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Question: The bond dissociation energy of \({C_2}(599\;kJ\;mo{l^{ - 1}})\) decreases slightly on forming \(C_2...

The bond dissociation energy of C2(599  kJ  mol−1){C_2}(599\;kJ\;mo{l^{ - 1}}) decreases slightly on forming C2+(513  kJ  mol−1)C_2^ + (513\;kJ\;mo{l^{ - 1}}) increases greatly on forming C2−(818  kJ  mol−1)C_2^ - (818\;kJ\;mo{l^{ - 1}}). State if the given statement is true or false?

Explanation

Solution

As we know that Carbon possess 1212 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 1212 electrons and they are arranged in bonding as well as anti-bonding molecular orbitals having the electronic configuration as:
(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2)(π2py2)(σ2pz0){(\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=Nb−Na2B.O = \dfrac{{{N_b} - {N_a}}}{2}
⇒B.O=8−42=2\Rightarrow B.O = \dfrac{{8 - 4}}{2} = 2
So the bond order of C2{C_2} is found to be 22.
Now, we can calculate the bond order of C2+C_2^ + where one electron is removed from the p-orbital, therefore it only possesses 1111 electrons. So the electronic configuration is (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2)(π2py1)(σ2pz0){(\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 C2+C_2^ + will be:
⇒B.O=7−42=1.5\Rightarrow B.O = \dfrac{{7 - 4}}{2} = 1.5
Hence, the bond order of C2+C_2^ + is found to be 1.51.5.
Similarly, we can calculate the bond order of C2−C_2^ - where we can see that one electron is added to the carbon molecular orbital, thus the total number of electrons results into 1313 electrons and the electronic configuration becomes (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px2)(π2py2)(σ2pz1){(\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:
⇒B.O=9−42=2.5\Rightarrow B.O = \dfrac{{9 - 4}}{2} = 2.5
Hence, the bond order of C2−C_2^ - is found to be 2.52.5.
Thus we can say from the above explanation that the bond order in C2+C_2^ + is less than that of C2−C_2^ - and thus the bond dissociation energy is also less in C2+C_2^ + when it is formed from C2{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.