The hybridizations of atomic orbitals of nitr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To determine the hybridization of the central nitrogen atom,we use the formula: $	ext{Number of hybrid orbitals} = \frac{1}{2} [V + M - C + A]$,w

Vedclass · Accepted Answer

$sp$,$sp^2$ and $sp^3$

Vedclass · Answer

(C) To determine the hybridization of the central nitrogen atom,we use the formula: $	ext{Number of hybrid orbitals} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1)$ For $NO_2^+$: $V = 5$,$M = 0$,$C = 1$,$A = 0$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hence,$sp$ hybridization.$2)$ For $NO_3^-$: $V = 5$,$M = 0$,$C = 0$,$A = 1$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. Hence,$sp^2$ hybridization.$3)$ For $NH_4^+$: $V = 5$,$M = 4$,$C = 1$,$A = 0$. $	ext{Hybrid orbitals} = \frac{1}{2} [5 + 4 - 1 + 0] = 4$. Hence,$sp^3$ hybridization.Therefore,the hybridizations are $sp$,$sp^2$,and $sp^3$ respectively.

The hybridizations of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$ and $NH_4^+$ respectively are

Similar Questions

In compound $X$,all the bond angles are exactly $109^o 28'$,$X$ is

The species in which the central atom uses $sp^2$ hybrid orbitals in its bonding is

The order of electronegativity of carbon in $sp$,$sp^{2}$ and $sp^{3}$ hybridised states follows:

What are the requirements for the formation of hybrid orbitals?

In which of the following combinations does the hybridisation of the central atom $(^*)$ $NOT$ change?

Vedclass Test Series

Exam Paper Generator

Online Exam Module