Calculate individual and average oxidation number (if required) of the marked element and also draw the structure of the following compounds or molecules.
(1) Na 2 S 2 O 3 (2) Na 2 S 4 O 6
(3) H 2 S O 5 (4) H 2 S 2 O 8
(5) H 2 S 2 O 7 (6) S 8
(7) H N O 4 (8) C 3 O 2
(9) Os O 4 (10) P H 3
(11) Cr O 4 2– (12) Cr 2 O 7 2– (13) Cr O 2 Cl 2 (14) C rO 5
(15) Na 2 H P O 4 (16) Fe S 2
(17) C 6 H 12 O 6 (18) N H 4 N O 3
Text Solution
Verified by Experts1
(1) +2 (6, –2) (2) +5/2(5, 5, 0, 0)
(3) +6 (4) +6 (+6, +6)
(5) +6 (+6, +6) (6) 0
(7) +5 (8) 4/3 (+ 2, +2, 0)
(9) +8 (10) –3
(11) +6 (12) +6 (+6, +6) (13) +6 (14) +6
(15) +5 (16) +2
(17) 0 (18) –3, +5
(1) Na 2 S 2 O 3 (Sodium thiosulphate).
2 × 1 + 2x + 3 × (– 2) = 0.
(+2) + 2x + (–6) = 0.
2x = (+6) + (–2).
2x = + 4.
x = + 2. (average oxidation number).
(2) Na 2 S 4 O 6 (Sodium tetra-thionate)
2 × (+1) + 4x + 6 × (– 2) = 0.
4x = (+12) + (–2).
4x = +10.
x = + 
= + 
. (average oxidation number).
(3) H 2 SO 5 (Caro's acid)
2 × (+1) + x + 2 × (–1) + 2 × (–1) + 3 × (– 2) = 0.
(+2) + x + (–2) + (–6) = 0.
x = +6. (average oxidation number).
(4) H 2 S 2 O 8 (Marshal's acid).
2 × (+1) + 2x + 2 × (–1) + 6 × (–2) = 0.
(+2) + 2x + (–2) + (–12) = 0.
2x = +12.
x = +6. (average oxidation number).
(5) H 2 S 2 O 7 (Pyro sulphuric acid).
2 × (+1) + 2x + 7 × (–2) = 0.
(+2) + 2x = + 14.
2x = (+14) + (–2).
2x = +12.
x = + 6.(average oxidation number).
(6) S 8 (Crown sulphur).
Oxidation no. of element in homogeneous molecule will be zero.
(7) HNO 4 (Peroxynitric acid)
+1 + x + 2 × (–1) + 2 × (–2) = 0.
x = (+6) + (–1).
x = +5.
(8) C 3 O 2 (Carbon suboxide)
3x + 2 × (–2) = 0.
3x = + 4.
x = 4/3.
(9) O 5 O 4 (Osmium tetra oxide)
x + 4 × (–2) = 0.
x = +8.
(10) PH 3 (phosphene)
x + 3 × (+1) = 0.
x = – 3.
(11) CrO 4
2– (cromate ion)
x + 4 × (–2) = – 2.
x = (+8) + (–2).
x = + 6.
(12) Cr 2 O 7 2– (dichromate ion)
2 × x + 7 × (–2) = – 2.
2x = (+14) + (–2).
2x = + 12.
x = + 6.
(13) CrO 2 Cl 2 (cromyl chloride)
x + 2 × (–2) + 2 × (–1) = 0.
x + (–4) + (–2) = 0.
x = + 6.
(14) CrO 5 (chromium per oxide)
x + 4 × (–1) + 1 × (–2) = 0.
x = + 6.
(15) Na 2 HPO 4 (disodium hydrogen phosphate)
(2 × +1) + (–1) + x + 4 × (–2) = 0.
(+2) + (+1) + x + (–8) = 0.
x = (+8) +
x = +5.
(16) FeS 2 (Ferrous disulphide or Fool's gold or Iron pyrite)
x + 2x – 1 = 0.
x = +2.
(17) C 6 H 12 O 6 (Glucose or Fructose)
6 × x + 12 × + 6 × (–2) = 0.
6x + (+12) + (–12) = 0.
x = 
= 0.
(18) NH 4 NO 3 (Ammonium nitrate)
NH 4 NO 3 → [NH 4 ] + [NO 3 ] – .
NH 4 +
x + 4 × (+1) = +1.
x = – 3.
NO 3 –
x + 3 × (–2) = – 1.
x = + 5.
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