Question

The magnetic nature of boron molecule is same as the magnetic nature of

• A. nitrogen molecule
• B. carbon molecule
• C. unipositive nitrogen molecule
• D. oxide ion

Answer 1

Answer: (C)

$B _{2}(5+5=10)=\sigma 1 s^{2}, \stackrel{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2} \pi 2 p_{x}^{1}, \approx \pi 2 p_{y}^{1}$

(Since, two unpaired electrons are present, the boron molecule is paramagnetic.)

(a) $N _{2}(7+7=14)=\sigma 1 s^{2},{ }^{*} \sigma 1 s^{2}, \sigma 2 s^{2}, ^{*}{\sigma} 2 s^{2} \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2} \sigma 2 p_{z}^{2}$

(No unpaired electron is present, so diamagnetic.)

(b) $C_{2}(6+6=12)=\sigma 1 s^{2}, ^{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, * 2 s^{2} \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}$

(No unpaired electron, so diamagnetic)

(c) $N _{2}^{+}(7+7-1=13)=\sigma 1 s^{2}, \stackrel{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \stackrel{*}{\sigma} 2 s^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \sigma 2 p_{z}^{1}$

(One unpaired electron, so paramagnetic).

(d) $O _{2}^{2-}(8+8+2=18)=\sigma 1 s^{2}, \stackrel{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}$

$^{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}$

(No unpaired electron, so diamagnetic).

Thus, the magnetic nature of boron molecule is same as that of unipositive nitrogen molecule.