The percentage ionic character of the HBr mol | vedclass

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(B) The theoretical dipole moment $(\mu_{calc})$ for $100 \%$ ionic character is calculated using the formula $\mu = q 	imes d$. Given $q = 4.8 	ime

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$7$

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(B) The theoretical dipole moment $(\mu_{calc})$ for $100 \%$ ionic character is calculated using the formula $\mu = q 	imes d$. Given $q = 4.8 	imes 10^{-10} \ esu$ (charge of an electron) and $d = 187.5 \ pm = 187.5 	imes 10^{-10} \ cm$. $\mu_{calc} = 4.8 	imes 10^{-10} \ esu 	imes 187.5 	imes 10^{-10} \ cm = 900 	imes 10^{-20} \ esu \ cm = 9 	imes 10^{-18} \ esu \ cm = 9 \ D$. Given observed dipole moment $\mu_{obs} = 0.63 \ D$. Percentage ionic character $= (\mu_{obs} / \mu_{calc}) 	imes 100$. Percentage ionic character $= (0.63 / 9) 	imes 100 = 7 \%$.

List-$I$	List-$II$
$(1)$ ${\rm{HBr}}$	$(A)$ $0.23$
$(2)$ ${{\rm{H}}_2}{\rm{S}}$	$(B)$ $1.04$
$(3)$ ${\rm{N}}{{\rm{F}}_3}$	$(C)$ $1.07$
$(4)$ ${\rm{CC}}{{\rm{l}}_4}$	$(D)$ $0.79$
$(5)$ ${\rm{CHC}}{{\rm{l}}_3}$	$(E)$ $0.00$
	$(F)$ $0.95$

The percentage ionic character of the $HBr$ molecule, if the dipole moment is $0.63 \ D$ and $HBr$ bond length is $187.5 \ pm$, is:

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