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What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? a.`Kul(I)_(3)` b. `H_(2)ul(S)_(4)O_(6)` c. `ul(Fe)_(3)O_(4)` d. `ul(C )H_(3)ul(C )H_(2)OH` e. `ul(C )H_(3)ul(C )OOH` |
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Answer» In `KI_(3)`, since the oxidation numbers of `K` is `+1`, therefore, the average oxidation number of iodine `= -1//3`. But the oxidation number cannot be fractional. Therefore, we must consider its structure, `K^(+)[I-I larrI]^(Θ)`. Here a coordinate bond is formed between `I_(2)` molecule and `I^(Θ)` ion. The oxidation number of two iodine atoms forming the `I_(2)` molecule is zero while that of iodine forming the coordinate bond is `-1`. b. By conventional method: `overset(+1)(H_(2))overset(x)(S_(4))overset(-2)(O_(6)) or 2(+1)+4x+6(-2)=0` `x=+2.5` (wrong). But it is wrong beacuse all the four `S` atoms cannot be in the same oxidation state. By chemical bonding method, the structure of `H_(2)S_(4)O_(6)` is shown below: `H-Ooverset(+5)(-)underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)overset(+5)(-)underset(O)underset(||)overset(O)overset(||)(S)-OH` The oxidation number of each of the `S` atoms linked with each other in the middle is zero while that of each of the remaining two `S` atoms is `+5`. c. By conventional method. `overset(x)(Fe_(3))overset(-2)(O_(4))` or `3x+4(-2)=0` or `x=8//3` By stoichiometry `Fe_(3)O_(4) -= overset(+2)(Fe)overset(-2)(O).overset(+3)(Fe_(2))overset(-2)(O_(3))` Fe has oxidation number of `+2` and `+3`. d. By conventional method. `CH_(3)CH_(2)OH= overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(O)` or `2x+6(+1)+1(-2)=0` or `x=-2` By chemical bonding `C_(2)` is attached to three `H` atoms (less electronegative than carbon) `Hoverset(2)(-)underset(H)underset(|)overset(H)overset(|)(C )overset(1)(-)underset(H)underset(|)overset(H)overset(|)(C )-OH` and one `CH_(2)OH` group (more electronegative than carbon). Therefore, oxidation number of `C_(3)=3(+1)-x+1(-1)=0` or `x=-2` `C_(1)` is, however, attached to one `OH` (oxidation number `-1)` and one `CH_(3)` (oxidation number `=+1)` of `C_(1)=+1+2(+1)+x+1(-1)=0` group, `:.` Oxidation number of `C_(1)=+1+2(+1)+x+1(-1)=0` `x=-2` e. By conventional method: `CH_(3)COOH=overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(O_(2)) or 2x+4-4=0` or `x=0` By chemical bonding method: `C_(2)` is attached to three `H` atoms (less electronegative then carbon) `Hoverset(2)(-)underset(H)underset(|)overset(H)overset(|)(C )overset(1)(-)overset(O)overset(||)(C )-OH` and one `-COOH` group (more electronegative tan carbon. Therefore, oxidation number of `C_(2)=3(+1)+x+1(-1)=0` or `x=-2` `C_(1)` is, however, attached to one oxygen atom by a double bond, one `OH` (oxidation number `=-1)` and one `CH_(3)` (oxidation number `=+1)` group. Therefore, oxidation number of `C_(1)=+1+x+1(-2)+1(-1)=0` or `x=+2` |
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