The $$H-Cl$$ bond length is $$1.275$$ Ao (charge$$ e $$-$$ $$4.8$$ × $$10-10$$ $$esu). lf μ $$=$$ $$1.02$$ D, then HCI is

Correct option is 283% covalent

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The H-Cl bond length is 1.275 $\overset{o}{A}$ (charge e - 4.8 × 10^-10 esu). lf $μ$ = 1.02 D, then HCI is

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100% ionic

83% covalent

50% covalent

40% ionic

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Correct option is B

The percent ionic nature= Observed μ Theoretical μ×100=1.021.275×4.8×100= l7 % ionic or = 83% covalent.

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The H-Cl bond length is 1.275 Ao (charge e - 4.8 × 10-10 esu). lf μ = 1.02 D, then HCI is

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Ready to Test Your Skills?

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The H-Cl bond length is 1.275 $\overset{o}{A}$ (charge e - 4.8 × 10^-10 esu). lf $μ$ = 1.02 D, then HCI is