Question

Match List I (Molecules) with List II (Bond order) and select the correct answer using the codes.

List - I	List - II
I. $L{i}_2$	A. 3
II. $N_2$	B. 1.5
III. $B{e}_2$	C. 1.0
IV. $O_2$	D. 0
	E. 2

• A. I - B, II - C, III - A, IV - E
• B. I - C, II - A, III - D, IV - E
• C. I - D, II - A, III - E, IV - C
• D. I - C, II - B, III - E, IV - A

Answer 1

The correct option is B I - C, II - A, III - D, IV - E

Bond order = $\frac{1}{2}$ (number of electrons in bonding MO- number of electrons in anti-bonding MO)

I. $L{i}_2$: ${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2\frac{1}{2}(2-2+2)=1$

Hence, option C

II. $N_2$: ${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2{(\sigma 2\ast s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4\frac{1}{2}\left[(2+2+2+4)-(2+2)\right]=3$

Hence, option A

III. $B{e}_2$: ${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2{(\sigma 2\ast s)}^2\frac{1}{2}\left[(2+2)-(2+2)\right]=0$

Hence, option D

IV.$O_2$: ${\left(\sigma 1s\right)}^2{(\sigma \ast 1s)}^2{\left(\sigma 2s\right)}^2{(\sigma 2\ast s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{(\pi \ast 2p)}^2\frac{1}{2}\left[(2+2+2+4)-(2+2+2)\right]=2$

Hence, option E

Hence, the answer is I-C,II-A,III-D,IV-E