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

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

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2

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(a) Na2PO4

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

Na+1H+12PxO-24

Then, we have Hence, the oxidation number of P is +5.

(b) Na+1H+1SxO-24

Then, we have Hence, the oxidation number of S is + 6.

(c) H+14Px2O-27

Then, we have Hence, the oxidation number of P is + 5.

(d) K+12MnxO-24

Then, we have Hence, the oxidation number of Mn is + 6.

(e) Ca+2Ox2

Then, we have Hence, the oxidation number of O is – 1.

(f) Na+1BxH-14

Then, we have Hence, the oxidation number of B is + 3

(g) H+2Sx2O-72

Then, we have Hence, the oxidation number of S is + 6.

(h) K+1Al3+(SxO2-4).12H+12O-2

Then, we have 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 Hence, the oxidation number of S is + 6.