Question

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

Answer 1

Hint:When a species does not have any unpaired electrons, it is usually diamagnetic. These species repel magnetic fields.

Complete step by step answer:
We have to write the electronic configuration of each element.
Electronic configuration of $$H_2(2)=\sigma {(1s)}^2$$
There are no unpaired electrons, hence, it is diamagnetic.
Electronic configuration of $$H{e}_2^{+}=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1$$
There is 1 unpaired electron in an antibonding orbital. Hence, it is paramagnetic.
Electronic configuration of $$Li(2)(6)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2$$
There are no unpaired electrons. Hence, Lithium is diamagnetic.
Electronic configuration of $$B{e}_2(8)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2$$
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of $$B{e}_2(10)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2,\pi {(2p_x)}^1,\pi {(2p_y)}^1$$
There are 2 unpaired electrons. Hence, it is paramagnetic.
Electronic configuration of $$C_2(12)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2,\pi {(2p_x)}^2,\pi {(2p_y)}^2$$
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of $$N_2(14)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2,\pi {(2p_x)}^2,\pi {(2p_y)}^2,\pi {(2p_z)}^2$$
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of $$O_2(17)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2,\sigma {(2p_x)}^2,\pi {(2p_y)}^2,\sigma {(2p_z)}^2,{\pi }^{\ast }{(2p_x)}^2,{\pi }^{\ast }{(2p_y)}^1$$
There is 1 unpaired electron, so it is paramagnetic.
Electronic configuration of $$F_2(18)=\sigma {(1s)}^2,{\sigma }^{\ast }{(1s)}^1,\sigma {(2s)}^2,{\sigma }^{\ast }{(2s)}^2,\sigma {(2p_z)}^2,\pi {(2p_x)}^2,\sigma {(2p_y)}^2,{\pi }^{\ast }{(2p_x)}^2,{\pi }^{\ast }{(2p_y)}^2$$
It has no unpaired electron and is diamagnetic.
Hence, we can conclude that$$H{e}_2^{+}$$,$$B{e}_2(10)$$,$$O_2^{-}$$ are paramagnetic in nature.

Additional Information: Where unpaired electrons in the valence shell are present, then the molecule becomes paramagnetic. That is, if the molecule is kept in a magnetic field it feels strong attraction towards the field. The magnetic field passes through the molecule. In case of a diamagnetic molecule, when it is kept in a magnetic field, it gets repelled by the field. Does not allow the field to pass through the molecule.

Note:
The number of unpaired electrons in the valence shell gives the atoms magnetic property. In case of molecules we should know the molecular orbital configuration.