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Chemistry Question on Molecular Orbital Theory

The molecule which has highest bond order is

A

C2C_2

B

N2N_2

C

B2B_2

D

O2O_2

Answer

N2N_2

Explanation

Solution

(a)C2(12)=s1s2,s∗1s2,s2s2,s∗2s2,π2px2≈π2py2(a) C_2(12)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y
       Bond order=Nb−Na2=8−42=2\, \, \, \, \, \, \, Bond\, order=\frac{N_b-N_a}{2}=\frac{8-4}{2}=2
(b)N(14)=s1s2,s∗1s2,s2s2,s∗2s2,π2px2≈π2py2,π∗2px2(b) N (14)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi^* 2p^2_x
                 Bond order=10−42=3\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-4}{2}=3
(c)B2(10)=s1s2,s∗1s2,s2s2,s∗2s2,π2px1≈π2py1(c) B2(10)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^1_x \approx \pi 2p^1_y
                 Bond order=6−42=1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{6-4}{2}=1
(d)02(16)=s1s2,s∗1s2,s2s2,s∗2s2,π2px2≈π2py2,π2px1≈π2p(d) 02 (16)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi 2p^1_x \approx \pi 2p
                 Bond order=10−62=2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-6}{2}=2
(d)02(16)=s1s2,s∗1s2,s2s2,s∗2s2,π2px2≈π2py2,π2px2(d) 02 (16)=s 1s^2, s^* 1s^2, s2s^2, s^* 2s^2,\pi 2p^2_x \approx \pi 2p^2_y,\pi 2p^2_x
                 Bond order=10−82=1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, Bond\, order=\frac{10-8}{2}=1
Hence, N2N_2 has the highest bond order.