Assign oxidation numbers 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

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Nikhil Reddy

(a) 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 Then, we have Hence, the oxidation number of P is +5. (b) Then, we have Hence, the oxidation number of S is + 6. (c) Then, we have Hence, the oxidation...
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(a) 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 Then, we have Hence, the oxidation number of P is +5. (b) Then, we have Hence, the oxidation number of S is + 6. (c) Then, we have Hence, the oxidation number of P is + 5. (d) Then, we have Hence, the oxidation number of Mn is + 6. (e) Then, we have Hence, the oxidation number of O is – 1. (f) Then, we have Hence, the oxidation number of B is + 3. (g) Then, we have Hence, the oxidation number of S is + 6. (h) 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. read less
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