The hybridisations of N-orbitals in NO_3^{-}, | vedclass

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

Question

(A) The hybridisation can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons of the centr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The hybridisation can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,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.For $NO_3^{-}$: $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^2$ hybridisation.For $NO_2^{-}$: $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$,which corresponds to $sp^2$ hybridisation.For $NH_4^{+}$: $H = \frac{1}{2} [5 + 4 - 1 + 0] = 4$,which corresponds to $sp^3$ hybridisation.Thus,the hybridisations are $sp^2, sp^2$ and $sp^3$ respectively.

The hybridisations of $N$-orbitals in $NO_3^{-}$,$NO_2^{-}$ and $NH_4^{+}$ respectively are

Similar Questions

Choose the molecules in which hybridization occurs in the ground state?
$(i)$ $BCl_3$ $(ii)$ $NH_3$
$(iii)$ $PCl_3$ $(iv)$ $BeF_2$
The correct answer is

$BF_3$ is a planar and electron-deficient compound. The hybridization and the number of electrons around the central atom,respectively,are:

Total number of vacant orbitals in the valence shell of sulphur when it undergoes the formation of $SF_4$.

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

Explain the formation of bonds using $sp^2$ hybrid orbitals with an example.

Vedclass Test Series

Exam Paper Generator

Online Exam Module