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Redox Reactions · Question 23

AIPMT 2003MCQEasy📈 Topic in 25% papers+4−1
1

The oxidation states of sulphur in the anions SO32-, S2O42- and S2O62- follow the order

  1. S2O42- < SO32- < S2O62-
  2. SO32- < S2O42- < S2O62-
  3. S2O42- < S2O62- < SO32-
  4. S2O62- < S2O42- < SO32-

Answer

(1) S2O42- < SO32- < S2O62-

Solution

Let the oxidation number of S be x (O = −2).

SO32-: x + 3(-2) = -2 x = +4

S2O42-: 2x + 4(-2) = -2 x = +3

S2O62-: 2x + 6(-2) = -2 x = +5

S2O42- (+3) < SO32- (+4) < S2O62- (+5)

Answer: (1)

AIPMT 2003

Keep practising — Redox Reactions

2

Which reaction is not feasible?

  1. 2KI + Br2 2KBr + I2
  2. 2KBr + I2 2KI + Br2
  3. 2KBr + Cl2 2KCl + Br2
  4. 2H2O + 2F2 4HF + O2
AIPMT 2002
3

Oxidation numbers of A, B, C are +2, +5 and -2 respectively. Possible formula of compound is

  1. A2(BC2)2
  2. A3(BC4)2
  3. A2(BC3)2
  4. A3(B2C)2
AIPMT 2000

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