Question

What is the bond order of $CN^{-}$?

Answer 1

Hint: Bond order is defined as half the difference between the number of electrons in the bonding molecular orbitals and the number of electrons in the antibonding molecular orbitals. Cyanide ion has a total of 14 electrons and so considering the bonding and antibonding orbitals from the configuration we can calculate the bond order.

Complete answer:
- Bond order of a compound defined as half the difference between the number of electrons in the bonding molecular orbitals of the compound which is represented by ‘a’ and t electrons in the antibonding molecular orbitals of the compound which is represented by ‘b’.
$BO=\dfrac{1}{2} (a-b)$
- Bond order of a compound also represents the number of covalent bonds in a compound. It represents the strength of the bonds and is positive for stable bonds. It finds its applications in the valence bond theory.
- The electrons in the bonding orbitals stabilise the compound and the electrons in the antibonding orbitals make the compound less stable.
For the above question, $CN^{-}$ has a total of 14 electrons which include 6 of carbon atoms and 7 of nitrogen atoms and it has a one extra electron due to the negative charge on the compound.
To calculate the bond order, we require the configuration which is
$(\sigma 1s)^{2}(\sigma^{\ast } 1s)^{2}(\sigma 2s)^{2}(\sigma^{\ast } 2s)^{2}(\sigma 2p_{z})^{2}\pi 2p^{2}_{x}=\pi 2p^{2}_{y}$
The electrons in bonding orbitals = 10
The electrons in antibonding orbitals = 4
Thus,
$BO=\dfrac{10-4}{2} =3$
Therefore, the bond order of cyanide is 3.

Note:
Cyanide is a chemical compound with a triple bond between carbon and nitrogen atoms. It forms salts with potassium and sodium which are highly toxic. It is extremely harmful for the cell of the human body as they prevent them from absorbing oxygen and it also affects the heart, brain and other organs.