Let's analyze each option:

Concept: Bond length is influenced by factors such as bond order, hybridization, and electronegativity differences. Generally, a higher bond order leads to a shorter bond length. Also, as the s-character of the hybrid orbital forming the bond increases, the bond length decreases.

Why (C) is correct:

In $\mathrm{P}{\mathrm{H}}_3$, the phosphorus atom is $\mathrm{s}{\mathrm{p}}^3$ hybridized, but due to the presence of a lone pair, the bond angles are less than 109.5° (approximately 93.5°). The P-H bonds are formed by the overlap of an $\mathrm{s}{\mathrm{p}}^3$ orbital of P and a 1s orbital of H. The s-character in the bonding orbitals is effectively lower than 25% due to Bent's rule or the preference for lone pairs to occupy orbitals with more s-character.

In $\mathrm{P}{\mathrm{H}}_4^{+}$, the phosphorus atom is $\mathrm{s}{\mathrm{p}}^3$ hybridized and has a perfect tetrahedral geometry with bond angles of 109.5°. All four P-H bonds are equivalent and are formed by the overlap of an $\mathrm{s}{\mathrm{p}}^3$ orbital of P (with 25% s-character) and a 1s orbital of H. Since the effective s-character in the P-H bonds of $\mathrm{P}{\mathrm{H}}_4^{+}$ is higher than in $\mathrm{P}{\mathrm{H}}_3$, the P-H bond length in $\mathrm{P}{\mathrm{H}}_4^{+}$ is shorter than in $\mathrm{P}{\mathrm{H}}_3$. Therefore, P–H bond length in $\mathrm{P}{\mathrm{H}}_3>\mathrm{P}{\mathrm{H}}_4^{+}$.

Option Analysis:

• A) B–F bond length in $\mathrm{B}{\mathrm{F}}_3>\mathrm{B}{\mathrm{F}}_4^{-}$: In $\mathrm{B}{\mathrm{F}}_3$, B is $\mathrm{s}{\mathrm{p}}^2$ hybridized, and there is partial double bond character due to back-bonding (p-p overlap between F and B), making the B-F bond shorter. In $\mathrm{B}{\mathrm{F}}_4^{-}$, B is $\mathrm{s}{\mathrm{p}}^3$ hybridized, and all B-F bonds are single bonds. Thus, B–F bond length in $\mathrm{B}{\mathrm{F}}_3<\mathrm{B}{\mathrm{F}}_4^{-}$. So, this statement is incorrect.
• B) N–H bond length in $\mathrm{N}{\mathrm{H}}_3>\mathrm{N}{\mathrm{H}}_2^{-}$: In $\mathrm{N}{\mathrm{H}}_3$, N is $\mathrm{s}{\mathrm{p}}^3$ hybridized with one lone pair. In $\mathrm{N}{\mathrm{H}}_2^{-}$, N is $\mathrm{s}{\mathrm{p}}^3$ hybridized with two lone pairs. The increased electron-electron repulsion from two lone pairs in $\mathrm{N}{\mathrm{H}}_2^{-}$ would tend to increase the bond length slightly compared to $\mathrm{N}{\mathrm{H}}_3$. However, the N-H bond length in $\mathrm{N}{\mathrm{H}}_3$ is approximately 101.7 pm, and in $\mathrm{N}{\mathrm{H}}_2^{-}$ it is approximately 103.0 pm. So, N–H bond length in $\mathrm{N}{\mathrm{H}}_3<\mathrm{N}{\mathrm{H}}_2^{-}$. This statement is incorrect.
• D) C–H bond length in $\mathrm{C}{\mathrm{H}}_3^{+}>\mathrm{C}{\mathrm{H}}_4$: In $\mathrm{C}{\mathrm{H}}_3^{+}$, C is $\mathrm{s}{\mathrm{p}}^2$ hybridized (trigonal planar). In $\mathrm{C}{\mathrm{H}}_4$, C is $\mathrm{s}{\mathrm{p}}^3$ hybridized (tetrahedral). Since $\mathrm{s}{\mathrm{p}}^2$ orbitals have more s-character (33.3%) than $\mathrm{s}{\mathrm{p}}^3$ orbitals (25%), the C-H bond in $\mathrm{C}{\mathrm{H}}_3^{+}$ is shorter than in $\mathrm{C}{\mathrm{H}}_4$. Thus, C–H bond length in $\mathrm{C}{\mathrm{H}}_3^{+}<\mathrm{C}{\mathrm{H}}_4$. This statement is incorrect.

Correct Answer: (C)