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STD 11 - 4. Chemical bonding and molecular structure — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 4. Chemical bonding and molecular structure4 Marks
MCQ
In $NO_2$ molecule $N$ atom undergoes in
  • A
    $sp^3$ hybridization
  • ✓
    $sp^2$ hybridization
  • C
    $sp$ hybridization
  • D
    $sp^2d$ hybridization
✓

Answer

Correct option: B.
$sp^2$ hybridization
b
$NO _2$ involves an $sp ^2$ type of hybridization. The most simple way to determine the hybridization of $NO _2$ is by drawing the Lewis structure and counting the number of bonds and lone electron pairs around the nitrogen atom. You will find that in nitrogen dioxide there are $2$ sigma bonds and $1$ lone electron pair. Now if we apply the hybridization rule then it states that if the sum of the number of sigma bonds, lone pair of electrons, and odd electrons is equal to three then the hybridization is $sp ^2$.

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$\begin{array}{*{20}{l}}
  {(u){H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHC{H_3}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (v){\mkern 1mu} C{H_2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} CHCl{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (w){\mkern 1mu} C{H_3}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{C} H_2^ + {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} (x){\mkern 1mu} H - C \equiv C - H} \\ 
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\end{array}$

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