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Question: Which of the following molecules is/are diamagnetic? A) \(L{i_2}\) B) \({B_2}\) C) \({C_2}\)...

Which of the following molecules is/are diamagnetic?
A) Li2L{i_2}
B) B2{B_2}
C) C2{C_2}
D) N2{N_2}

Explanation

Solution

We know that all substances show some kind of attraction. Attractive materials are grouped by their mass susceptibility. Ferromagnetism is liable for the majority of the impacts of attraction experienced in regular day to day existence, yet there are really a few kinds of attraction. Paramagnetic substances, like aluminum and oxygen, are pitifully drawn to an applied attractive field; diamagnetic substances, like copper and carbon, are feebly repulsed; while ant ferromagnetic materials, for example, chromium and twist glasses, have a more intricate relationship with an attractive field. The power of a magnet on paramagnetic, diamagnetic, and antiferromagnetic materials is normally too frail to even think about being felt and can be recognized exclusively by research facility instruments, so in regular daily existence, these substances are frequently portrayed as non-attractive.

Complete answer:
Any two electrons share a similar orbital; their twist quantum numbers must be unique. All in all, one of the electrons must be "turn up,” while the other electron is "turn down,”. This is significant with regards to deciding the absolute twist in an electron orbital. To choose whether electron turns drop, add their twist quantum numbers together. At whatever point two electrons are combined together in an orbital or their absolute spin is  0\;0 , they are called diamagnetic electrons.
The diamagnetic compounds among the given options are C2{C_2} andN2{N_2}. Hence options C and D are correct.
Now, we can see the explanation for the answers:
The molecular electronic configuration of carbon and nitrogen are as follows,
C2:KK(σ2s)2(σ∗2s)2(π2py)2(π2pz) 2{C_2}:KK{\left( {\sigma 2s} \right)^2}{({\sigma ^ * }2s)^2}{\left( {{\pi _{2p}}_y} \right)^2}\left( {{\pi _{2p}}_z} \right){{\text{ }}^2}
N2: KK(σ2s)2(σ∗2s)2(σ2px)2(π2py)2(π2pz)2{N_2}{\text{: }}KK{\left( {\sigma 2s} \right)^2}{({\sigma ^ * }2s)^2}{\left( {{\sigma _{2p}}_x} \right)^2}{\left( {{\pi _{2py}}} \right)^2}{\left( {{\pi _{2pz}}} \right)^2}
All the electrons in the above molecular electronic configurations are paired up hence they are diamagnetic.

Note:
We have to remember that the electrons that are distant from everyone else in an orbital are called paramagnetic electrons. Recall that if an electron is distant from everyone else in an orbital, the orbital has a net twist, in light of the fact that the twist of the solitary electron doesn't get counteracted. In the event that even one orbital has a net twist, the whole molecule will have a net twist. In this manner, a molecule is viewed as paramagnetic when it contains in any event one paramagnetic electron. At the end of the day, a molecule might have 1010 matched electrons, yet as long as it additionally has one unpaired electron; it is as yet viewed as a paramagnetic iota.