Question

Which of the following molecules has the shortest bond length of $C-O$ bond?
A) $C{H}_{3}OH$
B) $H_{2}CO$
C) $CO$
D) $N{a}_{2}C{O}_3$

Answer 1

Hint: We recognize that the space between 2 atoms taking part in a very bond, referred to as the bond length, is determined experimentally. The bond length decreases across an amount within the table and will increase down a group. As a result of down the group, the atoms of the weather are increasing in size because of a lot of variety of lepton shells. Since the distance of the nucleus to the outer electron of the opposite atom is longer, bond length will be longer.

Complete step by step answer:
We know that the bond order is the number of electrons pairs shared between two atoms in the formation of the bond. The quantity of energy necessary to break a bond is called bond dissociation energy. Thus greater the bond order greater is the bond energy.
$\text{Bond order}\propto \text{Bond energy}\propto {\displaystyle \frac{1}{\text{Bond length}}}$
As the bond order decreases, the bond decreases too. Because of that triple bonds between atoms are shorter than double bonds because it requires more energy to completely break all three bonds than to break two. But triple bonds are also stronger than double bonds. The same is with the double bonds and single bonds.
We can draw the structure of CO as,

In $$CO$$, $C-O$ bond gets triple bond character in one in all the resonating structures. Therefore it's the shortest bond length of $C-O$ bond. Therefore, the option C is correct.
Now we can draw the structure of $C{H}_{3}OH$ as,

However in A exceedingly $C-O$ has a single bond character. Therefore, the option A is wrong.
Let we can draw the structure of $H_{2}CO$ as,

In B, $C-O$bond has double character. Therefore, option B is wrong.
We can draw the structure of $N{a}_{2}C{O}_3$ as,

In D, $C-O$ bonds have bond length larger than covalent bond but below single $C-O$ bond.
Therefore, the option D is wrong.

So, the correct answer is Option C.

Note: As we know that the bond order decreases, the bond length increases. A large amount of energy has to use large energy to triple bond than double bond. That is why triple bonds are shorter.
Example:
The ground state configuration of ${H_2}^{+}$ molecular ion is $\sigma 1s^1$
Bond order $={\displaystyle \frac{\left(N_b\right)-\left(N_a\right)}{2}}$
Bond order $={\displaystyle \frac{\left(1\right)-\left(0\right)}{2}}=0.5$
The ground state configuration of ${H_2}^{+}$ ion in first excited state is $(\sigma 1s{)}^1$
Bond order $\text{ = }{\displaystyle \frac{\left({\text{N}}_{\text{b}}\right)\text{ - }\left({\text{N}}_{\text{a}}\right)}{\text{2}}}$
Bond order $={\displaystyle \frac{\left(1\right)-\left(0\right)}{2}}=0.5$
The ion ${H_2}^{+}$in the first excited molecule is unstable because the bond order is negative.
The electronic configuration hydrogen atom is $1s^1$. Because this is an atomic orbital it is higher in energy to molecular orbitals, Hydrogen atom is higher in energy than the hydrogen molecule and the separated $H^{+}$ ion does not have an electron, so it is stable.