Assign oxidation numbers to the underlined elements in each of the following species: $NaH_2\underline PO_4$ $NaH\underline SO_4$ $H_4\underline P_2O_7$ $K_2\underline {Mn}O_4$ $Ca\underline O_2$ $Na\underline BH_4$ $H_2\underline S_2O_7$ $KAl(\underline SO_4)_2.12 H_2O$

Question

Assign oxidation numbers to the underlined elements in each of the following species:  $NaH_2\underline PO_4$   $NaH\underline SO_4$   $H_4\underline P_2O_7$   $K_2\underline {Mn}O_4$   $Ca\underline O_2$   $Na\underline BH_4$   $H_2\underline S_2O_7$   $KAl(\underline SO_4)_2.12 H_2O$

cdquestions Admin · Accepted Answer

(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  $+5$ . (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  $+ 6$ . (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  $+ 5$ . (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  $+ 6$ . (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  $- 1$ . (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  $+ 3$ . (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  $+ 6$ . (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  $+ 6$ .

Assign oxidation numbers to the underlined elements in each of the following species:
\(NaH_2\underline PO_4\)
\(NaH\underline SO_4\)
\(H_4\underline P_2O_7\)
\(K_2\underline {Mn}O_4\)
\(Ca\underline O_2\)
\(Na\underline BH_4\)
\(H_2\underline S_2O_7\)
\(KAl(\underline SO_4)_2.12 H_2O\)

Solution and Explanation

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Redox Reactions

Redox Reaction:

Types of Redox Reactions:

Combination Reaction:

Decomposition Reaction:

Displacement Reaction:

Disproportionation Reaction:

Assign oxidation numbers to the underlined elements in each of the following species: \(NaH_2\underline PO_4\) \(NaH\underline SO_4\) \(H_4\underline P_2O_7\) \(K_2\underline {Mn}O_4\) \(Ca\underline O_2\) \(Na\underline BH_4\) \(H_2\underline S_2O_7\) \(KAl(\underline SO_4)_2.12 H_2O\)