Question

Order these species according to increasing \[C - F\] bond length: \[C{F^ + },{\text{ }}CF,{\text{ }}C{F^ - }\]
A: Bond lengths $C{F^ + } < CF < C{F^ - }$
B: Bond lengths $C{F^ + } = CF = C{F^ - }$
C: Bond lengths $CF < C{F^ + } < C{F^ - }$
D: Bond lengths \[C{F^ - } < CF < C{F^ + }\]

Answer 1

Hint Bond order refers to the number of chemical bonds that exist between the pair of atoms. On the other hand, bond length refers to the average distance between the nuclei of the two bonded atoms in any molecule. Both of them i.e. bond order as well as bond length are used to determine the strength and type of the covalent bonds between atoms.

Complete step by step answer:
 Bond length and bond order are both inversely proportional to each other. That means when bond length decreases, bond order is increased.
To solve the given question, we will follow the molecular orbital (MO) theory which uses a linear combination of the atomic orbitals to represent the molecular orbitals resulting from the bonds between the atoms. They are divided into three types which are bonding, non-bonding and antibonding.
The MO electronic configuration of \[CF\] (15e¯) can be written as \[\sigma 1{s^2}{,^ * }\sigma 1{s^2},{\text{ }}\sigma 2{s^2}{,^ * }\sigma 2{s^2},{\text{ }}\sigma 2{p^2}\]
\[\pi 2p_x^2 = \pi 2p_y^2,\pi 2p_x^1\]
Bond order (BO) can be estimated as follows:
$Bond{\text{ }}order = \dfrac{{\left( {No.{\text{ }}of{\text{ }}electrons{\text{ }}in{\text{ }}antibonding{\text{ }}MO} \right) - \left( {No.{\text{ }}of{\text{ }}electrons{\text{ }}in{\text{ }}bonding{\text{ }}MO} \right)}}{2}$
\[Bond{\text{ }}order{\text{ }}\left( {BO} \right){\text{ }}of{\text{ }}CF = \dfrac{{10 - 5}}{2} = 2.5\]
Similarly, we will calculate the bond lengths of \[C{F^ + }\]and \[C{F^ - }\]:
In \[C{F^ + }\], one electron is less in antibonding MO compared to \[CF\], therefore, \[BO{\text{ }}of{\text{ }}C{F^ + } = \dfrac{{10 - 4}}{2} = 3\]
In \[C{F^ - }\], one electron is more in antibonding MO compared to \[CF\], therefore, \[BO{\text{ }}of{\text{ }}C{F^ - } = \dfrac{{10 - 6}}{2} = 2\]
Thus, the order of BO is \[C{F^ + } > CF > C{F^ - }\]
We know that: \[Bond{\text{ }}length \propto \dfrac{1}{{Bond{\text{ }}order}}\]
As a result, the order of bond length would be: \[C{F^ + } < CF < C{F^ - }\]
Therefore, option (A) is correct.

Note: Always remember that the stronger the bond is, shorter will be the bond length. This means that single bonds are always longer than the double bonds which are again longer than the triple bonds. The order of bond length is triple bond < double bond < single bond.