Answer
Please see the answer below.
Work Step by Step
a)
$ ClF $
The oxidation number of Fluorine in all compounds is -1.
Let the oxidation number of Cl be ‘x’.
$ -1 + x = 0$
Then x = +1
Hence oxidation number of Cl in $ClF$ = +1
b)
$ IF_{7} $
The oxidation number of Fluorine in all compounds is -1.
Let the oxidation number of I be ‘x’.
$ (7\times-1) + x = 0$
Then x = +7
Hence oxidation number of I in $IF$ = +7.
c)
$ CH_{4} $
Oxidation number of hydrogen in all cases except in hydrides is +1.
Let the oxidation number of C be ‘x’.
$ (4\times+1) + x = 0$
Then x = +4
Hence oxidation number of C in $CH_{4}$ = +4.
d)
$ C_{2}H_{2} $
Oxidation number of hydrogen in all cases except in hydrides is +1.
Let the oxidation number of C be ‘x’.
$ (2\times+1) +(2\times x) = 0$
Then x = -1
Hence oxidation number of C in $C_{2}H_{2}$ = -1
e)
$ C_{2}H_{4} $
The oxidation number of hydrogen in all cases except in hydrides is +1.
Let the oxidation number of C be ‘x’.
$ (4\times+1) +(2\times x) = 0$
Then x = -2
Hence oxidation number of C in $C_{2}H_{4}$ = -2.
f)
$K_{2}CrO_{4}$
The oxidation number of Potassium in all compounds is +1.
The oxidation number of oxygen in all cases except in peroxides ans super oxides is -2.
Let the oxidation number of Cr be ‘x’.
$ (2\times+1) +(4\times -2)+ x = 0$
Then x = +6
Hence oxidation number of Cr in $K_{2}CrO_{4}$ = +6.
g)
$K_{2}Cr_{2}O_{7}$
The oxidation number of Potassium in all compounds is +1.
The oxidation number of oxygen in all cases except in peroxides ans super oxides is -2.
Let the oxidation number of Cr be ‘x’.
$ (2\times+1) +(7\times -2)+( 2\times x) = 0$
Then x = +6
Hence oxidation number of Cr in $K_{2}Cr_{2}O_{7}$ = +6.
h)
$KMnO_{4}$
The oxidation number of Potassium in all compounds is +1.
The oxidation number of oxygen in all cases except in peroxides ans super oxides is -2.
Let the oxidation number of Mn be ‘x’.
$+1 +(4\times -2)+ x = 0$
Then x = +7
Hence oxidation number of Mn in $KMnO_{4}$ = +7.
i) $NaHCO_{3}$
The oxidation number of Sodium in all compounds is +1.
The oxidation number of hydrogen in all cases except in hydrides is +1.
The oxidation number of oxygen in all cases except in peroxides ans super oxides is -2.
Let the oxidation number of C be ‘x’.
$+1 + (+1) +(3\times -2)+ x = 0$
Then x = +4
Hence oxidation number of C in $NaHCO_{3}$ = +4.
j)
$Li_{2}$ does not exist.
The oxidation number of Lithium in all compounds is +1.
k)
$NaIO_{3}$
The oxidation number of Sodium in all compounds is +1.
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Let the oxidation number of I be ‘x’.
$+1 + +(3\times -2)+ x = 0$
Then x = +5
Hence oxidation number of C in $NaIO_{3}$ = +5.
l) $KO_{2}$
The oxidation number of Potassium in all compounds is +1.
Let the oxidation number of O be ‘x’.
$+1 + +(2\times x) = 0$
Then x = - 1/2
Hence oxidation number of O in $KO_{2}$ = -1/2.
m) $ PF_{6}^{-} $
The oxidation number of Flourine in all compounds is -1.
Let the oxidation number of P be ‘x’.
$ (6\times-1) + x = -1$
Then x = +5
Hence oxidation number of P in $PF_{6}^{-}$ = +5.
n)
$KAuCl_{4}$
The oxidation number of Potassium in all compounds is +1.
The oxidation number of Chlorine in all cases except in the compounds with Fluorine is -1.
Let the oxidation number of Au be ‘x’.
$+1 +(4\times -1)+ x = 0$
Then x = +3
Hence oxidation number of Au in $KAuCl_{4}$ = +3.
