In which of the following molecules is the N | vedclass

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(C) $1$. In the nitronium ion $(NO_2^ )$,the $N$ atom is $sp$ hybridized (linear geometry).$2$. In the nitrate ion $(NO_3^-)$,the $N$ atom is $sp^2$ h

Vedclass · Accepted Answer

Trisilylamine $(N(SiH_3)_3)$

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(C) $1$. In the nitronium ion $(NO_2^ )$,the $N$ atom is $sp$ hybridized (linear geometry).$2$. In the nitrate ion $(NO_3^-)$,the $N$ atom is $sp^2$ hybridized (trigonal planar geometry).$3$. In trisilylamine $(N(SiH_3)_3)$,the $N$ atom is $sp^3$ hybridized (pyramidal geometry) because the lone pair on $N$ is donated into the empty $d$-orbitals of $Si$ ($p\pi-d\pi$ back bonding),but the steric arrangement remains $sp^3$.$4$. In borazine $(B_3N_3H_6)$,the $N$ atoms are $sp^2$ hybridized (planar ring structure).$5$. Therefore,the $N$ atom is not $sp^2$ hybridized in both the nitronium ion $(sp)$ and trisilylamine $(sp^3)$. However,in the context of standard multiple-choice questions where only one answer is expected,trisilylamine is the classic example of $sp^3$ hybridization due to the lone pair involvement in back bonding.

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 $[GeCl_6]^{2-}$

In which of the following molecules is the $N$ atom not $sp^2$ hybridized?

Similar Questions

Which hybrid orbitals are used by carbon atom in the following molecules?
$(a)$ $CH_3-CH_3$
$(b)$ $CH_3-CH=CH_2$
$(c)$ $CH_3-CH_2-OH$
$(d)$ $CH_3-CHO$
$(e)$ $CH_3COOH$

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

The hybridization states of the central atoms of the ions $I_3^-$,$ICl_4^-$ and $ICl_2^-$ are respectively

The structures of $BCl_3$ and $BCl_3 \cdot NH_3$,respectively are

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