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

Hint: 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.