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p-Block Elements (Group 15–18) · Question 7

JEE Advanced 2024 Paper 1MCQEasy📈 Topic in 17% papers+4−1
1

At room temperature, disproportionation of an aqueous solution of in situ generated nitrous acid (HNO2) gives the species

  1. H3O+, NO3- and NO
  2. H3O+, NO3- and NO2
  3. H3O+, NO- and NO2
  4. H3O+, NO3- and N2O

Answer

(1) H3O+, NO3- and NO

Solution

In HNO2, N is in +3 state; it disproportionates to +5 (NO3-) and +2 (NO).

3HNO2 HNO3 + 2NO + H2O

In water: 3HNO2 H3O+ + NO3- + 2NO

Electron balance: 1 N gains +2 (3→5), 2 N lose 1 each (3→2).

Answer: (A)

JEE Advanced 2024 Paper 1

Keep practising — p-Block Elements (Group 15–18)

2

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.

  1. P-5, Q-2, R-3, S-1
  2. P-5, Q-3, R-2, S-4
  3. P-4, Q-3, R-2, S-1
  4. P-4, Q-2, R-5, S-3
JEE Advanced 2024 Paper 1
3

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are

  1. [PF4]+[PF6]- and [PCl4]+[PF6]-
  2. [PCl4]+[PCl4F2]- and [PCl4]+[PF6]-
  3. PF3 and PCl3
  4. PF5 and PCl3
JEE Advanced 2024 Paper 2
4

The compound(s) having peroxide linkage is(are)

  1. H2S2O7
  2. H2S2O8
  3. H2S2O5
  4. H2SO5
JEE Advanced 2024 Paper 2

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