Formic acid is considered as a resonance hybrid of the four structures Which of the above order is correct for the stability of the four contributing structures :
$H - O | | C - O H \leftrightarrow H - O^{-} | C = + O H \leftrightarrow H -^{-} O - H \leftrightarrow H -^{+} O - H$
$I$ $I I$ $I I I$ $I V$

I > I I > I I I > I V

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I > I I > I V > I I I

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I > I I I > I I > I V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I > I V > I I I > I I

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Solution

The correct option is A $I > I I > I I I > I V$
The compound without charge is most stable. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets. Therefore $H C O O H$ is most stable.
The octet is complete and the number of pie bond is more in II.
A structure with a negative charge on the more electronegative atom will be more stable $+ C H - O O H$ .
A structure with a positive charge on the more electronegative atom with incomplete octet will not be stable $+ O H - C O H$ is least stable. the order is $I > I I > I I I > I V$

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Formic acid is considered as a resonance hybrid of the four structures Which of the above order is correct for the stability of the four contributing structures :H−O||C−OH↔H−O−|C=+OH↔H−O−|+C−O−H↔H−O+|−C−O−H I II III IV