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Question: Among \[{{H}_{2}},He_{2}^{+},L{{i}_{2}},B{{e}_{2}},{{B}_{2}},{{C}_{2}},{{N}_{2}},O_{2}^{-}\]and \[{{...

Among H2,He2+,Li2,Be2,B2,C2,N2,O2−{{H}_{2}},He_{2}^{+},L{{i}_{2}},B{{e}_{2}},{{B}_{2}},{{C}_{2}},{{N}_{2}},O_{2}^{-}and F2{{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)

Explanation

Solution

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 H2(2)=σ(1s)2{{H}_{2}}(2)=\sigma {{(1s)}^{2}}
There are no unpaired electrons, hence, it is diamagnetic.
Electronic configuration of He2+=σ(1s)2,σ∗(1s)1He_{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)=σ(1s)2,σ∗(1s)1,σ(2s)2Li(2)(6)=\sigma {{(1s)}^{2}},{{\sigma }^{*}}{{(1s)}^{1}},\sigma {{(2s)}^{2}}
There are no unpaired electrons. Hence, Lithium is diamagnetic.
Electronic configuration of Be2(8)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2B{{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 Be2(10)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2,π(2px)1,π(2py)1B{{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 C2(12)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2,π(2px)2,π(2py)2{{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 N2(14)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2,π(2px)2,π(2py)2,π(2pz)2{{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 O2(17)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2,σ(2px)2,π(2py)2,σ(2pz)2,π∗(2px)2,π∗(2py)1{{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 F2(18)=σ(1s)2,σ∗(1s)1,σ(2s)2,σ∗(2s)2,σ(2pz)2,π(2px)2,σ(2py)2,π∗(2px)2,π∗(2py)2{{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 thatHe2+He_{2}^{+},Be2(10)B{{e}_{2}}(10),O2−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.