Answer:
Option B
Explanation:
To identify the magnetic nature we need to check the molecular orbital configuration. If all orbitals are fully occupied, species is diamagnetic while when one or more molecular orbitals is/are singly occupied , species is paramagnetic
(a) $NO (7+8=15)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$
$=\pi2p_y^2,\pi2p_z^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^0$
One unpaired electron is present.
Hence , it is paramagnetic
(b) $ CO(6+8=14)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^2$
$=\pi2p_y^2,\sigma2p_z^2$
No unpaired electron is present
Hence , it is diamagenetic
(c) $O_{2}(8+8=16)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\sigma2p_z^2,\pi2p_x^2$
$=\pi2p_x^2,\pi^{*}2p_x^1$ $=\pi^{*}2p_y^1$
Two unpaired electrons are present. Hence it is paramagnetic
(d) $B_{2}(5+5)-\sigma1s^{2},\sigma^{*}1s^{2},\sigma2s^{2},\sigma^{*}2s^{2},\pi2p_x^1$
$=\pi2p_x^2,\pi^{*}2p_x^1$
$=\pi2p_y^1$
Two unpaired electrond are present. Hence it is paramagnetic
