​ ​the​ ​following​ ​results​ ​are​ ​obtained​ ​as shown ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​in​ ​the​ ​diagram​ ​below: ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Write​ ​the​ ​equation​ ​for​ ​the​ ​formation​ ​of​ ​liquid​ ​Y ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(b)​ ​Give​ ​a​ ​chemical​ ​test​ ​for​ ​liquid​ ​Y 6. Jane​ ​set-up​ ​the​ ​experiment​ ​as​ ​shown​ ​below​ ​to​ ​collect​ ​a​ ​gas.​ ​The​ ​wet​ ​sand​ ​was​ ​heated before heating​ ​Zinc granules (a)​ ​Complete​ ​the​ ​diagram​ ​for​ ​the​ ​laboratory​ ​preparation​ ​of​ ​the​ ​gas (b)​ ​Why​ ​was​ ​it​ ​necessary​ ​to​ ​heat​ ​wet​ ​sand​ ​before​ ​heating​ ​Zinc​ ​granules? 7. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Between​ ​N​ ​and​ ​M​ ​which​ ​part​ ​should​ ​be​ ​heated​ ​first?​ ​Explain (b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​occurring​ ​in​ ​the​ ​combustion​ ​tube. 8.
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7. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​(a)​ ​Between​ ​N​ ​and​ ​M​ ​which​ ​part​ ​should​ ​be​ ​heated​ ​first?​ ​Explain (b)​ ​Write​ ​a​ ​chemical​ ​equation​ ​for​ ​the​ ​reaction​ ​occurring​ ​in​ ​the​ ​combustion​ ​tube. 8. The​ ​set-up​ ​below​ ​was​ ​used​ ​to​ ​investigate​ ​electrolysis​ ​of​ ​a​ ​certain​ ​molten​ ​compound;- (a)​ ​Complete​ ​the​ ​circuit​ ​by​ ​drawing​ ​the​ ​cell​ ​in​ ​the​ ​gap​ ​left​ ​in​ ​the​ ​diagram (b)​ ​Write​ ​half-cell​ ​equation​ ​to​ ​show​ ​what​ ​happens​ ​at​ ​the​ ​cathode (c)​ ​Using​ ​an​ ​arrow​ ​show​ ​the​ ​direction​ ​of​ ​electron​ ​flow​ ​in​ ​the​ ​diagram​ ​above 9.Hydrogen​ ​can​ ​be​ ​prepared​ ​by​ ​reacting​ ​zinc​ ​with​ ​dilute​ ​hydrochloric​ ​acid. a)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​b)​ ​Name​ ​an​ ​appropriate​ ​drying​ ​agent​ ​for​ ​hydrogen​ ​gas. c)​ ​Explain​ ​why​ ​copper​ ​metal​ ​cannot​ ​be​ ​used​ ​to​ ​prepare​ ​hydrogen​ ​gas. d)​ ​Hydrogen​ ​burns​ ​in​ ​oxygen​ ​to​ ​form​ ​an​ ​oxide. ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction.
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c)​ ​Explain​ ​why​ ​copper​ ​metal​ ​cannot​ ​be​ ​used​ ​to​ ​prepare​ ​hydrogen​ ​gas. d)​ ​Hydrogen​ ​burns​ ​in​ ​oxygen​ ​to​ ​form​ ​an​ ​oxide. ​ ​ ​ ​(i)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction. ​ ​ ​ ​(ii)​ ​State​ ​two​ ​precautions​ ​that​ ​must​ ​be​ ​taken​ ​before​ ​the​ ​combustion​ ​begins​ ​and​ ​at​ ​the end​ ​of ​ ​ ​ ​ ​ ​ ​ ​ ​the​ ​combustion. e)​ ​Give​ ​two​ ​uses​ ​of​ ​hydrogen​ ​gas. f)​ ​When​ ​zinc​ ​is​ ​heated​ ​to​ ​redness​ ​in​ ​a​ ​current​ ​of​ ​steam,​ ​hydrogen​ ​gas​ ​is​ ​obtained.​ ​Write an ​ ​ ​ ​ ​equation​ ​for​ ​the​ ​reaction. g)​ ​Element​ ​Q​ ​reacts​ ​with​ ​dilute​ ​acids​ ​but​ ​not​ ​with​ ​cold​ ​water.​ ​Element​ ​R​ ​does​ ​not​ ​react with ​ ​ ​ ​ ​dilute​ ​acids.​ ​Elements​ ​S​ ​displaces​ ​element​ ​P​ ​from​ ​its​ ​oxide.​ ​P​ ​reacts​ ​with​ ​cold​ ​water. Arrange ​ ​ ​ ​ ​the​ ​four​ ​elements​ ​in​ ​order​ ​of​ ​their​ ​reactivity,​ ​starting​ ​with​ ​the​ ​most​ ​reactive. h)​ ​Explain​ ​how​ ​hydrogen​ ​is​ ​used​ ​in​ ​the​ ​manufacture​ ​of​ ​margarine. 10. a)​ ​The​ ​set-up​ ​below​ ​ ​is​ ​used​ ​to​ ​investigate​ ​the​ ​properties​ ​ ​of​ ​ ​hydrogen.
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h)​ ​Explain​ ​how​ ​hydrogen​ ​is​ ​used​ ​in​ ​the​ ​manufacture​ ​of​ ​margarine. 10. a)​ ​The​ ​set-up​ ​below​ ​ ​is​ ​used​ ​to​ ​investigate​ ​the​ ​properties​ ​ ​of​ ​ ​hydrogen. ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​i)​ ​On​ ​the​ ​diagram,​ ​indicate​ ​what​ ​should​ ​be​ ​done​ ​for​ ​the​ ​reaction​ ​to​ ​occur ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ii)​ ​Hydrogen​ ​gas​ ​is​ ​allowed​ ​to​ ​pass​ ​through​ ​the​ ​tube​ ​for​ ​some​ ​time​ ​before​ ​it​ ​is​ ​lit.​ ​Explain ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​iii)​ ​Write​ ​an​ ​equation​ ​for​ ​the​ ​reaction​ ​that​ ​occurs​ ​in​ ​the​ ​combustion​ ​tube ​ ​ ​ ​ ​ ​ ​ ​ ​ ​iv)​ ​When​ ​the​ ​reaction​ ​is​ ​complete,​ ​hydrogen​ ​gas​ ​is​ ​passed​ ​through​ ​the​ ​apparatus​ ​until​ ​they ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​cool​ ​down​ ​.​ ​ ​ ​Explain ​ ​ ​ ​ ​ ​ ​ ​ ​ ​v)​ ​What​ ​property​ ​of​ ​hydrogen​ ​is​ ​being​ ​investigated? ​ ​ ​ ​ ​ ​ ​ ​ ​vi)​ ​What​ ​observation​ ​confirms​ ​the​ ​property​ ​stated​ ​in​ ​(​v)​ ​above? ​ ​ ​ ​ ​ ​ ​ ​ ​vii)​ ​Why​ ​is​ ​zinc​ ​oxide​ ​not​ ​used​ ​to​ ​investigate​ ​this​ ​property​ ​of​ ​hydrogen​ ​gas? 11.
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​ ​ ​ ​ ​ ​ ​ ​ ​vi)​ ​What​ ​observation​ ​confirms​ ​the​ ​property​ ​stated​ ​in​ ​(​v)​ ​above? ​ ​ ​ ​ ​ ​ ​ ​ ​vii)​ ​Why​ ​is​ ​zinc​ ​oxide​ ​not​ ​used​ ​to​ ​investigate​ ​this​ ​property​ ​of​ ​hydrogen​ ​gas? 11. The​ ​set​ ​up​ ​below​ ​was​ ​used​ ​to​ ​collect​ ​gas​ ​K,​ ​produced​ ​by​ ​the​ ​reaction​ ​between​ ​water​ ​and ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​calcium​ ​metal. (a)​ ​Name​ ​gas​ ​K…………………………………………………………….. (b)​ ​At​ ​the​ ​end​ ​of​ ​the​ ​experiment,​ ​the​ ​solution​ ​in​ ​the​ ​beaker​ ​was​ ​found​ ​to​ ​be​ ​a​ ​weak​ ​base. Explain ​ ​ ​ ​ ​ ​ ​why​ ​the​ ​solution​ ​is​ ​a​ ​weak​ ​base
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ATOMIC STRUCTURE AND THE PERIODIC TABLE Preface The structure of the atom is extensively covered here. The periodic table is the arrangement of atoms of elements based on their atomic structure. Emphasis on the trends across and down the periodic table of atoms is important for the teacher facilitator. This work should be covered before “Chemical bonding and structure”. The two complements each other in understanding periodicity. Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element 2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements.
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Candidate-user preparing for any secondary level chemistry from members of A.ATOMIC STRUCTURE The atom is the smallest particle of an element that take part in a chemical reaction.The atom is made up of three subatomic particle: (i)Protons (ii)Electrons (iii)Neutrons (i)Protons 1.The proton is positively charged 2.Is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons in a atom of an element is its Atomic number (ii)Electrons 1.The Electrons is negatively charged 2.Is found in fixed regions surrounding the centre of an atom called energy levels/orbitals. 3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element 2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20 3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes.
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3.It has a relative mass 1/1840 4.The number of protons and electrons in a atom of an element is always equal (iii)Neutrons 1.The Neutron is neither positively or negatively charged thus neutral. 2.Like protons it is found in the centre of an atom called nucleus 3.It has a relative mass 1 4.The number of protons and neutrons in a atom of an element is its Mass number Diagram showing the relative positions of protons ,electrons and neutrons in an atom of an element 2 Diagram showing the relative positions of protons, electrons and neutrons in an atom of Carbon The table below show atomic structure of the 1st twenty elements. Element Symbol Protons Electrons Neutrons Atomic number Mass number Hydrogen H 1 1 0 1 1 Helium He 2 2 2 2 4 Lithium Li 3 3 4 3 7 Beryllium Be 4 4 5 4 9 Boron B 5 5 6 5 11 Carbon C 6 6 6 6 12 Nitrogen N 7 7 7 7 14 Oxygen O 8 8 8 8 16 Fluorine F 9 9 10 9 19 Neon Ne 10 10 10 10 20 3 Sodium Na 11 11 12 11 23 Magnesium Mg 12 12 12 12 24 Aluminium Al 13 13 14 13 27 Silicon Si 14 14 14 14 28 Phosphorus P 15 15 16 15 31 Sulphur S 16 16 16 16 32 Chlorine Cl 17 17 18 17 35 Argon Ar 18 18 22 18 40 Potassium K 19 19 20 19 39 Calcium Ca 20 20 20 20 40 Most atoms of elements exist as isotopes. Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e.
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Isotopes are atoms of the same element, having the same number of protons/atomic number but different number of neutrons/mass number. By convention, isotopes are written with the mass number as superscript and the atomic number as subscript to the left of the chemical symbol of the element. i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40 4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) .
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i.e. mass number atomic number m n X symbol of element Below is the conventional method of writing the 1st twenty elements showing the mass numbers and atomic numbers; 2He 4Be 6C 147N 168O 10Ne 12Mg 14Si 16S 18Ar 20C The table below shows some common natural isotopes of some elements Element Isotopes Protons Electrons Neutrons Atomic number Mass number Hydrogen 1H(deuterium) 31H(Tritium) 1 1 1 1 1 1 0 2 3 1 1 1 1 2 17Cl 3717Cl 17 17 17 17 18 20 17 17 35 19K 4019K 19 19 19 19 20 21 19 19 39 40 4 4119K 19 19 22 19 8O 188O 8 8 8 8 8 10 8 8 16 92U 23892U 92 92 92 92 143 146 92 92 235 10Ne 10Ne 10 10 10 10 10 10 12 10 11 10 10 10 22 20 21 The mass of an average atom is very small (10-22 g).Masses of atoms are therefore expressed in relation to a chosen element. The atom recommended is 12C isotope whose mass is arbitrarily assigned as 12.000 atomic mass units(a.m.u) . All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples.
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All other atoms are compared to the mass of 12C isotope to give the relative at The relative atomic mass(RAM) is therefore defined as “the mass of average atom of an element compared to 1/12 an atom of 12C isotope whose mass is arbitrarily fixed as 12.000 atomic mass units(a.m.u) ” i.e; RAM = mass of atom of an element 1/12 of one atom of 12C isotope Accurate relative atomic masses (RAM) are got from the mass spectrometer. Mass spectrometer determines the isotopes of the element and their relative abundance/availability. Using the relative abundances/availability of the isotopes, the relative atomic mass (RAM) can be determined /calculated as in the below examples. a) Chlorine occurs as 75% 3517Cl and 25% 3717Cl isotopes. Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as; 5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope.
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Calculate the relative atomic mass of Chlorine. Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as; 5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope.
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Working 100 atoms of chlorine contains 75 atoms of 3517Cl isotopes 100 atoms of chlorine contains 75 atoms of 3717Cl isotopes Therefore; RAM of chlorine = ( 75/100 x 35) + 25/100 x 37 = 35.5 Note that: Relative atomic mass has no units More atoms of chlorine exist as 3517Cl(75%) than as 3717Cl(25%) therefore RAM is nearer to the more abundant isotope. b) Calculate the relative atomic mass of potassium given that it exist as; 5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms.
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b) Calculate the relative atomic mass of potassium given that it exist as; 5 93.1% 3919K , 0.01% 4019K , 6.89% 4119K , Working 100 atoms of potassium contains 93.1 atoms of 3919K isotopes 100 atoms of potassium contains 0.01 atoms of 4019K isotopes 100 atoms of potassium contains 6.89 atoms of 4119K isotopes Therefore; RAM of potassium = (93.1/100 x39) + (0.01/100 x 40) +(6.89 /100 x 39) = Note that: Relative atomic mass has no units More atoms of potassium exist as 3919K (93.1%) therefore RAM is nearer to the more abundant 3919K isotope. c) Calculate the relative atomic mass of Neon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, Working 100 atoms of Neon contains 90.92 atoms of 2010Ne isotopes 100 atoms of Neon contains 0.26 atoms of 2110Ne isotopes 100 atoms of Neon contains 8.82 atoms of 2210 Ne isotopes Therefore; RAM of Neon = (90.92/100 x20) + (0.26/100 x 21) +(8.82 /100 x 22) = Note that: Relative atomic mass has no units More atoms of Neon exist as 2010Ne (90.92%) therefore RAM is nearer to the more abundant 2010Ne isotope. d) Calculate the relative atomic mass of Argon given that it exist as; 90.92% 2010Ne , 0.26% 2110Ne , 8.82% 2210Ne, NB The relative atomic mass is a measure of the masses of atoms. The higher the relative atomic mass, the heavier the atom. Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus.
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Electrons are found in energy levels/orbital. An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons 6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure.
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An energy level is a fixed region around/surrounding the nucleus of an atom occupied by electrons of the same (potential) energy. By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons 6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]: 7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) 8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels 9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels 10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus 11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2 12 B.PERIODIC TABLE There are over 100 elements so far discovered.
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By convention energy levels are named 1,2,3… outwards from the region nearest to nucleus. Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons 6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]: 7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) 8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels 9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels 10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus 11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2 12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table.
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Each energy level is occupied by a fixed number of electrons: The 1st energy level is occupied by a maximum of two electrons 6 The 2nd energy level is occupied by a maximum of eight electrons The 3rd energy level is occupied by a maximum of eight electrons( or eighteen electrons if available) The 4th energy level is occupied by a maximum of eight electrons( or eighteen or thirty two electrons if available) This arrangement of electrons in an atom is called electron configuration / structure. By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]: 7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) 8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels 9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels 10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus 11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2 12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers.
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By convention the electron configuration / structure of an atom of an element can be shown in form of a diagram using either cross(x) or dot(●) to Practice examples drawing electronic configurations a)11H has - in nucleus1proton and 0 neutrons - 1 electron in the 1st energy levels thus: Nucleus Energy levels Electrons(represented by cross(x) Electronic structure of Hydrogen is thus: 1: b) 42He has - in nucleus 2 proton and 2 neutrons - 2 electron in the 1st energy levels thus: Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Helium is thus: 2: c) 73Li has - in nucleus 3 proton and 4 neutrons - 2 electron in the 1st energy levels -1 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Lithium is thus: 2:1 Kommentar [s1]: 7 d) 94Be has - in nucleus 4 proton and 5 neutrons - 2 electron in the 1st energy levels -2 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Beryllium is thus: 2:2 e) 115B has - in nucleus 5 proton and 6 neutrons - 2 electron in the 1st energy levels -3 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Boron is thus: 2:3 f) 126C has - in nucleus 6 proton and 6 neutrons - 2 electron in the 1st energy levels -4 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Carbon is thus: 2:4 g) 147N has - in nucleus 7 proton and 7 neutrons - 2 electron in the 1st energy levels -5 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) 8 Electronic structure of Nitrogen is thus: 2:5 h) 168O has - in nucleus 8 proton and 8 neutrons - 2 electron in the 1st energy levels -6 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Oxygen is thus: 2:6 i) 199F has - in nucleus 9 proton and 10 neutrons - 2 electron in the 1st energy levels -7 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Fluorine is thus: 2:7 i) 2010Ne has - in nucleus 10 proton and 10 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels thus Nucleus Energy levels Electrons (represented by cross(x) Electronic structure of Neon is thus: 2:8 j) 2311Na has - in nucleus 11 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels 9 -1 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sodium is thus: 2:8:1 k) 2412Mg has - in nucleus 12 proton and 12 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -2 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Magnesium is thus: 2:8:2 l) 2713Al has - in nucleus 13 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -3 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Aluminium is thus: 2:8:3 m) 2814Si has - in nucleus 14 proton and 14 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -4 electron in the 3rd energy levels thus Nucleus Energy levels 10 Electrons (represented by dot(.) Electronic structure of Silicon is thus: 2:8:4 n) 3115P has - in nucleus 14 proton and 15 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -5 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Phosphorus is thus: 2:8:5 o) 3216S has - in nucleus 16 proton and 16 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -6 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Sulphur is thus: 2:8:6 p) 3517Cl has - in nucleus 18 proton and 17 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -7 electron in the 3rd energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Chlorine is thus: 2:8:7 p) 4018Ar has - in nucleus 22 proton and 18 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels thus Nucleus 11 Energy levels Electrons (represented by dot(.) Electronic structure of Argon is thus: 2:8:8 q) 3919K has - in nucleus 20 proton and 19 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -1 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Potassium is thus: 2:8:8:1 r) 4020Ca has - in nucleus 20 proton and 20 neutrons - 2 electron in the 1st energy levels -8 electron in the 2nd energy levels -8 electron in the 3rd energy levels -2 electron in the 4th energy levels thus Nucleus Energy levels Electrons (represented by dot(.) Electronic structure of Calcium is thus: 2:8:8:2 12 B.PERIODIC TABLE There are over 100 elements so far discovered. Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev.
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Scientists have tried to group them together in a periodic table. A periodic table is a horizontal and vertical arrangement of elements according to their atomic numbers. This table was successfully arranged in 1913 by the British scientist Henry Moseley from the previous work of the Russian Scientist Dmitri Mendeleev. The horizontal arrangement forms period. Atoms in the same period have the same the same number of energy levels in their electronic structure. i.e. The number of energy levels in the electronic configuration of an element determine the period to which the element is in the periodic table. e.g. Which period of the periodic table are the following isotopes/elements/atoms? a) 126C Electron structure 2:4 => 2 energy levels used thus Period 2 b) 2311Na Electron structure 2:8:1 => 3 energy levels used thus Period 3 c) 3919K Electron structure 2:8:8:1 => 4 energy levels used thus Period 4 d) 11H Electron structure 1: => 1 energy level used thus Period 1 The vertical arrangement of elements forms a group. Atoms in the same have the same the same group have the same number of outer energy level electrons as per their electronic structure. i.e. The number of electrons in the outer energy level an element determine the group to which the element is ,in the periodic table. 13 a) 126C Electron structure 2:4 => 4 electrons in outer energy level thus Group IV b) 2311C Electron structure 2:8:1 => 1 electron in outer energy level thus Group I c) 3919K Electron structure 2:8:8:1=>1 electron in outer energy level thus Group I d) 11H Electron structure 1: => 1 electron in outer energy level thus Group I By convention; (i)Periods are named using English numerals 1,2,3,4,… (ii)Groups are named using Roman numerals I,II,III,IV,… There are eighteen groups in a standard periodic table. There are seven periods in a standard periodic table. When an atom has maximum number of electrons in its outer energy level, it is said to be stable. When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable.
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When an atom has no maximum number of electrons in its outer energy level, it is said to be unstable. All stable atoms are in group 8/18 of the periodic table. All other elements are unstable. THE STANDARD PERIODIC TABLE OF ELEMENTS 14 All unstable atoms/isotopes try to be stable through chemical reactions. A chemical reaction involves gaining or losing outer electrons (electron transfer) .When electron transfer take place, an ion is formed. An ion is formed when an unstable atom gain or donate electrons in its outer energy level inorder to be stable. Whether an atom gain or donate electrons depend on the relative energy required to donate or gain extra electrons i.e. Examples 1. 199 F has electronic structure/configuration 2:7. It can donate the seven outer electrons to have stable electronic structure/configuration 2:. It can gain one extra electron to have stable electronic structure/configuration 2:8. Gaining requires less energy, and thus Fluorine reacts by gaining one extra electrons. 2. 2313 Al has electronic structure/configuration 2:8:3 It can donate the three outer electrons to have stable electronic structure/configuration 2:8. It can gain five extra electrons to have stable electronic structure/configuration 2:8:8. Donating requires less energy, and thus Aluminium reacts by donating its three outer electrons. Elements with less than four electrons in the outer energy level donates /lose the outer electrons to be stable and form a positively charged ion called cation. A cation therefore has more protons(positive charge) than electrons(negative charge) Generally metals usually form cation Elements with more than four electrons in the outer energy level gain /acquire extra electrons in the outer energy level to be stable and form a negatively charged ion called anion. An anion therefore has less protons(positive charge) than electrons(negative charge) Generally non metals usually form anion. Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains) 15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3.
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Except Hydrogen The charge carried by an ion is equal to the number of electrons gained/acquired or donated/lost. Examples of ion formation 1.11H H -> H+ + e (atom) (monovalent cation) (electrons donated/lost) Electronic configuration 1: (No electrons remains) 15 2. 2713 Al Al -> Al3+ + 3e (atom) (trivalent cation) (3 electrons donated/lost) Electron 2:8:3 2:8 structure (unstable) (stable) 3. 2311 Na Na -> Na+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 4. 2412Mg Mg -> Mg2+ + 2e (atom) (cation) ( 2 electrons donated/lost) Electron 2:8:1 2:8 structure (unstable) (stable) 5. 168O O + 2e -> O2- (atom) ( 2 electrons gained/acquired) (anion) Electron 2:6 2:8 structure (unstable) (stable) 6. 147N N + 3e -> N3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 7. 3115P P + 3e -> P3- (atom) ( 3 electrons gained/acquired) (anion) Electron 2:5 2:8 structure (unstable) (stable) 8. 199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl 16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation.
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199F F + e -> F- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:7 2:8 structure (unstable) (stable) 9. 3517Cl 16 Cl + e -> Cl- (atom) ( 1 electrons gained/acquired) (anion) Electron 2:8:7 2:8:8 structure (unstable) (stable) 3. 3919 K K -> K+ + e (atom) (cation) ( 1 electrons donated/lost) Electron 2:8:8:1 2:8:8 structure (unstable) (stable) When an element donate/loses its outer electrons ,the process is called oxidation. When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4 17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound.
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When an element acquires/gains extra electrons in its outer energy level,the process is called reduction.The charge carried by an atom, cation or anion is its oxidation state. Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4 17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions.
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Table showing the oxidation states of some isotopes Element Symbol of element / isotopes Charge of ion Oxidation state Hydrogen 1H(deuterium) 31H(Tritium) H+ H+ H+ +1 +1 +1 Chlorine 17Cl Cl- Cl- -1 -1 Potassium 19K 4119K K+ K+ K+ +1 +1 +1 Oxygen 8O O2- O2- -2 -2 Magnesium 2412Mg Mg2+ +2 sodium 2311Na Na+ +1 Copper Cu Cu+ Cu2+ +1 +2 Iron Fe2+ Fe3+ +2 +3 Lead Pb2+ Pb4+ +2 +4 17 Manganese Mn2+ Mn7+ +2 +7 Chromium Cr3+ Cr6+ +3 +6 Sulphur S4+ S6+ +4 +6 Carbon C2+ C4+ +2 +4 Note : Some elements can exist in more than one oxidation state.They are said to have variable oxidation state. Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae.
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Roman capital numeral is used to indicate the oxidation state of an element with a variable oxidation state in a compound. Examples: (i) Copper (I) means Cu+ as in Copper(I)oxide (ii) Copper (II) means Cu2+ as in Copper(II)oxide (iii) Iron (II) means Fe2+ as in Iron(II)sulphide (iv) Iron (III) means Fe3+ as in Iron(III)chloride (iv) Sulphur(VI)mean S6+ as in Iron(III)sulphate(VI) (v) Sulphur(VI)mean S6+ as in sulphur(VI)oxide (vi) Sulphur(IV)mean S4+ as in sulphur(IV)oxide (vii) Sulphur(IV)mean S4+ as in sodium sulphate(IV) (ix) Carbon(IV)mean C4+ as in carbon(IV)oxide (x) Carbon(IV)mean C4+ as in Lead(II)carbonate(IV) (xi) Carbon(II)mean C2+ as in carbon(II)oxide (xii) Manganese(IV)mean Mn4+ as in Manganese(IV)oxide A compound is a combination of two or more elements in fixed proportions. The ratio of the atoms making a compound is called the chemical formulae. Elements combine together to form a compound depending on their combining power. The combining power of atoms in an element is called Valency.Valency of an element is equal to the number of: (i)hydrogen atoms that an atom of element can combine with or displace. (ii)electrons gained /acquired in outer energy level by non metals to be stable/attain duplet/octet. (iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions 18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals.
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(iii)electrons donated/lost by outer energy level of metals to be stable/attain octet/duplet. (iv)charges carried by ions/cations/ions 18 Group of atoms that react as a unit during chemical reactions are called radicals.Elements with variable oxidation state also have more than one valency. Table showing the valency of common radicals. Radical name Chemical formulae Combining power / Valency Ammonium NH4 + 1 Hydroxide OH- 1 Nitrate(V) NO3 - 1 Hydrogen carbonate HCO3- 1 Hydrogen sulphate(VI) HSO4- 1 Hydrogen sulphate(IV) HSO3- 1 Manganate(VII) MnO4- 1 Chromate(VI) CrO42- 2 Dichromate(VI) Cr2O72- 2 Sulphate(VI) SO42- 2 Sulphate(IV) SO32- 2 Carbonate(IV) CO32- 2 Phosphate(V) PO42- 3 Table showing the valency of some common metal and non metals Element/metal Valency Element/non metal Valency Hydrogen 1 Florine 1 Lithium 1 Chlorine 1 Beryllium 2 Bromine 1 Boron 3 Iodine 1 Sodium 1 Carbon 4 Magnesium 2 Nitrogen 3 Aluminium 3 Oxygen 2 Potassium 1 Phosphorus 3 Calcium 2 Zinc 2 Barium 2 Mercury 2 Iron 2 and 3 Copper 1 and 2 Manganese 2 and 4 Lead 2 and 4 19 From the valency of elements , the chemical formular of a compound can be derived using the following procedure: (i)Identify the elements and radicals making the compound (ii)Write the symbol/formular of the elements making the compound starting with the metallic element (iii)Assign the valency of each element /radical as superscript. (iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula.
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(iv)Interchange/exchange the valencies of each element as subscript. (v)Divide by the smallest/lowest valency to derive the smallest whole number ratios Ignore a valency of 1. This is the chemical formula. Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1 20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen.
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Practice examples Write the chemical formula of (a)Aluminium oxide Elements making compound Aluminium Oxygen Symbol of elements/radicals in compound Al O Assign valencies as superscript Al3 O2 Exchange/Interchange the valencies as subscript Al2 O3 Divide by smallest valency to get whole number - - Chemical formula of Aluminium oxide is thus: Al2 O3 This means:2atoms of Aluminium combine with 3 atoms of Oxygen (b)Sodium oxide Elements making compound Sodium Oxygen Symbol of elements/radicals in compound Na O Assign valencies as superscript Na1 O2 Exchange/Interchange the valencies as subscript Na2 O1 Divide by smallest valency to get whole number - - Chemical formula of Sodium oxide is thus: Na2 O This means:2atoms of Sodium combine with 1 atom of Oxygen (c)Calcium oxide Elements making compound Calcium Oxygen Symbol of elements/radicals in compound Ca O Assign valencies as superscript Ca2 O2 Exchange/Interchange the valencies as subscript Ca2 O2 Divide by two to get smallest whole number ratio Ca1 O1 20 Chemical formula of Calcium oxide is thus: CaO This means:1 atom of calcium combine with 1 atom of Oxygen. (d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen.
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(d)Lead(IV)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb4 O2 Exchange/Interchange the valencies as subscript Pb2 O4 Divide by two to get smallest whole number ratio Pb1 O2 Chemical formula of Lead(IV) oxide is thus: PbO2 This means:1 atom of lead combine with 2 atoms of Oxygen. (e)Lead(II)oxide Elements making compound Lead Oxygen Symbol of elements/radicals in compound Pb O Assign valencies as superscript Pb2 O2 Exchange/Interchange the valencies as subscript Pb2 O2 Divide by two to get smallest whole number ratio Pb1 O1 Chemical formula of Lead(II) oxide is thus: PbO This means:1 atom of lead combine with 1 atom of Oxygen. (e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1 21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical.
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(e)Iron(III)oxide Elements making compound Iron Oxygen Symbol of elements/radicals in compound Fe O Assign valencies as superscript Fe3 O2 Exchange/Interchange the valencies as subscript Fe2 O3 Divide by two to get smallest whole number ratio - - Chemical formula of Iron(III) oxide is thus: Fe2O3 This means:2 atom of lead combine with 3 atom of Oxygen. (f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1 21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical.
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(f)Iron(II)sulphate(VI) Elements making compound Iron sulphate(VI) Symbol of elements/radicals in compound Fe SO4 Assign valencies as superscript Fe2 SO4 2 Exchange/Interchange the valencies as subscript Fe2 SO4 2 Divide by two to get smallest whole number ratio Fe1 SO4 1 21 Chemical formula of Iron(II) sulphate(VI) is thus: FeSO4 This means:1 atom of Iron combine with 1 sulphate(VI) radical. (g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical.
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(g)Copper(II)sulphate(VI) Elements making compound Copper sulphate(VI) Symbol of elements/radicals in compound Cu SO4 Assign valencies as superscript Cu2 SO4 2 Exchange/Interchange the valencies as subscript Cu2 SO4 2 Divide by two to get smallest whole number ratio Cu1 SO4 1 Chemical formula of Cu(II)sulphate(VI) is thus: CuSO4 This means:1 atom of Copper combine with 1 sulphate(VI) radical. (h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - - 22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical.
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(h)Aluminium sulphate(VI) Elements making compound Aluminium sulphate(VI) Symbol of elements/radicals in compound Al SO4 Assign valencies as superscript Al3 SO4 2 Exchange/Interchange the valencies as subscript Al2 SO4 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al2(SO4)3 This means:2 atom of Aluminium combine with 3 sulphate(VI) radical. (i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - - 22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical.
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(i)Aluminium nitrate(V) Elements making compound Aluminium nitrate(V) Symbol of elements/radicals in compound Al NO3 Assign valencies as superscript Al3 NO3 1 Exchange/Interchange the valencies as subscript Al1 NO3 3 Divide by two to get smallest whole number ratio - - Chemical formula of Aluminium sulphate(VI) is thus: Al (NO3)3 This means:1 atom of Aluminium combine with 3 nitrate(V) radical. (j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - - 22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical.
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(j)Potassium manganate(VII) Elements making compound Potassium manganate(VII) Symbol of elements/radicals in compound K MnO4 Assign valencies as superscript K 1 MnO4 1 Exchange/Interchange the valencies as subscript K1 MnO4 1 Divide by two to get smallest whole number ratio - - 22 Chemical formula of Potassium manganate(VII) is thus: KMnO4 This means:1 atom of Potassium combine with 4 manganate(VII) radical. (k)Sodium dichromate(VI) Elements making compound Sodium dichromate(VI) Symbol of elements/radicals in compound Na Cr2O7 Assign valencies as superscript Na 1 Cr2O7 2 Exchange/Interchange the valencies as subscript Na2 Cr2O7 1 Divide by two to get smallest whole number ratio - - Chemical formula of Sodium dichromate(VI) is thus: Na2 Cr2O7 This means:2 atom of Sodium combine with 1 dichromate(VI) radical. (l)Calcium hydrogen carbonate Elements making compound Calcium Hydrogen carbonate Symbol of elements/radicals in compound Ca CO3 Assign valencies as superscript Ca 2 HCO3 1 Exchange/Interchange the valencies as subscript Ca1 HCO3 2 Divide by two to get smallest whole number ratio - - Chemical formula of Calcium hydrogen carbonate is thus: Ca(HCO3)2 This means:1 atom of Calcium combine with 2 hydrogen carbonate radical. (l)Magnesium hydrogen sulphate(VI) Elements making compound Magnesium Hydrogen sulphate(VI) Symbol of elements/radicals in compound Mg HSO4 Assign valencies as superscript Mg 2 HSO4 1 Exchange/Interchange the valencies as subscript Mg1 HSO4 2 Divide by two to get smallest whole number ratio - - Chemical formula of Magnesium hydrogen sulphate(VI) is thus: Mg(HSO4)2 This means:1 atom of Magnesium combine with 2 hydrogen sulphate(VI) radical. Compounds are formed from chemical reactions. A chemical reaction is formed when atoms of the reactants break free to bond again and form products. A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations: 23 1. Write the word equation 2.
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A chemical reaction is a statement showing the movement of reactants to form products. The following procedure is used in writing a chemical equations: 23 1. Write the word equation 2. Write the correct chemical formula for each of the reactants and products 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. This is called balancing. Do not change the chemical formula of the products/reactants. 5. Assign in brackets, the physical state/state symbols of the reactants and products after each chemical formula as: (i) (s) for solids (ii) (l) for liquids (iii) (g) for gas (iv) (aq) for aqueous/dissolved in water to make a solution. Practice examples Write a balanced chemical equation for the following (a) Hydrogen gas is prepared from reacting Zinc granules with dilute hydrochloric acid. Procedure 1. Write the word equation Zinc + Hydrochloric acid -> Zinc chloride + hydrogen gas 2. Write the correct chemical formula for each of the reactants and products Zn + HCl -> ZnCl2 + H2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of Zn on the reactant side is equal to product side One atom of H in HCl on the reactant side is not equal to two atoms in H2 on product side. One atom of Cl in HCl on the reactant side is not equal to two atoms in ZnCl2 on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “2” to get “2” Hydrogen and “2” Chlorine on product and reactant side. Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g) 24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1.
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Zn + 2 HCl -> ZnCl2 + H2 5. Assign in brackets, the physical state/state symbols . Zn(s) + 2 HCl(aq) -> ZnCl2 (aq) + H2(g) 24 (b) Oxygen gas is prepared from decomposition of Hydrogen peroxide solution to water Procedure 1. Write the word equation Hydrogen peroxide -> Water + oxygen gas 2. Write the correct chemical formula for each of the reactants and products H2O2 -> H2O + O2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. Number of atoms of H on the reactant side is equal to product side Two atom of O in H2O2 on the reactant side is not equal to three atoms (one in H2O and two in O2) on product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply H2O2 by “2” to get “4” Hydrogen and “4” Oxygen on reactants Multiply H2O by “2” to get “4” Hydrogen and “2” Oxygen on product side When the “2” Oxygen in O2 and the“2” in H2O are added on product side they are equal to the“4” Oxygen on reactants side. 2H2O2 -> 2H2O + O2 5. Assign in brackets, the physical state/state symbols . 2H2O2(aq) -> 2H2O(l) + O2(g) (c) Chlorine gas is prepared from Potassium manganate(VII) reacting with hydrochloric acid to form potassium chloride solution, manganese(II) chloride solution,water and chlorine gas. Procedure 1. Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side.
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Write the word equation Potassium manganate(VII) + Hydrochloric acid -> potassium chloride + manganese(II) chloride + chlorine +water 2. Write the correct chemical formula for each of the reactants and products KMnO4 + HCl -> KCl + MnCl2 +H2O + Cl2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 25 Number of atoms of K and Mn on the reactant side is equal to product side Two atom of H in H2O on the product side is not equal to one atom on reactant side. Four atom of O in KMnO4 is not equal to one in H2O One atom of Cl in HCl on reactant side is not equal to three (one in H2O and two in Cl2) 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. Multiply HCl by “16” to get “16” Hydrogen and “16” Chlorine on reactants Multiply KMnO4 by “2” to get “2” Potassium and “2” manganese, “2 x4 =8” Oxygen on reactant side. Balance the product side to get: 2 KMnO4 +16 HCl -> 2 KCl + 2 MnCl2 +8 H2O + 5 Cl2 5. Assign in brackets, the physical state/state symbols . 2KMnO4(s) +16 HCl(aq)-> 2 KCl (aq) + 2MnCl2(aq)+8 H2O(l)+5 Cl2(g) (d)Carbon(IV)oxide gas is prepared from Calcium carbonate reacting with hydrochloric acid to form calcium chloride solution, water and carbon(IV)oxide gas. Procedure 1. Write the word equation Calcium carbonate + Hydrochloric acid -> calcium chloride solution+ water +carbon(IV)oxide 2. Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4.
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Write the correct chemical formula for each of the reactants and products CaCO3 + HCl -> CaCl2 +H2O + CO2 3. Check if the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side. 4. Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple if the number of atoms on one side is not equal. 5. Assign in brackets, the physical state/state symbols . CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) (d)Sodium hydroxide solution neutralizes hydrochloric acid to form salt and water. 26 NaOH(aq) + HCl(aq) -> NaCl (aq) + H2O(l) (e)Sodium reacts with water to form sodium hydroxide and hydrogen gas. 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) (f)Calcium reacts withwater to form calcium hydroxide and hydrogen gas Ca(s) + 2H2O(l) -> Ca(OH)2(aq) + H2(g) (g)Copper(II)Oxide solid reacts with dilute hydrochloric acid to form copper(II)chloride and water. CuO(s) + 2HCl(aq) -> CuCl2(aq) + H2O(l) (h)Hydrogen sulphide reacts with Oxygen to form sulphur(IV)Oxide and water. 2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water.
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2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(l) (i)Magnesium reacts with steam to form Magnesium Oxide and Hydrogen gas. Mg(s) + 2H2O(g) -> MgO(s) + H2(g) (j)Ethane(C2H6) gas burns in air to form Carbon(IV)Oxide and water. 2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l) (k)Ethene(C2H4) gas burns in air to form Carbon(IV)Oxide and water. C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l) (l)Ethyne(C2H2) gas burns in air to form Carbon(IV)Oxide and water. 2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(l) 27 C.PERIODICITY OF CHEMICAL FAMILES/DOWN THE GROUP. The number of valence electrons and the number of occupied energy levels in an atom of an element determine the position of an element in the periodic table.i.e The number of occupied energy levels determine the Period and the valence electrons determine the Group. Elements in the same group have similar physical and chemical properties. The trends in physical and chemical properties of elements in the same group vary down the group. Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level.
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Elements in the same group thus constitute a chemical family. (a) Group I elements: Alkali metals Group I elements are called Alkali metals except Hydrogen which is a non metal. The alkali metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Lithium Li 3 2:1 Li+ 1 Sodium Na 11 2:8:1 Na+ 1 Potassium K 19 2:8:8:1 K+ 1 Rubidium Rb 37 2:8:18:8:1 Rb+ 1 Caesium Cs 55 2:8:18:18:8:1 Cs+ 1 Francium Fr 87 2:8:18:32:18:8:1 Fr+ 1 All alkali metals atom has one electron in the outer energy level. They therefore are monovalent. They donate /lose the outer electron to have oxidation state M+ The number of energy levels increases down the group from Lithium to Francium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size of Potassium is bigger/larger than that of sodium because Potassium has more/4 energy levels than sodium (3 energy levels). Atomic and ionic radius The distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron/s is called atomic radius. Atomic radius is measured in nanometers(n).The higher /bigger the atomic radius the bigger /larger the atomic size. The distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron/s is called ionic radius. Ionic radius is also 28 measured in nanometers(n).The higher /bigger the ionic radius the bigger /larger the size of the ion. Atomic radius and ionic radius depend on the number of energy levels occupied by electrons. The more the number of energy levels the bigger/larger the atomic /ionic radius. e.g. The atomic radius of Francium is bigger/larger than that of sodium because Francium has more/7 energy levels than sodium (3 energy levels). Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level.
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Atomic radius and ionic radius of alkali metals increase down the group as the number of energy levels increases. The atomic radius of alkali metals is bigger than the ionic radius. This is because alkali metals react by losing/donating the outer electron and hence lose the outer energy level. Table showing the atomic and ionic radius of some alkali metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Lithium Li 3 0.133 0.060 Sodium Na 11 0.157 0.095 Potassium K 19 0.203 0.133 The atomic radius of sodium is 0.157nM .The ionic radius of Na+ is 0.095nM. This is because sodium reacts by donating/losing the outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity The ease of donating/losing electrons is called electropositivity. All alkali metals are electropositive. Electropositivity increase as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and can be lost/ donated easily/with ease. Francium is the most electropositive element in the periodic table because it has the highest/biggest atomic radius. Ionization energy The minimum amount of energy required to remove an electron from an atom of element in its gaseous state is called 1st ionization energy. The SI unit of ionization energy is kilojoules per mole/kJmole-1 .Ionization energy depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer electrons/energy level and thus the lower the ionization energy. For alkali metals the 1st ionization energy decrease down the group as 29 the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium.
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e.g. The 1st ionization energy of sodium is 496 kJmole-1 while that of potassium is 419 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from sodium to Potassium. It requires therefore less energy to donate/lose outer electrons in Potassium than in sodium. Physical properties Soft/Easy to cut: Alkali metals are soft and easy to cut with a knife. The softness and ease of cutting increase down the group from Lithium to Francium. This is because an increase in atomic radius, decreases the strength of metallic bond and the packing of the metallic structure Appearance: Alkali metals have a shiny grey metallic luster when freshly cut. The surface rapidly/quickly tarnishes on exposure to air. This is because the metal surface rapidly/quickly reacts with elements of air/oxygen. Melting and boiling points: Alkali metals have a relatively low melting/boiling point than common metals like Iron. This is because alkali metals use only one delocalized electron to form a weak metallic bond/structure. Electrical/thermal conductivity: Alkali metals are good thermal and electrical conductors. Metals conduct using the outer mobile delocalized electrons. The delocalized electrons move randomly within the metallic structure. Summary of some physical properties of the 1st three alkali metals Alkali metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conductivity 1st ionization energy Lithium Silvery white Not easy 180 1330 Good 520 Sodium Shiny grey Easy 98 890 Good 496 Potassium Shiny grey Very easy 64 774 Good 419 Chemical properties (i)Reaction with air/oxygen On exposure to air, alkali metals reacts with the elements in the air. Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate.
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Example On exposure to air, Sodium first reacts with Oxygen to form sodium oxide. 4Na(s) + O2(g) -> 2Na2O(s) The sodium oxide formed further reacts with water/moisture in the air to form sodium hydroxide solution. 30 Na2O(s) + H2O(l) -> 2NaOH(aq) Sodium hydroxide solution reacts with carbon(IV)oxide in the air to form sodium carbonate. 2NaOH(aq) + CO2(g) -> Na2CO3(g) + H2O(l) (ii)Burning in air/oxygen Lithium burns in air with a crimson/deep red flame to form Lithium oxide 4Li (s) + O2(g) -> 2Li2O(s) Sodium burns in air with a yellow flame to form sodium oxide 4Na (s) + O2(g) -> 2Na2O(s) Sodium burns in oxygen with a yellow flame to form sodium peroxide 2Na (s) + O2(g) -> Na2O2 (s) Potassium burns in air with a lilac/purple flame to form potassium oxide 4K (s) + O2(g) -> 2K2O (s) (iii) Reaction with water: Experiment Measure 500 cm3 of water into a beaker. Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium.
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Put three drops of phenolphthalein indicator. Put about 0.5g of Lithium metal into the beaker. Determine the pH of final product Repeat the experiment using about 0.1 g of Sodium and Potassium. Caution: Keep a distance Observations Alkali metal Observations Comparative speed/rate of the reaction Lithium -Metal floats in water -rapid effervescence/fizzing/bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Moderately vigorous Sodium -Metal floats in water -very rapid effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn Very vigorous 31 phenolphthalein indicator pink -pH of solution = 12/13/14 Potassium -Metal floats in water -explosive effervescence /fizzing /bubbling -colourless gas produced (that extinguishes burning splint with explosion /“pop” sound) -resulting solution turn phenolphthalein indicator pink -pH of solution = 12/13/14 Explosive/burst into flames Explanation Alkali metals are less dense than water. They therefore float in water.They react with water to form a strongly alkaline solution of their hydroxides and producing hydrogen gas. The rate of this reaction increase down the group. i.e. Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions.
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Potassium is more reactive than sodium .Sodium is more reactive than Lithium. The reactivity increases as electropositivity increases of the alkali increases. This is because as the atomic radius increases , the ease of donating/losing outer electron increase during chemical reactions. Chemical equations 2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g) 2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g) 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) 2Rb(s) + 2H2O(l) -> 2RbOH(aq) + H2(g) 2Cs(s) + 2H2O(l) -> 2CsOH(aq) + H2(g) 2Fr(s) + 2H2O(l) -> 2FrOH(aq) + H2(g) Reactivity increase down the group (iv) Reaction with chlorine: Experiment Cut about 0.5g of sodium into a deflagrating spoon with a lid cover. Introduce it on a Bunsen flame until it catches fire. Quickly and carefully lower it into a gas jar containing dry chlorine to cover the gas jar. Repeat with about 0.5g of Lithium. Caution: This experiment should be done in fume chamber because chlorine is poisonous /toxic. Observation Sodium metal continues to burn with a yellow flame forming white solid/fumes. 32 Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus.
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32 Lithium metal continues to burn with a crimson flame forming white solid / fumes. Alkali metal react with chlorine gas to form the corresponding metal chlorides. The reactivity increase as electropositivity increase down the group from Lithium to Francium.The ease of donating/losing the outer electrons increase as the atomic radius increase and the outer electron is less attracted to the nucleus. Chemical equations 2Li(s) + Cl2(g) -> 2LiCl(s) 2Na(s) + Cl2(g) -> 2NaCl(s) 2K(s) + Cl2(g) -> 2KCl(s) 2Rb(s) + Cl2(g) -> 2RbCl(s) 2Cs(s) + Cl2(g) -> 2CsCl(s) 2Fr(s) + Cl2(g) -> 2FrCl(s) Reactivity increase down the group The table below shows some compounds of the 1st three alkali metals Some uses of alkali metals include: (i)Sodium is used in making sodium cyanide for extracting gold from gold ore. (ii)Sodium chloride is used in seasoning food. (iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors. (iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents.
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(ii)Sodium chloride is used in seasoning food. (iii)Molten mixture of sodium and potassium is used as coolant in nuclear reactors. (iv)Sodium is used in making sodium hydroxide used in making soapy and soapless detergents. Lithium sodium Potassium Hydroxide LiOH NaOH KOH Oxide Li2O Na2O K2O Sulphide Li2S Na2S K2S Chloride LiCl NaCl KCl Carbonate Li2CO3 Na2CO3 K2CO3 Nitrate(V) LiNO3 NaNO3 KNO3 Nitrate(III) - NaNO2 KNO2 Sulphate(VI) Li2SO4 Na2SO4 K2SO4 Sulphate(IV) - Na2SO3 K2SO3 Hydrogen carbonate - NaHCO3 KHCO3 Hydrogen sulphate(VI) - NaHSO4 KHSO4 Hydrogen sulphate(IV) - NaHSO3 KHSO3 Phosphate - Na3PO4 K3PO4 Manganate(VI) - NaMnO4 KMnO4 Dichromate(VI) - Na2Cr2O7 K2Cr2O7 Chromate(VI) - Na2CrO4 K2CrO4 33 (v)Sodium is used as a reducing agent for the extraction of titanium from Titanium(IV)chloride. (vi)Lithium is used in making special high strength glasses (vii)Lithium compounds are used to make dry cells in mobile phones and computer laptops. Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals . The alkaline earth metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Beryllium Be 4 2:2 Be2+ 2 Magnesium Mg 12 2:8:2 Mg2+ 2 Calcium Ca 20 2:8:8:2 Ca2+ 2 Strontium Sr 38 2:8:18:8:2 Sr2+ 2 Barium Ba 56 2:8:18:18:8:2 Ba2+ 2 Radium Ra 88 2:8:18:32:18:8:2 Ra2+ 2 All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent.
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Group II elements: Alkaline earth metals Group II elements are called Alkaline earth metals . The alkaline earth metals include: Element Symbol Atomic number Electron structure Oxidation state Valency Beryllium Be 4 2:2 Be2+ 2 Magnesium Mg 12 2:8:2 Mg2+ 2 Calcium Ca 20 2:8:8:2 Ca2+ 2 Strontium Sr 38 2:8:18:8:2 Sr2+ 2 Barium Ba 56 2:8:18:18:8:2 Ba2+ 2 Radium Ra 88 2:8:18:32:18:8:2 Ra2+ 2 All alkaline earth metal atoms have two electrons in the outer energy level. They therefore are divalent. They donate /lose the two outer electrons to have oxidation state M2+ The number of energy levels increases down the group from Beryllium to Radium. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Calcium is bigger/larger than that of Magnesium because Calcium has more/4 energy levels than Magnesium (3 energy levels). Atomic radius and ionic radius of alkaline earth metals increase down the group as the number of energy levels increases. The atomic radius of alkaline earth metals is bigger than the ionic radius. This is because they react by losing/donating the two outer electrons and hence lose the outer energy level. Table showing the atomic and ionic radius of the 1st three alkaline earth metals Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Beryllium Be 4 0.089 0.031 Magnesium Mg 12 0.136 0.065 34 Calcium Ca 20 0.174 0.099 The atomic radius of Magnesium is 0.136nM .The ionic radius of Mg2+ is 0.065nM. This is because Magnesium reacts by donating/losing the two outer electrons and hence the outer energy level. The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size.
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The remaining electrons/energy levels experience more effective / greater nuclear attraction/pull towards the nucleus reducing the atomic radius. Electropositivity All alkaline earth metals are also electropositive like alkali metals. The electropositivity increase with increase in atomic radius/size. Calcium is more electropositive than Magnesium. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The two outer electrons in calcium experience less nuclear attraction and can be lost/ donated easily/with ease because of the higher/bigger atomic radius. Ionization energy For alkaline earth metals the 1st ionization energy decrease down the group as the atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. e.g. The 1st ionization energy of Magnesium is 900 kJmole-1 while that of Calcium is 590 kJmole-1 .This is because atomic radius increase and thus effective nuclear attraction on outer energy level electrons decrease down the group from magnesium to calcium. It requires therefore less energy to donate/lose outer electron in calcium than in magnesium. The minimum amount of energy required to remove a second electron from an ion of an element in its gaseous state is called the 2nd ionization energy. The 2nd ionization energy is always higher /bigger than the 1st ionization energy. This because once an electron is donated /lost form an atom, the overall effective nuclear attraction on the remaining electrons/energy level increase. Removing a second electron from the ion require therefore more energy than the first electron. The atomic radius of alkali metals is higher/bigger than that of alkaline earth metals.This is because across/along the period from left to right there is an increase in nuclear charge from additional number of protons and still additional number of electrons entering the same energy level. Increase in nuclear charge increases the effective nuclear attraction on the outer energy level which pulls it closer to the nucleus. e.g. Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on 35 the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus. Physical properties Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals.
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Atomic radius of Sodium (0.157nM) is higher than that of Magnesium (0.137nM). This is because Magnesium has more effective nuclear attraction on 35 the outer energy level than Sodium hence pulls outer energy level more nearer to its nucleus. Physical properties Soft/Easy to cut: Alkaline earth metals are not soft and easy to cut with a knife like alkali metals. This is because of the decrease in atomic radius of corresponding alkaline earth metal, increases the strength of metallic bond and the packing of the metallic structure. Alkaline earth metals are (i)ductile(able to form wire/thin long rods) (ii)malleable(able to be hammered into sheet/long thin plates) (iii)have high tensile strength(able to be coiled without breaking/ not brittle/withstand stress) Appearance: Alkali earth metals have a shiny grey metallic luster when their surface is freshly polished /scrubbed. The surface slowly tarnishes on exposure to air. This is because the metal surface slowly undergoes oxidation to form an oxide. This oxide layer should be removed before using the alkaline earth metals. Melting and boiling points: Alkaline earth metals have a relatively high melting/ boiling point than alkali metals. This is because alkali metals use only one delocalized electron to form a weaker metallic bond/structure. Alkaline earth metals use two delocalized electrons to form a stronger metallic bond /structure. The melting and boiling points decrease down the group as the atomic radius/size increase reducing the strength of metallic bond and packing of the metallic structure. e.g. Beryllium has a melting point of 1280oC. Magnesium has a melting point of 650oC.Beryllium has a smaller atomic radius/size than magnesium .The strength of metallic bond and packing of the metallic structure is thus stronger in beryllium. Electrical/thermal conductivity: Alkaline earth metals are good thermal and electrical conductors. The two delocalized valence electrons move randomly within the metallic structure. Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g. Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium.
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Electrical conductivity increase down the group as the atomic radius/size increase making the delocalized outer electrons less attracted to nucleus. Alkaline earth metals are better thermal and electrical conductors than alkali metals because they have more/two outer delocalized electrons.e.g. Magnesium is a better conductor than sodium because it has more/two delocalized electrons than sodium. The more delocalized electrons the better the electrical conductor. 36 Calcium is a better conductor than magnesium. Calcium has bigger/larger atomic radius than magnesium because the delocalized electrons are less attracted to the nucleus of calcium and thus more free /mobile and thus better the electrical conductor Summary of some physical properties of the 1st three alkaline earth metals Alkaline earth metal Appearance Ease of cutting Melting point (oC) Boiling point (oC) Conduct- ivity 1st ionization energy 2nd ionization energy Beryllium Shiny grey Not easy 1280 3450 Good 900 1800 Magnesium Shiny grey Not Easy 650 1110 Good 736 1450 calcium Shiny grey Not easy 850 1140 Good 590 970 Chemical properties (i)Reaction with air/oxygen On exposure to air, the surface of alkaline earth metals is slowly oxidized to its oxide on prolonged exposure to air. Example On exposure to air, the surface of magnesium ribbon is oxidized to form a thin film of Magnesium oxide . 2Mg(s) + O2(g) -> 2MgO(s) (ii)Burning in air/oxygen Experiment Hold a about 2cm length of Magnesium ribbon on a Bunsen flame. Stop heating when it catches fire/start burning. Caution: Do not look directly at the flame Put the products of burning into 100cm3 beaker. Add about 5cm3 of distilled water. Swirl. Test the mixture using litmus papers. Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride.
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Swirl. Test the mixture using litmus papers. Repeat with Calcium Observations -Magnesium burns with a bright blindening flame -White solid /ash produced -Solid dissolves in water to form a colourless solution -Blue litmus paper remain blue -Red litmus paper turns blue -colourless gas with pungent smell of urine Explanation Magnesium burns in air with a bright blindening flame to form a mixture of Magnesium oxide and Magnesium nitride. 37 2Mg (s) + O2(g) -> 2MgO(s) 3Mg (s) + N2 (g) -> Mg3N2 (s) Magnesium oxide dissolves in water to form magnesium hydroxide. MgO(s) + H2O (l) -> Mg(OH)2(aq) Magnesium nitride dissolves in water to form magnesium hydroxide and produce ammonia gas. Mg3N2 (s) + 6H2O(l) -> 3Mg(OH)2(aq) + 2NH3 (g) Magnesium hydroxide and ammonia are weakly alkaline with pH 8/9/10/11 and turns red litmus paper blue. Calcium burns in air with faint orange/red flame to form a mixture of both Calcium oxide and calcium nitride. 2Ca (s) + O2(g) -> 2CaO(s) 3Ca (s) + N2 (g) -> Ca3N2 (s) Calcium oxide dissolves in water to form calcium hydroxide. CaO(s) + H2O(l) -> Ca(OH)2(aq) Calcium nitride dissolves in water to form calcium hydroxide and produce ammonia gas. Ca3N2 (s) + 6H2O(l) -> 3Ca(OH)2(aq) + 2NH3 (g) Calcium hydroxide is also weakly alkaline solution with pH 8/9/10/11 and turns red litmus paper blue. (iii)Reaction with water Experiment Measure 50 cm3 of distilled water into a beaker. Scrub/polish with sand paper 1cm length of Magnesium ribbon Place it in the water. Test the product-mixture with blue and red litmus papers. Repeat with Calcium metal. Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed.
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Repeat with Calcium metal. Observations -Surface of magnesium covered by bubbles of colourless gas. -Colourless solution formed. -Effervescence/bubbles/fizzing takes place in Calcium. -Red litmus paper turns blue. -Blue litmus paper remains blue. Explanations Magnesium slowly reacts with cold water to form Magnesium hydroxide and bubbles of Hydrogen gas that stick on the surface of the ribbon. Mg(s) + 2H2O (l) -> Mg(OH)2(aq) + H2 (g) 38 Calcium moderately reacts with cold water to form Calcium hydroxide and produce a steady stream of Hydrogen gas. Ca(s) + 2H2O (l) -> Ca(OH)2(aq) + H2 (g) (iv)Reaction with water vapour/steam Experiment Put some cotton wool soaked in water/wet sand in a long boiling tube. Coil a well polished magnesium ribbon into the boiling tube. Ensure the coil touches the side of the boiling tube. Heat the cotton wool/sand slightly then strongly heat the Magnesium ribbon . Set up of apparatus Observations -Magnesium glows red hot then burns with a blindening flame. -Magnesium continues to glow/burning even without more heating. -White solid/residue. -colourless gas collected over water. Explanation On heating wet sand, steam is generated which drives out the air that would otherwise react with /oxidize the ribbon. Magnesium burns in steam/water vapour generating enough heat that ensures the reaction goes to completion even without further heating. White Magnesium oxide is formed and hydrogen gas is evolved. To prevent suck back, the delivery tube should be removed from the water before heating is stopped at the end of the experiment. Mg(s) + H2O (l) -> MgO(s) + H2 (g) 39 (v)Reaction with chlorine gas. Experiment Lower slowly a burning magnesium ribbon/shavings into a gas jar containing Chlorine gas. Repeat with a hot piece of calcium metal. Observation -Magnesium continues to burn in chlorine with a bright blindening flame. -Calcium continues to burn for a short time. -White solid formed . -Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid.
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-Pale green colour of chlorine fades. Explanation Magnesium continues to burn in chlorine gas forming white magnesium oxide solid. Mg(s) + Cl2 (g) -> MgCl2 (s) Calcium burns slightly in chlorine gas to form white calcium oxide solid. Calcium oxide formed coat unreacted Calcium stopping further reaction Ca(s) + Cl2 (g) -> CaCl2 (s) (v)Reaction with dilute acids. Experiment Place about 4.0cm3 of 0.1M dilute sulphuric(VI)acid into a test tube. Add about 1.0cm length of magnesium ribbon into the test tube. Cover the mouth of the test tube using a thumb. Release the gas and test the gas using a burning splint. Repeat with about 4.0cm3 of 0.1M dilute hydrochloric/nitric(V) acid. Repeat with 0.1g of Calcium in a beaker with all the above acid Caution: Keep distance when using calcium Observation -Effervescence/fizzing/bubbles with dilute sulphuric(VI) and nitric(V) acids -Little Effervescence/fizzing/bubbles with calcium and dilute sulphuric(VI) acid. -Colourless gas produced that extinguishes a burning splint with an explosion/ “pop” sound. -No gas is produced with Nitric(V)acid. -Colourless solution is formed. Explanation Dilute acids react with alkaline earth metals to form a salt and produce hydrogen gas. Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used.
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Nitric(V)acid is a strong oxidizing agent. It quickly oxidizes the hydrogen produced to water. Calcium is very reactive with dilute acids and thus a very small piece of very dilute acid should be used. Chemical equations 40 Mg(s) + H2SO4 (aq) -> MgSO4(aq) + H2 (g) Mg(s) + 2HNO3 (aq) -> Mg(NO3)2(aq) + H2 (g) Mg(s) + 2HCl (aq) -> MgCl2(aq) + H2 (g) Ca(s) + H2SO4 (aq) -> CaSO4(s) + H2 (g) (insoluble CaSO4(s) coat/cover Ca(s)) Ca(s) + 2HNO3 (aq) -> Ca(NO3)2(aq) + H2 (g) Ca(s) + 2HCl (aq) -> CaCl2(aq) + H2 (g) Ba(s) + H2SO4 (aq) -> BaSO4(s) + H2 (g) (insoluble BaSO4(s) coat/cover Ba(s)) Ba(s) + 2HNO3 (aq) -> Ba(NO3)2(aq) + H2 (g) Ba(s) + 2HCl (aq) -> BaCl2(aq) + H2 (g) The table below shows some compounds of some alkaline earth metals Some uses of alkaline earth metals include: (i)Magnesium hydroxide is a non-toxic/poisonous mild base used as an anti acid medicine to relieve stomach acidity. (ii)Making duralumin. Duralumin is an alloy of Magnesium and aluminium used for making aeroplane bodies because it is light. (iii) Making plaster of Paris-Calcium sulphate(VI) is used in hospitals to set a fractures bone. (iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building. (iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms. (v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth.
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(iii)Making cement-Calcium carbonate is mixed with clay and sand then heated to form cement for construction/building. (iv)Raise soil pH-Quicklime/calcium oxide is added to acidic soils to neutralize and raise the soil pH in agricultural farms. (v)As nitrogenous fertilizer-Calcium nitrate(V) is used as an agricultural fertilizer because plants require calcium for proper growth. Beryllium Magnesium Calcium Barium Hydroxide Be(OH)2 Mg(OH)2 Ca(OH)2 Ba(OH)2 Oxide BeO MgO CaO BaO Sulphide - MgS CaS BaS Chloride BeCl2 MgCl2 CaCl2 BaCl2 Carbonate BeCO3 MgCO3 CaCO3 BaCO3 Nitrate(V) Be(NO3)2 Mg(NO3)2 Ca(NO3)2 Ba(NO3)2 Sulphate(VI) BeSO4 MgSO4 CaSO4 BaSO4 Sulphate(IV) - - CaSO3 BaSO3 Hydrogen carbonate - Mg(HCO3)2 Ca(HCO3)2 - Hydrogen sulphate(VI) - Mg(HSO4)2 Ca(HSO4)2 - 41 (vi)In the blast furnace-Limestone is added to the blast furnace to produce more reducing agent and remove slag in the blast furnace for extraction of Iron. (c)Group VII elements: Halogens Group VII elements are called Halogens. They are all non metals. They include: Element Symbol Atomic number Electronicc configuration Charge of ion Valency State at Room Temperature Fluorine Chlorine Bromine Iodine Astatine F Cl Br I At 9 17 35 53 85 2:7 2:8:7 2:8:18:7 2:8:18:18:7 2:8:18:32:18:7 F- Cl- Br- I- At- 1 1 1 1 1 Pale yellow gas Pale green gas Red liquid Grey Solid Radioactive All halogen atoms have seven electrons in the outer energy level. They acquire/gain one electron in the outer energy level to be stable. They therefore are therefore monovalent .They exist in oxidation state X- The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g.
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They therefore are therefore monovalent .They exist in oxidation state X- The number of energy levels increases down the group from Fluorine to Astatine. The more the number of energy levels the bigger/larger the atomic size. e.g. The atomic size/radius of Chlorine is bigger/larger than that of Fluorine because Chlorine has more/3 energy levels than Fluorine (2 energy levels). Atomic radius and ionic radius of Halogens increase down the group as the number of energy levels increases. The atomic radius of Halogens is smaller than the ionic radius. This is because they react by gaining/acquiring extra one electron in the outer energy level. The effective nuclear attraction on the more/extra electrons decreases. The incoming extra electron is also repelled causing the outer energy level to expand to reduce the repulsion and accommodate more electrons. Table showing the atomic and ionic radius of four Halogens 42 Element Symbol Atomic number Atomic radius(nM) Ionic radius(nM) Fluorine F 9 0.064 0.136 Chlorine Cl 17 0.099 0.181 Bromine Br 35 0.114 0.195 Iodine I 53 0.133 0.216 The atomic radius of Chlorine is 0.099nM .The ionic radius of Cl- is 0.181nM. This is because Chlorine atom/molecule reacts by gaining/acquiring extra one electrons. The more/extra electrons/energy level experience less effective nuclear attraction /pull towards the nucleus .The outer enegy level expand/increase to reduce the repulsion of the existing and incoming gained /acquired electrons. Electronegativity The ease of gaining/acquiring extra electrons is called electronegativity. All halogens are electronegative. Electronegativity decreases as atomic radius increase. This is because the effective nuclear attraction on outer electrons decreases with increase in atomic radius. The outer electrons experience less nuclear attraction and thus ease of gaining/acquiring extra electrons decrease. It is measured using Pauling’s scale. Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron. Table showing the electronegativity of the halogens.
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It is measured using Pauling’s scale. Where Fluorine with Pauling scale 4.0 is the most electronegative element and thus the highest tendency to acquire/gain extra electron. Table showing the electronegativity of the halogens. Halogen F Cl Br I At Electronegativity (Pauling scale) 4.0 3.0 2.8 2.5 2.2 The electronegativity of the halogens decrease down the group from fluorine to Astatine. This is because atomic radius increases down the group and thus decrease electron – attracting power down the group from fluorine to astatine. Fluorine is the most electronegative element in the periodic table because it has the small atomic radius. Electron affinity The minimum amount of energy required to gain/acquire an extra electron by an atom of element in its gaseous state is called 1st electron affinity. The SI unit of electron affinity is kilojoules per mole/kJmole-1 . Electron affinity depend on atomic radius. The higher the atomic radius, the less effective the nuclear attraction on outer energy level electrons and thus the lower the electron affinity. For halogens the 1st electron affinity decrease down the group as the 43 atomic radius increase and the effective nuclear attraction on outer energy level electrons decrease. Due to its small size/atomic radius Fluorine shows exceptionally low electron affinity. This is because a lot of energy is required to overcome the high repulsion of the existing and incoming electrons. Table showing the election affinity of halogens for the process X + e -> X- The higher the electron affinity the more stable theion.i.e Cl- is a more stable ion than Br- because it has a more negative / exothermic electron affinity than Br- Electron affinity is different from: (i) Ionization energy. Ionization energy is the energy required to lose/donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state. (ii) Electronegativity. -Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process: X(g) + e -> X-(g) Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.
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(ii) Electronegativity. -Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process: X(g) + e -> X-(g) Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions. It does not involve use of energy but theoretical arbitrary Pauling’ scale of measurements. Physical properties State at room temperature Fluorine and Chlorine are gases, Bromine is a liquid and Iodine is a solid. Astatine is radioactive . All halogens exist as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces. Melting/Boiling point The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius. Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid. Iodine sublimes when heated to form (caution: highly toxic/poisonous) purple vapour. Halogen F Cl Br I Electron affinity kJmole-1 -333 -364 -342 -295 44 This is because Iodine molecules are held together by weak van-derwaals/intermolecular forces which require little heat energy to break. Electrical conductivity All Halogens are poor conductors of electricity because they have no free delocalized electrons. Solubility in polar and non-polar solvents All halogens are soluble in water(polar solvent). When a boiling tube containing either chlorine gas or bromine vapour is separately inverted in a beaker containing distilled water and tetrachloromethane (non-polar solvent), the level of solution in boiling tube rises in both water and tetrachloromethane. This is because halogen are soluble in both polar and non-polar solvents. Solubility of halogens in water/polar solvents decrease down the group. Solubility of halogens in non-polar solvent increase down the group. The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous.
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Solubility of halogens in non-polar solvent increase down the group. The level of water in chlorine is higher than in bromine and the level of tetrachloromethane in chlorine is lower than in bromine. Caution: Tetrachloromethane , Bromine vapour and Chlorine gas are all highly toxic/poisonous. Table showing the physical properties of Halogens Halogen Formula of molecule Electrical conductivity Solubility in water Melting point(oC) Boiling point(oC) Fluorine F2 Poor Insoluble/soluble in tetrachloromethane -238 -188 Chlorine Cl2 Poor Insoluble/soluble in tetrachloromethane -101 -35 Bromine Br2 Poor Insoluble/soluble in tetrachloromethane 7 59 Iodine I2 Poor Insoluble/soluble in tetrachloromethane 114 sublimes Chemical properties (i)Displacement Experiment Place separately in test tubes about 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions. Add 5 drops of chlorine water to each test tube: Repeat with 5 drops of bromine water instead of chlorine water 45 Observation Using Chlorine water -Yellow colour of chlorine water fades in all test tubes except with sodium chloride. -Coloured Solution formed. Using Bromine water Yellow colour of bromine water fades in test tubes containing sodium iodide. -Coloured Solution formed. Explanation The halogens displace each other from their solution. The more electronegative displace the less electronegative from their solution. Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine. Bromine is more electronegative than iodide but less 6than chlorine. On adding Bromine water, iodine is displaced from its solution but not chlorine.
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