Question

In which of the following option, covalent bond is having maximum s% character?

• A. S-H bond in H2S
• B. P-H bond in PH3
• C. N-H bond in NH3
• D. All have equal s% character

Answer 1

Answer: (C)

N-H bond in NH3

Answer 2

The percentage s-character in a hybrid orbital determines its properties. Higher s-character leads to stronger, shorter bonds and larger bond angles. Drago's rule (as described in the passage) states that for heavier elements (third period and below) with less electronegative substituents (electronegativity $\leq$ 2.5), hybridization becomes less significant, and bonding occurs primarily through p-orbitals, while lone pairs reside in s-orbitals. This means very low s-character in bonding orbitals.

• (a) S-H bond in H₂S: Sulfur (S) is a third-row element. Hydrogen (H) has an electronegativity of 2.20, which is $\leq$ 2.5. According to the passage, the S-H bonds are formed using almost pure p-orbitals, meaning very low s% character. The bond angle in H₂S is approximately 92°.
• (b) P-H bond in PH₃: Phosphorus (P) is a third-row element. Hydrogen (H) has an electronegativity of 2.20, which is $\leq$ 2.5. According to the passage, the P-H bonds are formed using almost pure p-orbitals, meaning very low s% character. The bond angle in PH₃ is approximately 93.5°.
• (c) N-H bond in NH₃: Nitrogen (N) is a second-row element. The rule given in the passage does not apply to second-row elements. Nitrogen in NH₃ undergoes sp³ hybridization. In sp³ hybrid orbitals, the s% character is 25%. The bond angle in NH₃ is approximately 107°.
• (d) All have equal s% character: This is incorrect as the central atoms are from different periods, and the hybridization/bonding modes differ.

The N-H bond in NH₃ has the highest s% character (approximately 25% from sp³ hybridization) compared to the S-H and P-H bonds, which have very low s% character (bonding via almost pure p-orbitals as per the passage).