Arrange the following compounds in increasing order of their dipole moment: [2025]

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Solution

Q.
Arrange the following compounds in increasing order of their dipole moment:

${HBr, H}_{2} {S, NF}_{3} {and CHCl}_{3}$ [2025]

1 ${NF}_{3} < HBr < H_{2} S < {CHCl}_{3}$

2 ${CHCl}_{3} < {NF}_{3} < HBr < H_{2} S$

3 $HBr < H_{2} S < {NF}_{3} < {CHCl}_{3}$

4 $H_{2} S < HBr < {NF}_{3} < {CHCl}_{3}$

Ans.
(1)

$\to_{Increasing value of μ}^{\begin{matrix} {NF}_{3} & HBr & H_{2} S & {CHCl}_{3} \\ μ & 0.24 & 0.79 & 0.95 & 1.04 \\ (in D) \end{matrix}}$

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Solution

Q. Arrange the following compounds in increasing order of their dipole moment: HBr, H2S, NF3 and CHCl3 [2025]

1 NF3<HBr<H2S<CHCl3

2 CHCl3<NF3<HBr<H2S

3 HBr<H2S<NF3<CHCl3

4 H2S<HBr<NF3<CHCl3

Ans. (1) →Increasing value of μNF3HBrH2SCHCl3μ0.240.790.951.04(in D)

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Q.
Arrange the following compounds in increasing order of their dipole moment:

${HBr, H}_{2} {S, NF}_{3} {and CHCl}_{3}$ [2025]

1 ${NF}_{3} < HBr < H_{2} S < {CHCl}_{3}$

2 ${CHCl}_{3} < {NF}_{3} < HBr < H_{2} S$

3 $HBr < H_{2} S < {NF}_{3} < {CHCl}_{3}$

4 $H_{2} S < HBr < {NF}_{3} < {CHCl}_{3}$

Ans.
(1)

$\to_{Increasing value of μ}^{\begin{matrix} {NF}_{3} & HBr & H_{2} S & {CHCl}_{3} \\ μ & 0.24 & 0.79 & 0.95 & 1.04 \\ (in D) \end{matrix}}$