The molecular formula of second homologue in the homologous series of monocarboxylic acid is

The correct answer is CH₃COOH.

A homologous series is a group of organic compounds that share the same functional group, exhibit similar chemical properties, and have a general chemical formula. Each successive member of the series differs from the previous one by a
$C{H}_2$
(methylene) unit.

The general formula for the homologous series of monocarboxylic acids (alkanoic acids) is represented as:
$C_n{H}_{2n+1}COOH$
where $n=0,1,2,3,...$

By substituting values of $n$ into the general formula, we get the members of this series in increasing order of molecular mass:
1. First homologue ($n=0$): $HCOOH$ (Methanoic acid / Formic acid)
2. Second homologue ($n=1$): $C{H}_{3}COOH$ (Ethanoic acid / Acetic acid)
3. Third homologue ($n=2$): $C{H}_{3}C{H}_{2}COOH$ (Propanoic acid)

Note: The structure shown in the attached image has a main chain with a carboxylic acid functional group and a methyl branch, which is:
$C{H}_3-CH\left(C{H}_3\right)-COOH$
This is 2-methylpropanoic acid (isobutyric acid), which is a branched structural isomer of a four-carbon carboxylic acid and is not the second member of the series.

Therefore, the molecular formula of the second homologue in the homologous series of monocarboxylic acid is CH₃COOH.