(A) In $NF_3$, the central nitrogen atom is bonded to three fluorine atoms by three $\sigma$-bonds and has one lone pair of electrons. To calculate th

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(A) In $NF_3$, the central nitrogen atom is bonded to three fluorine atoms by three $\sigma$-bonds and has one lone pair of electrons. To calculate the hybridisation, we use the formula: $\text{Steric Number} = \text{Number of } \sigma \text{-bonds} + \text{Number of lone pairs}$. $\text{Steric Number} = 3 + 1 = 4$. $A$ steric number of $4$ corresponds to $sp^3$ hybridisation.

Class 11 Chemistry Chemical Bonding and Molecular Structure Hybridisation

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The hybridisation of the central atom in $NF_3$ is:

WBJEE 2009 All WBJEE

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

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Class 11 Questions

Class 11

Chemistry Questions

Chapter

Chemical Bonding and Molecular Structure

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Hybridisation

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The hybridisations of the atomic orbitals of nitrogen in $NO_{2}^{-}$,$NO_{2}^{+}$ and $NH_{4}^{+}$ respectively are.

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Match the species given in Column-$I$ with the hybridizations given in Column-$II$.

Column-$I$ Column-$II$

$A$. Boron in $[B(OH)_4]^-$ $1$. $sp^2$

$B$. Aluminum in $[Al(H_2O)_6]^{3+}$ $2$. $sp^3$

$C$. Boron in $B_2H_6$ $3$. $sp^3d^2$

$D$. Carbon in Buckminsterfullerene

$E$. $Si$ in $SiO_4^{4-}$

$F$. $Ge$ in $[GeCl_6]^{2-}$

Column-$I$	Column-$II$
$A$. Boron in $[B(OH)_4]^-$	$1$. $sp^2$
$B$. Aluminum in $[Al(H_2O)_6]^{3+}$	$2$. $sp^3$
$C$. Boron in $B_2H_6$	$3$. $sp^3d^2$
$D$. Carbon in Buckminsterfullerene
$E$. $Si$ in $SiO_4^{4-}$
$F$. $Ge$ in $[GeCl_6]^{2-}$