Question

Order of the species $CF$, $C{F^ + }$and $C{F^ - }$ according to the increasing order of C-F bond length:
(A) $C{F^ + }$< $CF$< $C{F^ - }$
(B) $C{F^ + }$< $C{F^ - }$<
(C) $C{F^ - }$< $CF$< $C{F^ + }$
(D) $CF$< $C{F^ - }$< $C{F^ + }$

Answer 1

In order to find the increasing order of the bond length of the species $CF$, $C{F^ + }$ and $C{F^ - }$ of the C-F molecule, we must be knowing about the relationship between the bond length ad the bond order. We can calculate the bond order of the $CF$, $C{F^ + }$ and $C{F^ - }$, with that we can say the increasing order of bond length of those species

Complete Solution :
First of all, what is a molecular orbital? Molecular orbitals are said to be the linear combination of the atomic orbital. In order to find the total number of electrons present in the CF molecule, we must calculate the number of electrons in Carbon and Fluorine separately. Finally, we have to add these values.
Carbon atom has an atomic number 6; hence, six electrons will be present in it.
Fluorine atom has atomic number 9; hence it will contain six electrons in it.
The electronic configuration of carbon and Fluorine is given below:
$C = 6 = 1{s^2}2{s^2}2{p^2}$
$F = 9 = 1{s^2}2{s^2}2{p^5}$
Total electron present in the CF molecule = 6 + 9 = 15 electrons.

In the molecular orbital diagram of CF molecule, the electrons are arranged in the following order:
$\sigma 1{S^2},{\sigma ^ * }1{S^2},\sigma 2{S^2},{\sigma ^ * }2{S^2},\sigma 2P_X^2,\pi 2P_Y^2 = \pi 2P_Z^2,{\pi ^ * }2P_Y^1 = {\pi ^ * }2P_Z^0$
${\text{BondOrder = }}\dfrac{{{\text{number of electrons in bonding molecular orbital - number of electrons in antibonding molecular orbital}}}}{{\text{2}}}$

- Therefore, ${\text{bond order of CF = }}\dfrac{{{\text{10 - 5}}}}{{\text{2}}} = 2.5$
In the $C{F^ + }$ orbital there will be one electron less than CF, therefore total number of electrons in $C{F^ + }$=14

There will be one electron less in the antibonding orbital
${\text{bond order of C}}{{\text{F}}^ + }{\text{ = }}\dfrac{{{\text{10 - 4}}}}{{\text{2}}} = 3$
In the $C{F^ - }$ orbital there will be one electron more than CF, therefore total number of electrons in $C{F^ - }$=16

There will be one electron more in the antibonding orbital
${\text{bond order of C}}{{\text{F}}^ - }{\text{ = }}\dfrac{{{\text{10 - 6}}}}{{\text{2}}} = 2$
$CF$, $C{F^ + }$, $C{F^ - }$ are arranged in the increasing order of their bond order
$C{F^ - }$< $CF$< $C{F^ + }$

As we know that bond order is inversely proportional to the bond length.
Then order of $CF$, $C{F^ + }$, $C{F^ - }$ are arranged in the increasing order of the bond length as follows:
$C{F^ + }$< $CF$< $C{F^ - }$
So, the correct answer is “Option A”.

Additional Information.
The molecular orbitals are of three types, they are:
- Bonding Molecular orbital:
When there is an attractive interaction between the atomic orbitals of the atoms present in a molecule, it is called Bonding Molecular orbital.
- Antibonding Molecular orbital:
When the interaction between the atomic orbitals of the atoms present in a molecule is weak, it is called Antibonding molecular orbitals.
- Non-bonding Molecular orbital:
When we add or remove an electron from a molecular orbital, it won't change the energy of the molecule. Such Molecular orbitals are called Non-bonding molecular orbitals.

Note: We should always remember the relation between the bond order and bond length of a molecule. The bond order is always inversely proportional to the bond length, i.e. when the bond order of a molecule increases its bond length will decrease.