Question

Consider following statements
(i) Bond order of CO and ${N_2}$ is same
(ii) ${O_2}^ + $ is more stable than ${O_2}$
(iii) Bond order of $C{O^ + }$ and ${N_2}^ + $ are same
(iv) MOT explain diamagnetic character of oxygen
(v) HOMO of ${F_2}$ molecule is $\sigma _{2{p_z}}^*$
Correct statements are

Answer 1

Hint: The formula to find the bond order of the species is as below.
\[{\text{Bond order = }}\frac{{{\text{No}}{\text{. of electrons in bonding orbitals - No}}{\text{. of electrons in antibonding orbitals}}}}{2}\]

Complete step by step solution:
We will check all the given statements to check whether they are correct or not.
(i) The electronic configuration of molecular orbitals in CO is: $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_y}^2,\pi 2{p_z}^2,\sigma 2{p_x}^2$
- We know that orbitals which have * sign shows antibonding orbital. The formula to find the bond order is as below.
\[{\text{Bond order = }}\frac{{{\text{No}}{\text{. of electrons in bonding orbitals - No}}{\text{. of electrons in antibonding orbitals}}}}{2}\]
\[{\text{Bond order = }}\frac{{10 - 4}}{2} = 6\]
Now, for ${N_2}$ the electronic configuration of its molecular orbitals can be given as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_y}^2,\pi 2{p_z}^2,\sigma 2{p_x}^2$ .
Now, we put the available values into the formula of bond order, we get
\[{\text{Bond order = }}\frac{{10 - 4}}{2} = 6\]
Thus, the bond order of both species is the same and this statement is right.
(ii) We will see the molecular orbital configuration of both ${O_2}^ + $ and ${O_2}$ to find their bond order. Then we can say that the compound which has higher bond order, will be more stable.
M.O configuration of ${O_2}^ + $ : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2,\pi 2{p_z}^2,{\pi ^*}2{p_y}^1$
So, bond order = $\frac{{10 - 5}}{2} = 2.5$
M.O. configuration of ${O_2}$ : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2,\pi 2{p_z}^2,{\pi ^*}2{p_y}^1,{\pi ^*}2{p_z}^1$
So, bond order = $\frac{{10 - 6}}{2} = 2$
Thus, we can say that as ${O_2}^ + $ has higher bond order, it is a more stable species.
(iii) The M.O. configuration of $C{O^ + }$ is : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_y}^2,\pi 2{p_z}^2,\sigma 2{p_x}^1$
M.O. configuration of ${N_2}^ + $ is : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_y}^2,\pi 2{p_z}^2,\sigma 2{p_x}^1$
As, both species have the same M.O. configuration, they will have same bond order and it will be $\frac{{9 - 4}}{2} = 2.5$
(iv) No, according to MOT, actually ${O_2}$ is a paramagnetic molecule because it has two unpaired electrons in its structure. The M.O configuration of ${O_2}$ is $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2,\pi 2{p_z}^2,{\pi ^*}2{p_y}^1,{\pi ^*}2{p_z}^1$ .Thus, this statement is not correct.
(v) HOMO stands for highest occupied molecular orbital. We will take a look at MO configuration of ${F_2}$ to find its HOMO.
MO configuration of ${F_2}$ = $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2{p_z}^2,\pi 2{p_x}^2,\pi 2{p_y}^2,{\pi ^*}2{p_x}^2,{\pi ^*}2{p_y}^2$
Thus, we can see that the highest occupied molecular orbital is ${\pi ^*}2{p_y}$ orbital. Thus, the given statement is not correct.

Therefore, statements (i), (ii) and (iii) are correct.

Note: The compound which has an unpaired electron in its MO diagram is called a paramagnetic compound. The compound which does not have any unpaired electron in MO diagram is called diamagnetic compound.