A covalent molecule XY is found to have a dip | vedclass

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(B) The dipole moment of a purely ionic bond is calculated as $\mu_{calc} = q 	imes d$. Given bond length $d = 150 \ pm = 150 	imes 10^{-12} \ m$. T

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(B) The dipole moment of a purely ionic bond is calculated as $\mu_{calc} = q 	imes d$. Given bond length $d = 150 \ pm = 150 	imes 10^{-12} \ m$. The charge of an electron is $q = 1.6 	imes 10^{-19} \ C$. $\mu_{calc} = 150 	imes 10^{-12} \ m 	imes 1.6 	imes 10^{-19} \ C = 2.4 	imes 10^{-29} \ C \cdot m$. The percentage ionic character is given by $\frac{\mu_{observed}}{\mu_{calc}} 	imes 100$. Given $\mu_{observed} = 1.5 	imes 10^{-29} \ C \cdot m$. Percentage ionic character $= \frac{1.5 	imes 10^{-29}}{2.4 	imes 10^{-29}} 	imes 100 = 62.5\%$. Therefore, the correct option is $B$.

List-$I$ (Molecule)	List-$II$ (Dipole moment in $D$)
$A. HCl$	$I. 1.07$
$B. NH_3$	$II. 1.85$
$C. H_2O$	$III. 0.23$
$D. NF_3$	$IV. 1.47$

$A$ covalent molecule $XY$ is found to have a dipole moment of $1.5 \times 10^{-29} \ C \cdot m$ and a bond length of $150 \ pm$. The percent ionic character of the bond will be (in $\%$)

Similar Questions

Match the following:
List-$I$ (Molecule) List-$II$ (Dipole moment in $D$)
$A. HCl$ $I. 1.07$
$B. NH_3$ $II. 1.85$
$C. H_2O$ $III. 0.23$
$D. NF_3$ $IV. 1.47$
The correct answer is

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