Question

The increasing order with respect to dipole moment of the following molecules is: align* BF_3, H_2O, NF_3, NH_3 align*

• A. \(NF_3 \lt BF_3 \lt NH_3 \lt H_2O\)
• B. \(NF_3 \lt BF_3 \lt H_2O \lt NH_3\)
• C. \(BF_3 \lt NH_3 \lt H_2O \lt NF_3\)
• D. \(BF_3 \lt NF_3 \lt NH_3 \lt H_2O\)

Hint:

Remember these standard dipole moment trends: \[ \begin{aligned} BF_3 &: \text{Symmetrical, } \mu=0 \\ NF_3 &: \text{Bond moments oppose lone-pair moment} \\ NH_3 &: \text{Bond moments reinforce lone-pair moment} \\ H_2O &: \text{Bent structure with high resultant dipole moment} \end{aligned} \]

Answer 1

The Correct Option is D

Step 1: Recall what dipole moment depends on.
Dipole moment depends on the shape of the molecule and on whether the bond dipoles cancel or add up.

Step 2: Look at $BF_3$.
It is flat and symmetric (trigonal planar). The three $B-F$ dipoles point out at $120^\circ$ and cancel completely, so its net dipole is zero. It is the smallest.

Step 3: Look at $NF_3$.
Nitrogen has a lone pair, so it is pyramidal. But the $N-F$ bond dipoles point away from nitrogen and partly oppose the lone-pair dipole, so the total is small but not zero.

Step 4: Look at $NH_3$.
Also pyramidal, but here the $N-H$ dipoles point toward nitrogen, the same general way as the lone pair, so they add up. Its dipole is larger than $NF_3$.

Step 5: Look at $H_2O$.
Water is bent with two lone pairs. The bond dipoles and lone pairs reinforce strongly, giving the biggest dipole of the group.

Step 6: Order them.
So increasing dipole: $BF_3 \lt NF_3 \lt NH_3 \lt H_2O$.
\[ \boxed{BF_3 \lt NF_3 \lt NH_3 \lt H_2O} \]