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Question: Consider following statements (i) Bond order of CO and \({N_2}\) is same (ii) \({O_2}^ + \) is m...

Consider following statements
(i) Bond order of CO and N2{N_2} is same
(ii) O2+{O_2}^ + is more stable than O2{O_2}
(iii) Bond order of CO+C{O^ + } and N2+{N_2}^ + are same
(iv) MOT explain diamagnetic character of oxygen
(v) HOMO of F2{F_2} molecule is σ2pz∗\sigma _{2{p_z}}^*
Correct statements are

Explanation

Solution

The formula to find the bond order of the species is as below.
Bond order = No. of electrons in bonding orbitals - No. of electrons in antibonding orbitals2{\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: σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2py2,π2pz2,σ2px2\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.
Bond order = No. of electrons in bonding orbitals - No. of electrons in antibonding orbitals2{\text{Bond order = }}\frac{{{\text{No}}{\text{. of electrons in bonding orbitals - No}}{\text{. of electrons in antibonding orbitals}}}}{2}
Bond order = 10−42=6{\text{Bond order = }}\frac{{10 - 4}}{2} = 6
Now, for N2{N_2} the electronic configuration of its molecular orbitals can be given as σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2py2,π2pz2,σ2px2\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
Bond order = 10−42=6{\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 O2+{O_2}^ + and O2{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 O2+{O_2}^ + : σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2px2,π2py2,π2pz2,π∗2py1\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 = 10−52=2.5\frac{{10 - 5}}{2} = 2.5
M.O. configuration of O2{O_2} : σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2px2,π2py2,π2pz2,π∗2py1,π∗2pz1\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 = 10−62=2\frac{{10 - 6}}{2} = 2
Thus, we can say that as O2+{O_2}^ + has higher bond order, it is a more stable species.
(iii) The M.O. configuration of CO+C{O^ + } is : σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2py2,π2pz2,σ2px1\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 N2+{N_2}^ + is : σ1s2,σ∗1s2,σ2s2,σ∗2s2,π2py2,π2pz2,σ2px1\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 9−42=2.5\frac{{9 - 4}}{2} = 2.5
(iv) No, according to MOT, actually O2{O_2} is a paramagnetic molecule because it has two unpaired electrons in its structure. The M.O configuration of O2{O_2} is σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2px2,π2py2,π2pz2,π∗2py1,π∗2pz1\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 F2{F_2} to find its HOMO.
MO configuration of F2{F_2} = σ1s2,σ∗1s2,σ2s2,σ∗2s2,σ2pz2,π2px2,π2py2,π∗2px2,π∗2py2\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 π∗2py{\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.