The molecular formula of second homologue in the homologous series of mono carboxylic acids is :

We need to determine the molecular formula of the second homologue in the homologous series of monocarboxylic acids.

Key Concept: A homologous series is a group of organic compounds sharing the same functional group and general chemical properties, where each successive member differs from the previous one by a methylene ($$\text{-}CH_2\text{-}$$) unit. The general formula for the homologous series of open-chain monocarboxylic acids is $$C_nH_{2n}O_2$$ (where $$n \ge 1$$).

Step 1: Finding the first member of the series ($$n = 1$$)

The first homologue contains one carbon atom. This compound is methanoic acid (formic acid), having the structural formula $$HCOOH$$. Combining the atoms gives its molecular formula:

$$CH_2O_2$$

Step 2: Finding the second member of the series ($$n = 2$$)

The second homologue is formed by adding a $$\text{-}CH_2\text{-}$$ group to the first member, meaning it contains two carbon atoms ($$n = 2$$). This compound is ethanoic acid (acetic acid), having the structural formula $$CH_3COOH$$.

Step 3: Writing the final molecular formula

Counting the total number of carbon, hydrogen, and oxygen atoms in $$CH_3COOH$$ gives the molecular formula:

$$C_2H_4O_2$$

Therefore, the molecular formula of the second homologue is $$C_2H_4O_2$$, which corresponds to Option B.

Answer: Option B — $$C_2H_4O_2$$