The hybridizations of the atomic orbitals of nitrogen in NO2+, NO3^- and NH4+ respectively are

Correct answer: (C) sp, sp^2 and sp^3. Answer: (C) sp, sp^2 and sp^3 Solution: NO2+ is linear (2 sigma, no lone pair) → sp; NO3^- is trigonal planar (3 sigma regions) → sp^2; NH4+ is tetrahedral (4 sigma) → sp^3.

NEET 2016 Phase 2 · ChemistryHybridisationPrevious Year Question

The hybridizations of the atomic orbitals of nitrogen in $NO X_{2} X^{+}$ , $NO X_{3} X^{-}$ and $NH X_{4} X^{+}$ respectively are

Answer: (C) $s p$ , $s p^{2}$ and $s p^{3}$ . \textbf{Answer:} (C) $s p$ , $s p^{2}$ and $s p^{3}$ \textbf{Solution:} $NO X_{2} X^{+}$ is linear (2 sigma, no lone pair) → $s p$ ; $NO X_{3} X^{-}$ is trigonal planar (3 sigma regions) → $s p^{2}$ ; $NH X_{4} X^{+}$ is tetrahedral (4 sigma) → $s p^{3}$ .

A. $s p$ , $s p^{3}$ and $s p^{2}$
B. $s p^{3}$ , $s p^{2}$ and $s p$
C. $s p$ , $s p^{2}$ and $s p^{3}$ ✓
D. $s p^{2}$ , $s p$ and $s p^{3}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $s p$ , $s p^{2}$ and $s p^{3}$ \textbf{Solution:} $NO X_{2} X^{+}$ is linear (2 sigma, no lone pair) → $s p$ ; $NO X_{3} X^{-}$ is trigonal planar (3 sigma regions) → $s p^{2}$ ; $NH X_{4} X^{+}$ is tetrahedral (4 sigma) → $s p^{3}$ .

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The hybridizations of the atomic orbitals of nitrogen in NOX2​X+, NOX3​X− and NHX4​X+ respectively are

Solution & Explanation

More NEET PYQ solutions

The hybridizations of the atomic orbitals of nitrogen in $NO X_{2} X^{+}$ , $NO X_{3} X^{-}$ and $NH X_{4} X^{+}$ respectively are