Question

The four structures:
$H - (C \equiv O) - OH \leftrightarrow H - ( + C - {O^ - }) - O - H \leftrightarrow H - ( - C - O + ) - O - H$
Which of the following orders is correct for the stability of the four contributing structures.
(A) $I > II > III$
(B) $I > III > II$
(C) $III > I > II$
(D) $II > I > III$

Answer 1

Hint : Stability of an atom depends on the ionization energy. More the ionization energy, more is the stability, because when an atom is stable it requires more energy to remove electron from its outermost shell, making it unstable.For the above structures, we need to find out which one of the structure is ionized and which one is non-ionized. Non-ionized structures are more stable than the ionized one.

Complete Step By Step Answer:
Structure $III$ has a positive charge on oxygen and oxygen is more electronegative than carbon; while structure $I{\text{ and }}II$ are more stable compared to $III$ . This is because structure $I$ has no charge and structure $II$ has a negative charge on oxygen which makes it more stable than structure $III$ .
Therefore, option (a) $I > II > III$ is correct.

Note :
When an atom is positively charged, it means that the atom has less electrons and it makes the atom more unstable and makes it accept electrons from other sources which will make it stable. However, when an atom is negatively charged, it means that the atom possesses more electrons than it is supposed to. In this case, the atom will give away the extra electrons it possesses in its outermost shell in order to make itself stable.