Question

assign oxidation number to the under lined element.

a) K Mno4 b) Na BH4.​

Answer 1

Explanation:

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

Answer

(a) NaH2PO4

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

(b) NaHSO4

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

(c) H4P2O7

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

(d) K2MnO4

Let assume oxidation number of Mn is x.

Oxidation number of K = +1

Oxidation number of O = -2

Then we have:

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

⇒ 2 + x - 8 = 0

⇒ x - 6 = 0

⇒ x = +6

Hence, Oxidation number of Mn is +6

(e) CaO2

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

(f) NaBH4

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.

(g) H2S2O7

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.

(h) KAl(SO4)2.12 H2O

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.