Question

The common features among $CO$ , $C{N^ - }$ and $N{O^ + }$ are:
(A) Bond order three and isoelectronic
(B) Bond order three and weak field ligands
(C) Bond order two and $\pi$ - acceptors
(D) Bond order three and $\pi$ - donors
(E) Isoelectronic and strong field ligands

Answer 1

Bond order is nothing but the number of chemical bonds with the atoms that form a compound. This in turn also indicates the stability of the bond.
While ligands are nothing but ions or molecules that bond to the central atom while forming a compound. Ligands must have at least one donor with a lone pair of electrons such that it is capable of forming a covalent bond with the central atom. To know whether a ligand is in the weak field spectra or the strong field spectra you can always refer to the spectrochemical series.

Formulas used: We will be using the formula to find the bond order of a molecule from its molecular orbital configuration, $B.O. = \dfrac{1}{2}\left( {{N_b} - {N_a}} \right)$ Where $B.O.$ is the bond order, ${N_b}$ is the number of bonding electrons or the number of electrons in bonding molecular orbital, and ${N_a}$ is the number of antibonding electrons in the molecular orbit.

Complete Step by Step answer
Now we have 3 molecules/ions: $CO$ , $C{N^ - }$ and $N{O^ + }$ .
Let us start by calculating the total number of electrons in these compounds,
The number of electrons in $CO$ will be given by,
${N_{CO}}$ =number of electrons in Carbon atom + number of electrons in oxygen atom
${N_{CO}} = 6 + 7$
$\Rightarrow {N_{CO}} = 14$
Similarly, the number of electrons in $C{N^ - }$ will be, ${N_{C{N^ - }}} = 6 + 7 + 1$
$\Rightarrow {N_{C{N^ - }}} = 14$
The total number of electrons in $N{O^ + }$ will be, ${N_{N{O^ + }}} = 7 + 8 - 1$
$\Rightarrow {N_{N{O^ + }}} = 14$
Since all the three compounds have the same number of electrons, we can say that the 3 compounds are isoelectronic in nature.
Now considering the molecular configuration of the molecules/ions,
Molecular orbital configuration of $CO$ : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2p_x^2 \approx \pi 2p_y^2,\pi 2p_z^2$
Thus, the bond order of $CO$ is given by $B.O. = \dfrac{1}{2}\left( {{N_b} - {N_a}} \right)$ ,where ${N_b} = 10$ and ${N_a} = 4$ .
$\Rightarrow B.O. = \dfrac{1}{2}\left( {10 - 4} \right) = \dfrac{1}{2}\left( 6 \right) = 3$
Molecular orbital configuration of $C{N^ - }$ : $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2p_x^2\pi 2p_y^2,\pi 2p_z^2$
Thus, the bond order of $C{N^ - }$ is given with ${N_b} = 10$ and ${N_a} = 4$ .
$\Rightarrow B.O. = \dfrac{1}{2}\left( {10 - 4} \right) = \dfrac{1}{2}\left( 6 \right) = 3$
Similarly, for $N{O^ + }$ the M.O would be, $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2p_x^2 \approx \pi 2p_y^2,\pi 2p_z^2$
Thus, the bond order will be given by,
$B.O. = \dfrac{1}{2}\left( {10 - 4} \right) = \dfrac{1}{2}\left( 6 \right) = 3$
And now we can see that all the three molecules/ions have the same bond order of 3. Thus, they are similar to each other for the fact that the molecules have the same bond order of 3 and are isoelectronic with 14 electrons in its molecular orbits.
Hence, the correct answer is option A.

Note
If you are wondering how to find ${N_b}$ and ${N_a}$ , observe the M.O of the molecules/ions, just like electronic configuration of elements the number above the orbitals represent the number of electrons in them and the orbitals with a $*$ represents anti-bonding orbitals.