Question

number of antibonding electrons in n2+ is​

Answer 1

The total number of antibonding electrons in N2 and O2 molecules is 4 and 6 respectively. Hence,

N

2

:(σ1s)

2

(σ

∗

1s)

2

(σ2s)

2

(σ

∗

2s)

2

(σ2p

x

)

2

(π2p

y

)

2

(π2p

z

)

2

O

2

:(σ1s)

2

(σ

∗

1s)

2

(σ2s)

2

(σ

∗

2s)

2

(σ2p

x

)

2

(π2p

y

)

2

(π2p

z

)

2

(π

∗

2p

z

Answer 2

The number of antibonding electrons in $\[{N_2}^ + \]$ is​ 4.

Explanation:

• There are a total of 14 electrons in the nitrogen molecule.
• The electronic configuration of the nitrogen molecule can be given as$\[{N_2}:\,\,{\left( {\sigma 1s} \right)^2}\;{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^{2\;}}{\left( {{\sigma ^*}2s} \right)^2}\;{\left( {\pi 2Px} \right)^2}\;{\left( {\pi 2py} \right)^{2\;}}{\left( {\sigma 2pz} \right)^2}\]$
• It consists of 10 bonding electrons and 4 anti-bonding electrons.
• $\[{N_2}^ + \]$ will have a total of 13 electrons. One electron is removed from $\[{\left( {\sigma 2pz} \right)^2}\]$ bonding orbital.
• Therefore, the electronic configuration of $\[{N_2}^ + \]$ will be $\[{N_2}^ + :\,\,{\left( {\sigma 1s} \right)^2}\;{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^{2\;}}{\left( {{\sigma ^*}2s} \right)^2}\;{\left( {\pi 2Px} \right)^2}\;{\left( {\pi 2py} \right)^{2\;}}{\left( {\sigma 2pz} \right)^1}\]$
• As an electron is removed from a bonding orbital, the number of bonding electrons becomes 9 and the number of anti-bonding electrons remains the same, i.e. 4 electrons.
• Thus, the number of antibonding electrons in $\[{N_2}^ + \]$ is 4.