The H - Cl bond length is 1.275 mathringA (charge e=4.8 × 10-10 esu). If μ=1.02 D , then HCl is

Question

The H - Cl bond length is 1.275 mathringA (charge e=4.8 × 10-10 esu). If μ=1.02 D , then HCl is (A) 100 % ionic (B) 83 % covalent (C) 50 % covalent (D) 40 % ionic

Tardigrade Admin · Accepted Answer

The percent ionic nature =( text Observed μ/ text Theoretical μ) × 100=(1.02/1.275 × 4.8) × 100 =17 % ionic or =83 % covalent.

Q. The $H - Cl$ bond length is $1.275 \overset{˚}{A}$ (charge $e = 4.8 \times 1 0^{- 10}$ esu). If $μ = 1.02 D,$ then $H Cl$ is

100 % ionic

83 % covalent

50 % covalent

40 % ionic

The percent ionic nature

$= \frac{Observed μ}{Theoretical μ} \times 100 = \frac{1.02}{1.275 \times 4.8} \times 100$

$= 17%$ ionic or $= 83%$ covalent.

Q. The H−Cl bond length is 1.275A˚ (charge e=4.8×10−10 esu). If μ=1.02D, then HCl is