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What are the oxidation number of the underlined elements in each of the followingand how do you rationalize your results ? (a) KunderlineI_(3) (b) H_(2)underlineS_(4)O_(6) ( c) underlineFe_(3)O_(4) (d) underlineCH_(3)underlineCH_(2)OH (e) underlineCH_(3)underlineCOOH |
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Answer» Solution :(a) `KunderlineI_(3)` = Oxidation number of K = +1 `THEREFORE` Oxidation number of `I=-1/3`, but oxidation number of I be in whole number. `K^(+)[I-IlarrI]^(-)` In `KI_(3),I_(2)andI^(-)` makes a covelant bond between them. Where oxidation number of `I_(2)=0`. `therefore` Oxidation number of I is -1. In `KI_(3)` respectively, oxidation number of 3 IODINE are 0, 0 and -1. (b) `H_(2)underlineS_(4)O_(6)` : Here, same oxidation number of four sulfur are not possible. `H-O-underset(O)underset(||)overset(O)overset(||)(S)-S-S-underset(O)underset(||)overset(O)overset(||)(S)-O-H` Here, 2, S - S bonding their oxidation number is zero. RATHER the two S left, their oxidation number is +1. ( c) `underlineFe_(3)O_(4)` : 3Fe + 4(O) = 0 3Fe + 4(-2) = 0 `x=8/3` Now, from the stoichiometry : `Fe_(3)O_(4)=FeO*Fe_(2)O_(3)` `FeOtoFe+(-2)=0` `thereforeFe=+2` `Fe_(2)O_(3)to2Fe+3(-2)=0` `therefore2Fe-6=0` `therefore2Fe=+6` `thereforeFe=+3` (d) `underlineCH_(3)underlineCH_(2)OH=C_(2)H_(6)O` `therefore2(C)+6(+1)+1(-2)=0` `therefore2(C)+6-2=0` `therefore2(C)+4=0` `thereforeC=-2` Structure : `H-underset(H)underset(|)overset(H)overset(|)(""^(2)C)-underset(H)underset(|)overset(H)overset(|)(C^(1))-OH` Here, `C_(2)` attached with three hydrogen atom and ALSO with the group `CH_(2)OH`. Oxidation number of `C_(2)=3(+1)+C_(2)+1(-1)=0` `thereforeC_(2)=-2` `C_(1)` attached with OH (Oxi. number -1) and - `CH_(3)` (Oxi. number +1). Oxidation number of `C_(1)=+1+2(+1)+C_(1)+1(-1)=0` `thereforeC_(1)=-2` (E) `underlineCH_(3)underlineCOOH` : x + 1 - 2 `CH_(3)COOH:C_(2)H_(4)O_(2)` `therefore2x+4(+1)+2(-2)=0` `therefore2x+4-4=0` `thereforex=0` Structure : `H-underset(H)underset(|)overset(H)overset(|)(""^(2)C)-underset(H)underset(||)overset(1)(C)-OH` Here, `C_(2)` is attached to three hydrogen atoms and with group -COOH. Oxidation number of `C_(2)=3(+1)+C_(2)+1(-1)=0` `thereforeC_(2)=-2` `C_(1)` is attached with oxygen with double bond and other bond with OH and one `CH_(3)` group (Oxi. number = +1). Therefore, Oxidation number of `C_(1)=+1+C_(1)+1(-2)+1(-1)=0` `thereforeC_(1)=+2` |
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