Formation of nitrogen molecule using molecular orbital theory
Answers
Answer:
Nitrogen molecule (N_{2})(N
2
): The electronic configuration of nitrogen \left ( Z= 7 \right )(Z=7) in ground state is 1s^{2}2s^{2}2p_{x}^{1}2p_{y}^{1}2p_{z}^{1}1s
2
2s
2
2p
x
1
2p
y
1
2p
z
1
. Therefore, the total number of electrons present in nitrogen molecule \left ( N_{2} \right )(N
2
) is 1414. These 1414 electrons can be accommodated in the various molecular orbitals in order of increasing energy.
N_{2}: KK\left ( \sigma _{2s} \right )^{2}\left ( \sigma {_{2s}}^{\star } \right )^{2}\left ( \pi _{2Px} \right )^{2}\left ( \pi _{2py} \right )^{2}\left ( \sigma _{2px}^{\star } \right )^{2}\left ( \pi _{2py}^{\star } \right )^{2}N
2
:KK(σ
2s
)
2
(σ
2s
⋆
)
2
(π
2Px
)
2
(π
2py
)
2
(σ
2px
⋆
)
2
(π
2py
⋆
)
2
Here \left ( \sigma _{1s} \right )^{2}\left ( \sigma _{1s}^{\star } \right )^{2}(σ
1s
)
2
(σ
1s
⋆
)
2
part of configuration is abbreviated as KKKK, which denotes the KK shells of the two atoms. In calculating bond order, we can ignore KKKK, as it includes two bonding and two antibonding electrons.
The molecular orbital energy level diagram of N_{2}N
2
is given in fig.
The bond order of N_{2}N
2
can be calculated as follows:
Here, N_{b}= 8N
b
=8 and N_{b}= 2N
b
=2
Bond order = \frac{N_{b}-N_{a}}{2}=\frac{8-2}{2}= 3=
2
N
b
−N
a
=
2
8−2
=3
Nature of bond: A bond order of 33 means that a triple bond is present in a molecule of nitrogen.
Diamagnetic nature: Since all the electrons in nitrogen are paired, it is diamagnetic in nature.