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

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O

(a) NaH${}_{2}$PO4

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

$⇒$$\stackrel{+1}{\mathrm{Na}}{\stackrel{+1}{\mathrm{H}}}_{2}\stackrel{\mathrm{x}}{\mathrm{P}}{\stackrel{-2}{\mathrm{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) $\mathrm{NaH}\overline{)\mathrm{S}}{\mathrm{O}}_{4}$

$\stackrel{+1}{\mathrm{Na}}\stackrel{+1}{\mathrm{H}}\stackrel{\mathrm{x}}{\mathrm{S}}\stackrel{-2}{{O}_{4}}$

Then, we have

1(+1)+1(+1)+1(x)+4(-2)=0

$⇒$1+1+x-8=0

$⇒$x=+6

Hence, the oxidation number of S is + 6.

(c) ${\mathrm{H}}_{4}{\overline{)\mathrm{P}}}_{2}{\mathrm{O}}_{7}$

${\stackrel{+1}{\mathrm{H}}}_{4}{\stackrel{x}{P}}_{2}{\stackrel{-2}{O}}_{7}$

Then, we have

4(+1)+2(x)+7(-2)=0

$⇒$4+2 x-14=0

$⇒$2x=+10

$⇒$x=+5

Hence, the oxidation number of P is + 5.

(d) ${\mathrm{K}}_{2}\overline{)\mathrm{M}}{\mathrm{nO}}_{4}$

${\stackrel{+1}{\mathrm{K}}}_{2}\stackrel{\mathrm{x}}{\mathrm{M}}\mathrm{n}{\stackrel{-2}{\mathrm{O}}}_{4}$

Then, we have

2(+1)+x+4(-2)=0

$⇒$2+x-8=0

$⇒$x=+6

Hence, the oxidation number of Mn is + 6.

(e) $\mathrm{Ca}{\overline{)\mathrm{O}}}_{2}$

$\stackrel{+2}{\mathrm{Ca}}{\stackrel{\mathrm{x}}{\mathrm{O}}}_{2}$

Then, we have

(+2)+2(x)=0

$⇒$2+2 x=0

$⇒$x=-1

Hence, the oxidation number of O is – 1.

(f) $\mathrm{Na}\overline{)\mathrm{B}}{\mathrm{H}}_{4}$

$\stackrel{+1}{\mathrm{Na}}\stackrel{\mathrm{x}}{\mathrm{B}}{\stackrel{-1}{\mathrm{H}}}_{4}$

Then, we have

1(+1)+1(x)+4(-1)=0

$⇒$1+x-4=0

$⇒$x=+3

Hence, the oxidation number of B is + 3.

(g) ${\mathrm{H}}_{2}{\overline{)\mathrm{S}}}_{2}{\mathrm{O}}_{7}$

${\stackrel{+1}{\mathrm{H}}}_{2}{\stackrel{\mathrm{x}}{\mathrm{S}}}_{2}{\stackrel{-2}{\mathrm{O}}}_{7}$

Then, we have

2(+1)+2(x)+7(-2)=0

$⇒$2+2x-14=0

$⇒$2x=12

$⇒$x=+6

Hence, the oxidation number of S is + 6.

(h) $\mathrm{KAl}{\left(\overline{)\mathrm{S}}{\mathrm{O}}_{4}\right)}_{2}·12{\mathrm{H}}_{2}\mathrm{O}$

Then, we have

1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2)=0

$⇒$1+3+2 x-16+24-24=0

$⇒$2x=12

$⇒$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+2 x-16=0

$⇒$2x=12

$⇒$x=+6

Hence, the oxidation number of S is + 6.