Let's analyze each statement regarding the $\overline{\mathrm{C}}{X}_3$ species (a carbanion).

Concept: The hybridization of the central carbon atom in a carbanion $\overline{\mathrm{C}}{X}_3$ is typically $s{p}^3$, leading to a pyramidal geometry. However, the s-character can vary depending on the electronegativity of the surrounding atoms (Bent's Rule).

Why (D) is incorrect:

• Statement A: If the electronegativity of the surrounding element $X$ is less than 2.5 (e.g., H, alkyl groups), the carbanion is typically $s{p}^3$ hybridized. In $s{p}^3$ hybridization, the s-character is 25%. However, due to the presence of a lone pair, the lone pair occupies an orbital with more s-character (Bent's Rule). This pushes the bonding orbitals to have less s-character, but the overall hybridization is still considered $s{p}^3$. The statement says "almost 33% S-character", which would imply $s{p}^2$ hybridization. For a carbanion, if X is less electronegative, the lone pair will reside in an orbital with higher s-character, leaving the bonding orbitals with slightly less s-character than 25%. However, the statement implies a significant increase in s-character in bonding orbitals, which is generally incorrect for less electronegative substituents. Let's re-evaluate this in context with other options. If the lone pair occupies an orbital with more s-character, the bonding orbitals will have less s-character. So, 33% s-character is incorrect for bonding orbitals. However, the question might be referring to the overall hybridization of the carbon atom, which is $s{p}^3$. Let's consider the general trend.
• Statement B: If the electronegativity of the surrounding element $X$ is greater than 2.5 (e.g., F, Cl), the lone pair on carbon will occupy an orbital with less s-character (as more electronegative atoms prefer orbitals with more p-character). This would leave the bonding orbitals with slightly more s-character than 25%. So, "almost 25% S-character" is plausible, as the deviation might be small, or it might be referring to the ideal $s{p}^3$ hybridization.
• Statement C: If $X$ is $F$, the species is $\overline{\mathrm{C}}{F}_3$. Due to the lone pair on carbon and the three C-F bonds, the central carbon is $s{p}^3$ hybridized. The geometry is pyramidal (like ammonia). Since the C-F bonds are polar and the molecule is not symmetrical (due to the lone pair), it will be polar. Thus, this statement is correct.
• Statement D: If $X$ is $H$, the species is $\overline{\mathrm{C}}{H}_3$. The central carbon atom has three bond pairs and one lone pair, making it $s{p}^3$ hybridized. The geometry is pyramidal (like ammonia), not planar. Also, due to the lone pair and the C-H bond polarity, the molecule is polar. Therefore, stating it is planar is incorrect.

Revisiting A and B: Bent's rule states that more electronegative substituents prefer hybrid orbitals with more p-character, and lone pairs prefer hybrid orbitals with more s-character. For $\overline{\mathrm{C}}{X}_3$, the carbon is $s{p}^3$ hybridized. If X is less electronegative, the lone pair will have more s-character, leaving the C-X bonds with less s-character (less than 25%). If X is more electronegative, the lone pair will have less s-character, leaving the C-X bonds with more s-character (more than 25%). Therefore, statement A (33% s-character for less electronegative X) and statement B (25% s-character for more electronegative X) are generally incorrect interpretations of Bent's rule for bonding orbitals. However, in the context of multiple-choice questions, we look for the most definitively incorrect statement.

Statement D clearly states that $\overline{\mathrm{C}}{H}_3$ is planar. Carbanions like $\overline{\mathrm{C}}{H}_3$ are known to be pyramidal due to $s{p}^3$ hybridization of the central carbon atom, with the lone pair occupying one of the $s{p}^3$ hybrid orbitals. Therefore, statement D is definitively incorrect.

Correct Answer: (D)