Question

Among \[{{H}_{2}},He_{2}^{+},L{{i}_{2}},B{{e}_{2}},{{B}_{2}},{{C}_{2}},{{N}_{2}},O_{2}^{-}\]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 \[He_{2}^{+}=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}}\]
There is 1 unpaired electron in an antibonding orbital. Hence, it is paramagnetic.
Electronic configuration of \[Li(2)(6)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}}\]
There are no unpaired electrons. Hence, Lithium is diamagnetic.
Electronic configuration of \[B{{e}_{2}}(8)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}}\]
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of \[B{{e}_{2}}(10)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}},\pi {{(2{{p}_{x}})}^{1}},\pi {{(2{{p}_{y}})}^{1}}\]
There are 2 unpaired electrons. Hence, it is paramagnetic.
Electronic configuration of \[{{C}_{2}}(12)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}},\pi {{(2{{p}_{x}})}^{2}},\pi {{(2{{p}_{y}})}^{2}}\]
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of \[{{N}_{2}}(14)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}},\pi {{(2{{p}_{x}})}^{2}},\pi {{(2{{p}_{y}})}^{2}},\pi {{(2{{p}_{z}})}^{2}}\]
There are no unpaired electrons. Hence, it is diamagnetic.
Electronic configuration of \[{{O}_{2}}(17)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}},\sigma {{(2{{p}_{x}})}^{2}},\pi {{(2{{p}_{y}})}^{2}},\sigma {{(2{{p}_{z}})}^{2}},{{\pi }^{*}}{{(2{{p}_{x}})}^{2}},{{\pi }^{*}}{{(2{{p}_{y}})}^{1}}\]
There is 1 unpaired electron, so it is paramagnetic.
Electronic configuration of \[{{F}_{2}}(18)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}},{{\sigma }^{*}}{{(2s)}^{2}},\sigma {{(2{{p}_{z}})}^{2}},\pi {{(2{{p}_{x}})}^{2}},\sigma {{(2{{p}_{y}})}^{2}},{{\pi }^{*}}{{(2{{p}_{x}})}^{2}},{{\pi }^{*}}{{(2{{p}_{y}})}^{2}}\]
It has no unpaired electron and is diamagnetic.
Hence, we can conclude that\[He_{2}^{+}\],\[B{{e}_{2}}(10)\],\[O_{2}^{-}\] 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.