Question

Among $H_2,H_2^{+},L{i}_2,B{e}_2,B_2,C_2,O_2^{-},$, $N_2$, and$F_2$ the number of diamagnetic species are: (Atomic numbers: $H=1,He=2,Li=3,Be=4,B=5,C=6,N=7,O=8,F=9)$

Answer 1

Molecular orbitals configurations-
$H_2\to 1s{\sigma }^2$
$H_2^{+}\to 1s{\sigma }^1$
$B{e}_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}$
$L{i}_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^2$
$B_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}(2p_x{\pi }^1=2p_y{\pi }^1)$
$C_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}(2p_x{\pi }^2=2p_y{\pi }^2)$
$N_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}(2p_x{\pi }^2=2p_y{\pi }^2)2p_z{\sigma }^2$
$O_2^{-}\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}2p_z{\sigma }^2(2p_x{\pi }^2=2p_y{\pi }^2)(2p_x{\pi }^{\ast 1}=2p_y{\pi }^{\ast 0})$
$F_2\to 1s{\sigma }^{2}1s{\sigma }^{\ast 2}2s{\sigma }^{2}2s{\sigma }^{\ast 2}2p_z{\sigma }^2(2p_x{\pi }^2=2p_y{\pi }^2)(2p_x{\pi }^{\ast 2}=2p_y{\pi }^{\ast 2})$

$Diamagneticspecies\to allelectronspaired$
$H_2,L{i}_2,B{e}_2,C_2,N_2$ and $F_2$ are diamagnetic species.

However, the bond order of $B{e}_2$ is zero thus $B{e}_2$ does not exist. So, the number of diamagnetic species is 5.