Which of the following statement regarding $NH_{3}$ and $NF_{3}$ is correct?
A$NH_{3}$ has pyramidal and $NF_{3}$ has trigonal planar shape.
BBond angle in $NH_{3}$ is smaller than $NF_{3}$ .
CResultant dipole moment of $NH_{3}$ is $(4.90 \times 10^{-30} \, \text{cm})$ and that of $NF_{3}$ is $(0.8 \times 10^{-30} \, \text{cm})$ .
DThey both are $sp^2$ hybridised.
Answer & Solution
Correct answer: C. Resultant dipole moment of $NH_{3}$ is $(4.90 \times 10^{-30} \, \text{cm})$ and that of $NF_{3}$ is $(0.8 \times 10^{-30} \, \text{cm})$ .
Both $NH_3$ and $NF_3$ are pyramidal and $sp^3$ hybridised, so A and D are incorrect. In $NF_3$, the bond moments oppose the lone-pair moment more strongly, so its net dipole moment is much smaller than that of $NH_3$; therefore B is also wrong because bond angle in $NF_3$ is smaller. Hence option C is correct.
Related questions
A student is told to name the arrangement that includes every electron pair around the cenA chemist wants to compare how tightly an ionic solid is held together. The quantity to usThat distinction matters because the arrangement counting all electron pairs and the arranWater also has four electron pairs on its central atom, but two of them are lone pairs. ItYet ammonia's measured H-N-H angle is a little under 109.5 degrees. The reason is that a lAmmonia has four electron pairs around nitrogen. Its electron-pair geometry is therefore:A central atom carries three regions of electron density and no lone pairs. The geometry aBeryllium fluoride has just two regions of electron density around the central atom. VSEPR