Question

The bond dissociation energy of ${C_2}(599\;kJ\;mo{l^{ - 1}})$ decreases slightly on forming $C_2^ + (513\;kJ\;mo{l^{ - 1}})$ increases greatly on forming $C_2^ - (818\;kJ\;mo{l^{ - 1}})$. State if the given statement is true or false?

Answer 1

Hint:As we know that Carbon possess $12$ electrons and they are arranged in bonding as well as anti-bonding molecular orbitals and we also know that the bond dissociation energy is directly proportional to the bond order of the molecule and greater the bond order, greater will be the dissociation energy of bond.

Complete step-by-step answer: As we know that carbon possess $12$ electrons and they are arranged in bonding as well as anti-bonding molecular orbitals having the electronic configuration as:
${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}(\pi 2p_x^2)(\pi 2p_y^2)(\sigma 2p_z^0)$
So we can calculate the bond order of the carbon diatom by counting the number of bonding electrons and antibonding electrons in the molecular orbitals and it is given as:
$B.O = \dfrac{{{N_b} - {N_a}}}{2}$
$ \Rightarrow B.O = \dfrac{{8 - 4}}{2} = 2$
So the bond order of ${C_2}$ is found to be $2$.
Now, we can calculate the bond order of $C_2^ + $ where one electron is removed from the p-orbital, therefore it only possesses $11$ electrons. So the electronic configuration is ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}(\pi 2p_x^2)(\pi 2p_y^1)(\sigma 2p_z^0)$ and thus the bond order of $C_2^ + $ will be:
$ \Rightarrow B.O = \dfrac{{7 - 4}}{2} = 1.5$
Hence, the bond order of $C_2^ + $ is found to be $1.5$.
Similarly, we can calculate the bond order of $C_2^ - $ where we can see that one electron is added to the carbon molecular orbital, thus the total number of electrons results into $13$ electrons and the electronic configuration becomes ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}(\pi 2p_x^2)(\pi 2p_y^2)(\sigma 2p_z^1)$ .
So the bond order will be:
$ \Rightarrow B.O = \dfrac{{9 - 4}}{2} = 2.5$
Hence, the bond order of $C_2^ - $ is found to be $2.5$.
Thus we can say from the above explanation that the bond order in $C_2^ + $ is less than that of $C_2^ - $ and thus the bond dissociation energy is also less in $C_2^ + $ when it is formed from ${C_2}$.

Therefore the given statement is correct or true.

Note:Always remember that bond dissociation energy is directly related to the bond order of the given molecule or compound. As the bond order increases of a given compound, the bond dissociation energy will always increase and if it decreases, the bond dissociation energy will also decrease.