Dipole moment of H 2 O is 1-85 D- If bond angle is 105- and O - H bond length is 0-94 AA- 3-23 - 10-10B- 1-5 - 10-10 esuC- 2-32 - 10-10 esuD- 4-5 - 10-15 esu

The correct option is A 3.23 × 10^ 10 From the bond angle and vector moment μ = e × l Thus e(charge) can be determined AB = BC cos 52.5^∘ =0.94 × 0.609 = 0.5 ...

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Standard XII

Chemistry

Bond Length

Dipole moment...

Question

Dipole moment of $H_{2} O$ is 1.85 D. If bond angle is $105^{\circ}$ and O - H bond length is $0.94 \circ A$

2.32 \times 10^{- 10} e s u

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3.23 \times 10^{- 10}

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1.5 \times 10^{- 10} e s u

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4.5 \times 10^{- 15} e s u

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Solution

The correct option is B $3.23 \times 10^{- 10}$

From the bond angle and vector moment $μ = e \times l$
Thus e(charge) can be determined
$A B = B C c o s {52.5}^{\circ}$
$= 0.94 \times 0.609 = 0.572$
$= 0.572 \times 10^{- 4} c m$
$μ = e \times A B$
$e = \frac{μ}{A B} = \frac{1.85 \times 10^{- 18} e s u c m}{0.572 \times 10^{- 4} c m} = 3.23 \times 10 - 10 e s u$

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Similar questions

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Dipole moment of H2O is 1.85 D. If bond angle is 105∘ and O - H bond length is 0.94 ∘A

The correct option is B 3.23×10−10 From the bond angle and vector moment μ=e×l Thus e(charge) can be determined AB=BC cos 52.5∘ =0.94×0.609=0.572 =0.572×10−4 cm μ=e×AB e=μAB=1.85×10−18 esu cm0.572×10−4 cm=3.23×10−10 esu

Dipole moment of $H_{2} O$ is 1.85 D. If bond angle is $105^{\circ}$ and O - H bond length is $0.94 \circ A$