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Question: The bond order of \(C_{2}^{+}\) is (A) 1 (B) 2 (C) \(\dfrac{3}{2}\) (D) \(\dfrac{1}{2}\)...

The bond order of C2+C_{2}^{+} is
(A) 1
(B) 2
(C) 32\dfrac{3}{2}
(D) 12\dfrac{1}{2}

Explanation

Solution

To solve this question, we first need to know what is bond order. The number of chemical bonds through which a pair of atoms are bonded is known as the bond order. The bond order of a molecule can be determined through the concept of molecular orbital theory.

Complete answer:
Now, to determine the bond order of C2+C_{2}^{+}, we first need to draw its molecular orbital diagram.

So, the electronic configuration of the C2+C_{2}^{+} molecule according to the molecular orbital theory is
C2+=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2py)2(π2pz)1C_{2}^{+}={{(\sigma 1s)}^{2}}{{(\sigma *1s)}^{2}}{{(\sigma 2s)}^{2}}{{(\sigma *2s)}^{2}}{{(\pi 2py)}^{2}}{{(\pi 2pz)}^{1}}
Where σ/π\sigma /\pi orbitals depict bonding molecular orbitals whereas π∗/σ∗\pi */\sigma * depict antibonding molecular orbitals.
Now, the bond order of a molecule is given by the formula
BO=12[Nb−Na]BO=\dfrac{1}{2}[{{N}_{b}}-{{N}_{a}}]
Where the number of electrons in the bonding orbitals is denoted by Nb{{N}_{b}} and the number of electrons in the antibonding orbitals is denoted by Na{{N}_{a}}.
From the molecular orbital diagram of the C2+C_{2}^{+} molecule we can see that it has 7 electrons in its bonding orbitals (Nb{{N}_{b}}) and 4 electrons in its antibonding orbitals (Na{{N}_{a}}).
So, its bond order will be

& B{{O}_{C_{2}^{+}}}=\dfrac{1}{2}[7-4] \\\ & B{{O}_{C_{2}^{+}}}=\dfrac{3}{2} \\\ \end{aligned}$$ **Hence the correct answer is option (C) $\dfrac{3}{2}$.** **Additional Information:** When a molecule has an unpaired electron in its orbital, it has a net spin value and hence is paramagnetic. Whereas if a molecule does not have an unpaired electron in its orbital, it has a 0 net spin value and hence is diamagnetic. In $C_{2}^{+}$, the $\pi 2pz$ orbital has an unpaired electron; hence the molecule is paramagnetic. **Note:** It should be noted that the stability of a bond can be indicated by bond order. The higher the bond order of a molecule, the more the atoms are held together tightly due to an increase in attraction between its electrons.