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.

$LIST$-$I$ (Hybridization)	$LIST$-$II$ (Orientation in Space)
$A$. $sp^3$	$I$. Trigonal bipyramidal
$B$. $dsp^2$	$II$. Octahedral
$C$. $sp^3 d$	$III$. Tetrahedral
$D$. $sp^3 d^2$	$IV$. Square planar

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

Similar Questions

The hybridization of the central atom will change when

Assuming pure $2s$ and $2p$ orbitals of carbon are used in forming $CH_4$ molecule,which of the following statements is false?

Which of the following is a linear molecule?

$C-H$ bond length is least in

Match List-$I$ with List-$II$ :

$LIST$-$I$ (Hybridization) $LIST$-$II$ (Orientation in Space)

$A$. $sp^3$ $I$. Trigonal bipyramidal

$B$. $dsp^2$ $II$. Octahedral

$C$. $sp^3 d$ $III$. Tetrahedral

$D$. $sp^3 d^2$ $IV$. Square planar

Choose the correct answer from the options given below :

Vedclass Test Series

Exam Paper Generator

Online Exam Module