Question

Assign oxidation number to the underlined elements in each of the following species.
(i)${NaH}_2\underset{–}{P}{O}_4$
(ii)$NaH\underset{–}{S}{O}_4$
(iii)$H_4{\underset{–}{P}}_2{O}_7$
(iv)$H_2{\underset{–}{S}}_2{O}_7$
(v)$Ca{\underset{–}{O}}_2$
(vi)$Na\underset{–}{B}{H}_4$
(vii)$KAl{\left({\underset{–}{SO}}_4\right)}_2.12H_{2}O$
(viii)$\underset{–}{C{r}_2}{O}_7^{2-}$

Answer 1

(i) Let assume oxidation number of P is x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have

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

⇒ 1 + 2 + x - 8 = 0

⇒ x - 5 = 0

⇒ x = + 5

Hence the oxidation number of P is +5

(ii) Let assume oxidation number of S is x.

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

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

⇒ 1 + 1 + x - 8 = 0

⇒ x-6 = 0

⇒ x = +6

Hence the oxidation number of S is +6

(iii) Let assume oxidation number of P is x.

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

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

⇒ 4 + 2x - 14 = 0

⇒ 2x - 10 = 0

⇒ 2x = +10

⇒ x = +5

Hence, Oxidation number of P is +5

(iv) Let assume oxidation number of S is x.

Oxidation number of O = -2

Oxidation number of H = +1

Then we have:

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

⇒ 2 + 2x - 14 = 0

⇒ 2x - 12 = 0

⇒ x = +6

Hence, Oxidation number of S is +6.

(v) Let assume oxidation number of O is x.

Oxidation number of Ca = +2

Then we have:

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

⇒ 2 + 2x = 0

⇒ 2x = -2

⇒ x = -1

Hence, Oxidation number of O is -1

(vi) Let assume oxidation number of B is x.

Oxidation number of Na = +1

Oxidation number of H = -1

Then we have:

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

⇒ 1 + x -4 = 0

⇒ x - 3 = 0

⇒ x = +3

Hence, Oxidation number of B is +3.

(vii) Let assume oxidation number of S is x.

Oxidation number of K = +1

Oxidation number of Al = +3

Oxidation number of O = -2

Oxidation number of H = +1

Then we have:

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

⇒ 1 + 3 + 2x -16 +24 -24 = 0

⇒ 2x - 12 = 0

⇒ 2x = +12

⇒ x = +6

Hence, Oxidation number of S is +6.

(viii) Let us assume oxidation number of Cr is x.

Oxidation number of O = -2

Then we have :

2(x) -2(7) = -2

x = +6

Hence, Oxidation number of Cr is +6.