Question

What is the correct dipole moment of $N{H}_3$ and $N{F}_3$ respectively?

• A. $4.90\times {10}^{-30}$ cm and $0.80\times {10}^{-30}$ cm
• B. $0.80\times {10}^{-30}$ cm and $4.90\times {10}^{-30}$ cm
• C. $4.90\times {10}^{-30}$ cm and $4.90\times {10}^{-30}$ cm
• D. $0.80\times {10}^{-30}$ cm and $0.80\times {10}^{-30}$ cm

Answer 1

The correct option is A $4.90\times {10}^{-30}$ cm and $0.80\times {10}^{-30}$ cm

$1Debye={10}^{-18}esucm$

As shown in the image all the vectors in $N{H}_3$ are in the same direction and in opposite direction in the case of $N{F}_3$.

so the resultant dipole moment will be more for $N{H}_3$.

The correct option is [A]