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What are th oxidation number of the underlined elements in each of the following and how do you rationalize your results ? (a) KI_(3) (b) H_(2)S_(4)O_(6) (c )Fe_(3)O_(4) (d) CH_(3)CH_(2)OH (e )CH_(3)COOH |
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Answer» Solution :(A) in `KI_(3)` since the oxidation number of K is +1 therefore the average oxidation number of iodine =-`1//3` but the oxidation numbercannot be fractional therefore we must consider its structure `K^(+)[I-lrarr]^(-)` here a coordinate bond is formed between `I_(2)` moleceul and `T^(-)` ion the oxidation number of two IODIN eatoms forming the `IO_(2)` molecure is zero while that of iodine forming the coordinate bond is -1 thus the O.N of there I atoms atoms in `KI_(3)` are 0,0 and -1 respectively (b) By conventional method O.N of S in `H_(2)S_(4)O_(6) =overset((+1)H_(2)overset(x)S_(4)overset(-2)O_(6) or 2(+1)` (b) By conventional method O.N of S in `H_(2)S_(2)O_(6)=H_(2)S_(4)O_(6)` or 2(+1) +4 x+6 (-2) =0 or x=+2.5 (wrong) but it is wrong because all the FOIU S atoms cannot be int he same oxidation state by chemical bonding method the structure of `H_(2)S_(4)O_(6)` is show below `H-O overset(+5)-underset(O)underset(||)overset(O)overset(||)S-overset(S)-overset(0)S-underset(O)underset(||)overset(O)overset(||)Soverset((+5)-OH` The O.N of each of the S atoms linked witheach other in the middle is zero whilethat of each of the remaining two S atoms is +5 (c ) By conventional method O.N of Fe in `Fe_(3) O_(4)^(-2)` therefore of Fe in `Fe_(3)O_(4)^2` and +3 (d) By conventional method O.N of Cin `H_(3)CH_(2)OH =C_(2)H_(6)O` or 2x+6(+1)+1 (-2) =0or x=-2 by chemical bonding `C_(2)` is ATTACHED to threeH atoms (less electronegative than carbon ) and one `CH_(2)OH` group (more electronegative than carbon) therefore `H overset(2)-underset(H)underset(|)overset(H)overset(|)C-underset(H)underset(|)overset(H)overset(|)C-OH` O.N of `C_(2)=3(+1)+x1(-1)=0 or x=-2` `C_(1)` is however attachedot one OH (O.N =-1) and one `CH_(3)`(O.N)=+1group (e ) By conventional method `CH_(3)COOH=C_(2)H_(4)O_(2) or 2x+4-4=0 or x=0` by chemical bonding method `C_(2)` is attached to three h atomsand one -COOH group (more electronative than carbon) `H-underset(H)underset(|)overset(H)overset(|)C-overset(O)overset(||)C-OH` n `C_(1)` is however attached to one oxygen atom by double bond one OH(O.N =-1) and one `CH_(3)(O.N=+1)` group therefore therefore O.N of `C_(1) =+1+x+1(-2)+1 =0 OH or x=+2` |
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