Question

Assign oxidation numbers to the underlined elements in each of the following species:

1. $NaH_2\underline PO_4$
2. $NaH\underline SO_4$
3. $H_4\underline P_2O_7$
4. $K_2\underline {Mn}O_4$
5. $Ca\underline O_2$
6. $Na\underline BH_4$
7. $H_2\underline S_2O_7$
8. $KAl(\underline SO_4)_2.12 H_2O$

Answer 1

(a) $NaH_2\underline PO_4$
Let the oxidation number of$P$ be $x$.
We know that,
Oxidation number of $Na$ = $+1$
Oxidation number of $H$ = $+1$
Oxidation number of $O$ = $-2$
$\Rightarrow \overset{+1}Na\overset{+1}H_2\overset{x}P\overset{-2}O_4$
Then, we have

$1(+1) + 2(+1) + 1(x) + 4(-2) = 0$
$1 + 2 + x -8 = 0$
$x = +5$
Hence, the oxidation number of $P$ is $+5$.

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(b) $NaH\underline SO_4$
$\overset{+1}Na\overset{+1}H\overset{x} S\overset{-2}O_4$
Then, we have
$1(+1) + 1(+1) + 1(x) + 4(-2) = 0$
\Rightarrow$$1 + 1 + x -8 = 0
\Rightarrow$$x = +6
Hence, the oxidation number of $S$ is $+ 6$.

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(c) $H_4\underline P_2O_7$
$\overset{+1} H_4\overset{x} P_2\overset{-2}O_7$
Then, we have
$4(+1) + 2(x) + 7(-2) = 0$
\Rightarrow$$4 + 2x – 14 = 0
\Rightarrow$$2x = +10
\Rightarrow$$x = +5
Hence, the oxidation number of $P$ is $+ 5$.

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(d) $K_2\underline {Mn}O_4$
$\overset{+1}K_2\overset{x} {Mn}\overset{-2}O_4$
Then, we have
$2(+1) + x + 4(-2) = 0$
\Rightarrow$$2 + x – 8 = 0
\Rightarrow$$x = +6
Hence, the oxidation number of $Mn$ is $+ 6$.

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(e) $Ca\underline O_2$
$\overset{+2}Ca\overset{x} O_2$
Then, we have
$(+2) + 2(x) = 0$
$\Rightarrow$ $2 + 2x = 0$
$\Rightarrow$ $2x = -2$
$\Rightarrow$ $x = -1$
Hence, the oxidation number of $O$ is $- 1$.

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(f) $Na\underline BH_4$
$\overset{+1}Na\overset{x} B\overset{-1}H_4$
Then, we have
$1(+1) + 1(x) + 4(-1) = 0$
$\Rightarrow$ $1 + x -4 = 0$
$\Rightarrow$ $x = +3$
Hence, the oxidation number of $B$ is $+ 3$.

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(g) $H_2\underline S_2O_7$
$\overset{+1}H_2\overset{x} S_2\overset{-2}O_7$
Then, we have
$2(+1) + 2(x) + 7(-2) = 0$
$\Rightarrow$ $2 + 2x – 14 = 0$
$\Rightarrow$ $2x = +12$
$\Rightarrow$ $x = +6$
Hence, the oxidation number of $S$ is $+ 6$.

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(h) $KAl(\underline SO_4)_2.12 H_2O$
$\overset{+1}K\overset{+3}Al\bigg(\overset{x} S\overset{2-}O_4\bigg)_2.12 \overset{+1}H_2\overset{-2}O$
Then, we have
$1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2) = 0$
$\Rightarrow$ $1 + 3 + 2x – 16 + 24 – 24 = 0$
$\Rightarrow$ $2x = +12$
$\Rightarrow$ $x = +6$
Or,
We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero.
Therefore, after ignoring the water molecule, we have:
$1(+1) + 1(+3) + 2(x) + 8(-2) = 0$
$1 + 3 + 2x -16 = 0$
$2x = 12$
$x = +6$
Hence, the oxidation number of $S$ is $+ 6$.