In the ALDOL CONDENSATION, two carbonyl compounds (containing $\alpha$-hydrogens) react in the presence of a base to give a $\beta$-hydroxy aldehyde or ketone. The KEY first step is

1. NCERT §8.4 describes the aldol condensation mechanism (base-catalysed version). 2. STEP 1: The base (e.g. $\mathrm{OH^-}$) deprotonates the $\alpha$-CARBON of one aldehyde/ketone, removing one $\alpha$-hydrogen. 3. The resulting carbanion is stabilised by delocalisation into the adjacent $\mathrm{C{=}O}$ — this is the ENOLATE ion. 4. STEP 2: The enolate carbon (nucleophilic) attacks the electrophilic carbon of a SECOND aldehyde/ketone molecule, forming a new C-C bond. 5. STEP 3: Protonation of the resulting alkoxide gives the aldol — a $\beta$-hydroxy carbonyl. (Further dehydration under heat gives an $\alpha,\beta$-unsaturated carbonyl — the 'condensation' product.) 6. Option B (homolytic radicals) is not the aldol mechanism. Option C is water addition, unrelated. Option D describes Cannizzaro, which competes with aldol for $\alpha$-H-FREE substrates. _Source: NCERT Class 12 Chemistry Part 2, Ch 8, §8.4 (Aldol condensation mechanism), p. 13._