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Question: The species having bond order different from that in CO is (A)\(N{O^ - }\) (B)\(N{O^ + }\) (C)...

The species having bond order different from that in CO is
(A)NON{O^ - }
(B)NO+N{O^ + }
(C)CNC{N^ - }
(D)N2{N_2}

Explanation

Solution

Bond order is defined as the half of the difference between the bonding and antibonding electrons. The species having the same number of electrons are called isoelectronic species. Also, the isoelectronic species are having the same bond order.

Complete step by step solution: We have been given four different species out of which we have to find the one whose bond order is different from that in carbon monoxide.
We have been given a CO molecule which contains fourteen electrons. We will write the bond order of this molecule. The electronic configuration is given as follows
CO=σ1s2σ1s2σ2s2π2px2π2py2σ2pz2 = \sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}\pi 2p_x^2 \approx \pi 2p_y^2\sigma 2p_z^2
Now BO=12(NBNA) = \dfrac{1}{2}({N_B} - {N_A})
Where BO is the bond order and NB{N_B},NA{N_A} are the number of electrons in bonding and antibonding orbitals respectively.
\RightarrowBO=1042=3 = \dfrac{{10 - 4}}{2} = 3
Now for NON{O^ - } the bond order is given as follows. The electronic configuration NON{O^ - } is given as follows
It is having sixteen electrons so the electronic configuration is as follows
NON{O^ - } =σ1s2σ2s2σ2s2σ2pz2π2px2π2py2π2px1π2py1 = \sigma 1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}\sigma 2p_z^2\pi 2p_x^2 \approx \pi 2p_y^2{\pi ^*}2p_x^1 \approx {\pi ^*}2p_y^1
Now BO=12(NBNA) = \dfrac{1}{2}({N_B} - {N_A})
\RightarrowBO=1062=2 = \dfrac{{10 - 6}}{2} = 2
Now NO+N{O^ + } is having the same number of electrons and has the same bond order as that of CO.
i.e.BO=3 = 3
Now cyanide ion(CNC{N^ - }) is having fourteen electrons and having the same bond order.
i.e.BO=3BO = 3
Also, nitrogen molecules (N2{N_2}) have fourteen electrons and have the same bond order.
i.e. BO=3 = 3
Therefore NO+,CN,N2N{O^ + },C{N^ - },{N_2} is having the same bond order as CO and NON{O^ - } has a different bond order compared to CO.

Hence, the correct answer is option A.

Note: Other than the isoelectronic species concept we can solve this problem by manually calculating the bond order of every species and then compare it with the bond order of CO to get the correct answer. Bond order in isoelectronic species is the same as bond order depends on the number of electrons in bonding and antibonding orbitals. Isoelectronic species have the same number of electrons and hence same bond order.