Consider the following carbanions.

Correct decreasing order of stability is:

I I > I I I > I V > I

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I I I > I V > I > I I

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I V > I > I I > I I I

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I > I I > I I I > I V

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Solution

The correct option is A $I I > I I I > I V > I$
Electron withdrawing group at ortho and para position in aromatic ring increases the stability of carbanion whereas electron donating group bonded at ortho and para position decreases the stability of carbanion.
$- N O_{2}$ group is -R showing group and is present at para position in $I I$ . $- O C H_{3}$ group is +R showing group and is present at para position in $I$ . $- C H_{3}$ group is also electron donating group through hyperconjugation and is present at para position in structure $I V$ . Hence order of stability of carbanions will be:
$I I > I I I > I V > I$

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