(B) The total number of valence electrons in $NH_3$ is $5 + (3 \times 1) = 8$. Using the formula for steric number $X = \frac{1}{2} (V + M - C + A)$,w

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(B) The total number of valence electrons in $NH_3$ is $5 + (3 \times 1) = 8$. Using the formula for steric number $X = \frac{1}{2} (V + M - C + A)$,where $V = 5$ (valence electrons of $N$),$M = 3$ (monovalent atoms),$C = 0$,and $A = 0$. $X = \frac{8}{2} = 4$. Since the steric number is $4$,the hybridization is $sp^3$.

Class 11 Chemistry Chemical Bonding and Molecular Structure Hybridisation

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What is the type of hybridization of $N$ in $NH_3$?

A
$sp^2$
B
$sp^3$
C
$dsp^2$
D
$sp$

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Class 11

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Chapter

Chemical Bonding and Molecular Structure

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Hybridisation

Chlorine atom,in its third excited state,reacts with fluorine to form a compound $X$. The formula and shape of $X$ are :

MediumAP EAMCET 2003

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Which of the following statements is correct in the context of the allene molecule,$C_3H_4$?

Difficult

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Among the following transformations,the hybridisation of the central atom remains unchanged in

DifficultKVPY 2019

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As the $s$-character of the hybrid orbital decreases,the bond angle ...........

Easy

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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 :

MediumJEE MAIN 2024

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$LIST$-$I$ (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

What is the type of hybridization of $N$ in $NH_3$?

Similar Questions

Chlorine atom,in its third excited state,reacts with fluorine to form a compound $X$. The formula and shape of $X$ are :

Which of the following statements is correct in the context of the allene molecule,$C_3H_4$?

Among the following transformations,the hybridisation of the central atom remains unchanged in

As the $s$-character of the hybrid orbital decreases,the bond angle ...........

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 :

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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 :