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)

$B_2(5+5=10)=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2,\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2\pi 2p_x^1,\approx \pi 2p_y^1$

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

(a) $N_2(7+7=14)=\sigma 1s^2,{}^{\ast }\sigma 1s^2,\sigma 2s^2{,}^{\ast }\sigma 2s^2\pi 2p_x^2\approx \pi 2p_y^2\sigma 2p_z^2$

(No unpaired electron is present, so diamagnetic.)

(b) $C_2(6+6=12)=\sigma 1s^2{,}^{\ast }\sigma 1s^2,\sigma 2s^2,\ast 2s^2\pi 2p_x^2\approx \pi 2p_y^2$

(No unpaired electron, so diamagnetic)

(c) $N_2^{+}(7+7-1=13)=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2,\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2$

$\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^1$

(One unpaired electron, so paramagnetic).

(d) $O_2^{2-}(8+8+2=18)=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2,\sigma 2s^2$

${}^{\ast }\sigma 2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^2,\pi 2p_y^2$

(No unpaired electron, so diamagnetic).

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