If the dipole moment of HI is 0.38 , D and th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The observed dipole moment $(\mu_{obs})$ is $0.38 \, D$. The bond length $(d)$ is $1.61 \, \mathring{A} = 1.61 	imes 10^{-10} \, m$. The theoreti

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The observed dipole moment $(\mu_{obs})$ is $0.38 \, D$. The bond length $(d)$ is $1.61 \, \mathring{A} = 1.61 	imes 10^{-10} \, m$. The theoretical dipole moment $(\mu_{theo})$ for $100 \%$ ionic character is calculated as $q 	imes d$,where $q = 1.602 	imes 10^{-19} \, C$. $\mu_{theo} = (1.602 	imes 10^{-19} \, C) 	imes (1.61 	imes 10^{-10} \, m) = 2.579 	imes 10^{-29} \, C \cdot m$. Converting to Debye $(1 \, D = 3.336 	imes 10^{-30} \, C \cdot m)$: $\mu_{theo} = \frac{2.579 	imes 10^{-29}}{3.336 	imes 10^{-30}} \approx 7.73 \, D$. The percentage ionic character is given by $\frac{\mu_{obs}}{\mu_{theo}} 	imes 100$. $	ext{Percentage ionic character} = \frac{0.38}{7.73} 	imes 100 \approx 4.91 \% \approx 5 \%$. Thus,the correct option is $A$.

If the dipole moment of $HI$ is $0.38 \, D$ and the bond length is $1.61 \, \mathring{A}$,what is the percentage ionic character of the $H^{\delta +} - I^{\delta -}$ covalent bond (in $\%$)?

Similar Questions

Number of compounds from the following with zero dipole moment is . . . . . .
$HF, H_2, H_2S, CO_2, NH_3, BF_3, CH_4, CHCl_3, SiF_4, H_2O, BeF_2$

Both $BF_3$ and $NH_3$ are covalent molecules. However,$BF_3$ is non-polar while $NH_3$ is polar. What is the reason for this?

The correct set of species with zero dipole moment is:
$(i) \, CO_2, \, (ii) \, COCl_2, \, (iii) \, CH_2Cl_2, \, (iv) \, BCl_3$

Arrange the following molecules from most to least polar ($I$,$II$,$III$,$IV$,$V$ respectively): $CH_4$,$CF_2Cl_2$,$CF_2H_2$,$CCl_4$,$CCl_2H_2$.

Explain why $O=C=O$ is non-polar while $R-O-R$ is polar?

Vedclass Test Series

Exam Paper Generator

Online Exam Module