Question

Among these species, the correct order of resonance energy is:
$I.\text{ HN=CH-}\overset{-}{\mathop{NH}}\,$
$II.\text{ O=CH-}\overset{-}{\mathop{N}}\,H$
$\text{III}\text{. O=CH-OH}$
A. I ˃ II ˃ III
B. II ˃ I ˃ III
C. III ˃ I ˃ II
D. III ˃ II ˃ I

Answer 1

Hint: Resonance structure is the representation of the delocalized electron in molecules which have either a double-bond or lone pair of electrons. More is the resonating structure, more will be the stability of a molecule because they are directly proportional to each other.

Complete Solution :
- In the given question, we have to explain the correct order of the resonance energy among the given options.
- Now, resonance energy tells us that the difference between the energy of the most stable resonating structure and the original structure. In the given compounds, let us draw the resonating first.
-In the first molecule, the positive charge i.e. present on the nitrogen will convert into a $\pi -bond$ and the double-bond between the NH and H will convert into a negative ion that is present on NH.
$HN=CH-N{{H}^{-}}\leftrightarrow N{{H}^{-}}-CH=NH$

-Now, in the second molecule, $O=CH-N{{H}^{-}}\leftrightarrow {{O}^{-}}-HC=NH$
-The resonating structure will be unstable because negative charge is present on the most electronegative atom which makes it unstable.

-Now, let's draw the resonating structure of the third molecule,
$O=CH-OH\leftrightarrow {{O}^{+}}-CH=O{{H}^{-}}$
-This molecule is also unstable because a positive charge on oxygen is highly unstable. Hence, make the molecule highly unstable.
-So, the order of the stability will be I ˃ II ˃ III
So, the correct answer is “Option A”.

Note: Always remember that the value of resonating energy will be negative. Students should always remember the rule of writing the resonating structure i.e. only lone pairs and $\pi -bond$ are delocalized in the resonating structure.