NEET 2019 Odisha · ChemistryDipole momentPrevious Year Question

Which of the following is the correct order of dipole moment?

Answer: (B) $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$ . \textbf{Answer:} (B) $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$ \textbf{Solution:} $BF X_{3}$ is trigonal planar and symmetric so $μ = 0$ ; in $NF X_{3}$ the bond dipoles oppose the lone-pair dipole (lower $μ$ ) whereas in $NH X_{3}$ they reinforce it (higher $μ$ ); $H X_{2} O$ (bent, $104. 5^{\circ}$ ) has the largest $μ$ ( $1.85 D$ ).

A. $NH X_{3}$ < $BF X_{3}$ < $NF X_{3}$ < $H X_{2} O$
B. $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$ ✓
C. $BF X_{3}$ < $NH X_{3}$ < $NF X_{3}$ < $H X_{2} O$
D. $H X_{2} O$ < $NF X_{3}$ < $NH X_{3}$ < $BF X_{3}$

Correct Answer

(B) $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$

Solution & Explanation

\textbf{Answer:} (B) $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$ \textbf{Solution:} $BF X_{3}$ is trigonal planar and symmetric so $μ = 0$ ; in $NF X_{3}$ the bond dipoles oppose the lone-pair dipole (lower $μ$ ) whereas in $NH X_{3}$ they reinforce it (higher $μ$ ); $H X_{2} O$ (bent, $104. 5^{\circ}$ ) has the largest $μ$ ( $1.85 D$ ). Hence $BF X_{3}$ < $NF X_{3}$ < $NH X_{3}$ < $H X_{2} O$ .

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