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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

In which of the following pairs of molecules/ions, the central atoms have sp2sp^2 hybridization?

A

NO2−NO_2^- and NH3NH_3

B

BF3BF_3 and NO2−NO_2^-

C

NH2−NH_2^- and H2OH_2O

D

BF3BF_3 and NH2−NH_2^-

Step-by-Step Solution

  1. Method: Determine the steric number (SNSN) using the formula: SN=12(V+M−C+A)SN = \frac{1}{2}(V + M - C + A), where VV is valence electrons, MM is monovalent atoms, CC is cationic charge, and AA is anionic charge. SN=3SN=3 corresponds to sp2sp^2 hybridization .
  2. Analyze BF3BF_3:
  • Boron (Group 13) has V=3V=3. Fluorine is monovalent (M=3M=3).
  • SN=12(3+3−0+0)=3SN = \frac{1}{2}(3 + 3 - 0 + 0) = 3. Hybridization is sp2sp^2.
  1. Analyze NO2−NO_2^- (Nitrite Ion):
  • Nitrogen (Group 15) has V=5V=5. Oxygen is divalent (M=0M=0). Charge (A=1A=1).
  • SN=12(5+0−0+1)=3SN = \frac{1}{2}(5 + 0 - 0 + 1) = 3. Hybridization is sp2sp^2.
  1. Analyze other species:
  • NH3NH_3: SN=12(5+3)=4→sp3SN = \frac{1}{2}(5 + 3) = 4 \rightarrow sp^3.
  • NH2−NH_2^-: SN=12(5+2+1)=4→sp3SN = \frac{1}{2}(5 + 2 + 1) = 4 \rightarrow sp^3.
  • H2OH_2O: SN=12(6+2)=4→sp3SN = \frac{1}{2}(6 + 2) = 4 \rightarrow sp^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 Structurefollowingmoleculesionscentralhybridization

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