The types of hybrid orbitals of nitrogen in N | vedclass

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

Question

(B) To determine the hybridization $(H)$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$Where $V$ = number of valence electrons of the central at

Vedclass · Accepted Answer

$sp, sp^2, sp^3$

Vedclass · Answer

(B) To determine the hybridization $(H)$,we use the formula: $H = \frac{1}{2}(V + M - C + A)$Where $V$ = number of valence electrons of the central atom,$M$ = number of monovalent atoms,$C$ = cationic charge,and $A$ = anionic charge.$1$. For $NO_2^{+}$:$H = \frac{1}{2}(5 + 0 - 1 + 0) = 2$,which corresponds to $sp$ hybridization.$2$. For $NO_3^{-}$:$H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$,which corresponds to $sp^2$ hybridization.$3$. For $NH_4^{+}$:$H = \frac{1}{2}(5 + 4 - 1 + 0) = 4$,which corresponds to $sp^3$ hybridization.Thus,the hybridization types are $sp, sp^2, sp^3$ respectively.

The types of hybrid orbitals of nitrogen in $NO_2^{+}$,$NO_3^{-}$ and $NH_4^{+}$ respectively are

Similar Questions

The $H-O-H$ bond angles in $H_3O^+$ are approximately $107^\circ$. The orbitals used by oxygen in these bonds are best described as

What type of overlap is involved in the formation of $C-H$ bonds in acetylene molecules?

The hybridisation of $Xe$ and the number of lone pairs of electrons on it in $XeF_6$ are

What type of hybridisation is involved in the central atom of hydrides of group $16$ elements?

Which of the following pairs has the same bond angle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module