Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\mathrm{N}{\mathrm{H}}_3$ has a larger H–N–H angle than the F–N–F angle in $\mathrm{N}{\mathrm{F}}_3$ because fluorine pulls bonding electron density away from nitrogen, so A is false. Both $\mathrm{C}{\mathrm{H}}_4$ and $\mathrm{C}{\mathrm{F}}_4$ are regular tetrahedra with equivalent bonds, as are $\mathrm{B}{\mathrm{H}}_4^{-}$ and $\mathrm{B}{\mathrm{F}}_4^{-}$, so B and D give tetrahedral bond angles. The H–P–H angle exceeds the H–As–H angle down group 15, so C is correct.