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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

The hybridisations of atomic orbitals of nitrogen in NO+^+, NO3−_3^- and NH3_3 respectively are:

A

sp, sp3^3 and sp2^2

B

sp2^2, sp3^3 and sp

C

sp, sp2^2 and sp3^3

D

sp2^2, sp and sp3^3

Step-by-Step Solution

  1. Formula for Hybridisation: The steric number (H) or number of hybrid orbitals can be calculated using the formula: H=12[V+M−C+A]H = \frac{1}{2} [V + M - C + A], where V = valence electrons of central atom, M = number of monovalent atoms, C = positive charge, A = negative charge.
  2. Analysis of NO+^+ (Nitrosonium ion):
  • Central atom N (V=5V=5). No monovalent atoms (M=0M=0). Cationic charge (C=1C=1).
  • H=12[5+0−1+0]=2H = \frac{1}{2} [5 + 0 - 1 + 0] = 2.
  • H=2H=2 corresponds to sp hybridisation .
  1. Analysis of NO3−_3^- (Nitrate ion):
  • Central atom N (V=5V=5). No monovalent atoms (M=0M=0). Anionic charge (A=1A=1).
  • H=12[5+0−0+1]=3H = \frac{1}{2} [5 + 0 - 0 + 1] = 3.
  • H=3H=3 corresponds to sp2^2 hybridisation .
  1. Analysis of NH3_3 (Ammonia):
  • Central atom N (V=5V=5). Three monovalent H atoms (M=3M=3).
  • H=12[5+3−0+0]=4H = \frac{1}{2} [5 + 3 - 0 + 0] = 4.
  • H=4H=4 corresponds to sp3^3 hybridisation .
  1. Conclusion: The correct order is sp, sp2^2, and sp3^3.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Bonding and Molecular Structurehybridisationsatomicorbitalsnitrogenrespectively

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