- ✓$NO$
- B$CO$
- C$C{N^ - }$
- D${O_2}$
According to Molecular orbital theory, the electronic configuration for given molecules as follow:
$A]$ The electronic configuration of $NO$ :
$\sigma 1 s ^2 \,\sigma^* 1 s ^2 \,\sigma 2 s ^2 \,\sigma^* 2 s ^2 \,\sigma 2 p _{ z }^2\, \pi 2 p _{ x }^2 \,\pi 2 p _{ y }^2\, \pi^* 2 p _{ x }^1$
Therefore $NO$ molecule has one unpaired electron in $\pi^* 2 p _{ x }^1$ orbital. Therefore, option A is answer.
Explanation for incorrect options :
$B]$ The electronic configuration of $CO$ :
$\sigma 1 s ^2\,\sigma^* 1 s ^2 \,\sigma 2 s ^2 \,\sigma^* 2 s ^2\, \sigma 2 p _{ z }^2\, \pi 2 p _{ x }^2 \,\pi 2 p _{ y }^2$
All electrons in $CO$ molecule is paired.
$C]$ The electronic configuration of $CN ^{-}$:
$\sigma 1 s ^2\, \sigma^* 1 s ^2 \,\sigma 2 s ^2 \,\sigma^* 2 s ^2 \,\sigma 2 p _z^2 \,\pi 2 p _{ x }^2 \,\pi 2 p _{ y }^2$
All electrons in $CN ^{-}$molecule is paired.
$D$ ] The electronic configuration of $O _2$ :
$\sigma 1 s ^2\, \sigma^* 1 s ^2\, \sigma 2 s ^2 \,\sigma^* 2 s ^2\, \sigma 2 p _z^2 \,\pi 2 p _{ x }^2 \,\pi 2 p _{ y }^2\, \pi^* 2 p _{ x }^1 \,\pi^* 2 p _{ y }^1$
Therefore $O _2$ molecule has $1-1$ unpaired electron in $\pi^* 2 p _{ x }^1$ and $\pi^* 2 p _{ y }^1$ orbital. That mean, $O _2$ has two unpaired electrons.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ The principal quantum number $'n'$ is a positive integer with values of $'n'=1,2,3, \ldots$.
$(B)$ The azimuthal quantum number $'l'$ for a given $'n'$ (principal quantum number) can have values as $'l'=0,1,2, \ldots n$
$(C)$ Magnetic orbital quantum number $'m _{l}'$ for a particular $'l'$ (azimuthal quantum number) has $(2 l$ + 1) values.
$(D)$ $\pm 1 / 2$ are the two possible orientations of electron spin.
$(E)$ For $l=5$, there will be a total of $9$ orbital.
Which of the above statements are correct?

(Molecular Weight of $\mathrm{HNO}_{3}=63$ )