The observed dipole moment of $H C l$ is $1.3 D$ . If the bond length of $H C l$ is $1.3 ˚ A$ , then percentage ionic character of $H - C l$ bond is nearly :

(Charge on electron $= 4.8 \times 10^{- 10} e s u$ )

50 %

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40 %

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30 %

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17 %

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Solution

The correct option is D $17 %$
$%$ ionic character $= \frac{μ P r a c t i c a l}{μ T h e o r e t i c a l} \times 100$

$μ P r a c t i c a l$ is given $= 1.3 D$

$μ T h e o r e t i c a l$ $=$ charge on electron in $e s u \times$ bond length

$= 4.8 \times 10^{- 10} \times 1.3 A^{o} = 6.24 \times 10^{- 10} A^{o}$

$%$ ionic character $= \frac{1.3}{6.24} \times 100 \approx 17 %$

Hence, the correct option is D.

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The observed dipole moment of HCl is 1.3D. If the bond length of HCl is 1.3˚A, then percentage ionic character of H−Cl bond is nearly :(Charge on electron =4.8×10−10esu)

The correct option is D 17%% ionic character =μPracticalμTheoretical×100μ Practical is given =1.3Dμ Theoretical = charge on electron in esu× bond length =4.8×10−10×1.3Ao=6.24×10−10Ao% ionic character=1.36.24×100≈17%Hence, the correct option is D.

The observed dipole moment of $H C l$ is $1.3 D$ . If the bond length of $H C l$ is $1.3 ˚ A$ , then percentage ionic character of $H - C l$ bond is nearly :

(Charge on electron $= 4.8 \times 10^{- 10} e s u$ )