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Assign oxidation number to the underlined element in each of the following species NaH_(2)PO_(4) (b) NaHSO_(4) (c )H_(4)P_(2)O_(7) (d)K_(2)MnO_(4) (e )CO_(2) (f) NaBH_(4) (g)H_(2)S_(2)O_(7) (h)KAI(SO_(4))_(2)12H_(2)O |
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Answer» Solution :(a) Let the oxidation number of p be x writing the oxidation number of each atom above its symbol we sum of oxidation numbr of various atoms in `NaH_(2)PO_(4)=1(+1)+2 (+1)+1(x)+4(-2)=x-5` but the sum of oxdation numberof various atoms in `NaH_(2)PO_(4)` (neutral) is ZERO Thus the oxidation nuber of p in `NaH_(2)PO_(4)=5` (b) `overset(+1)NA overset (x)H overset(x)S overset(-2) O_(4) therefore +1(+1)+x+4(+1)-2=0 or x=_6` Thus the oxidation number of S in `NaHSO_(4)=+5` (C ) `overset(+1)H_(4)overset(+1)P_(2)overset(x)S overset(-2)O_(4)therefore4(+1)+2(x)+7(-2)=0 or x=+5` (d)`K_(2)Mn O_(4)^(-2) therefore 2(+1)+1(x)+7(-2)=0 or x =+6` thus the oxidation numbr of Mn in `K_(2)MnIO_(4)=+7` (e ) let hte oxidation number of O be x since ca is an alkaline earth METAL therefore its oxidation number is +2 thus `CaO_(2)therefore +2+2(x)=0 or x=-1` (f) In `NaBH_(4)` H is present as hydride ion therefore its oxidation number ois -1 thus `Na BH_(4) therefore2(+1)+x+(-1)=0 or x=+3` thus the oxidation number of B in `NaBH_(4)=+3` (g) `H_(2)S_(2)O_(7)^(2-) therefore 2(+1)+2(x)+7(-2)=0 or x_6` thus the oxidation number of s in `H_(2)S_(2)O_(7)=+6` (h)`K AI(SO_(4))12(H_(2)O or +1+3+2x+8(-2)+12(2xx1-2)or x=+6` alternatively since `H_(2)O` is a neutral molecule therefore sum of oxidation numberof S `therefore+1+3+2+x-16=0or x=+6` thus the oxidation number of S in KaI `(SO_(4))12H_(2)O=+6` |
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