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Chemistry Question on Bond Parameters

The dipole moment of HBr is 1.6×10−301.6 \times 10^{-30} C-m and interatomic spacing is 1A˚1 \mathring{A}. The per cent ionic character of HBr is

A

7

B

10

C

15

D

27

Answer

10

Explanation

Solution

Charge of electron =1.6×10−19C= 1.6 \times 10^{-19}C Dipole moment of HBr =1.6×1.6×−30= 1.6 \times 1.6\times ^{-30} Inter-atomic spacing =1A˚= 1 \mathring{A}                =1×10−10m\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = 1 \times 10^{-10} m % ionic character in HBr                  =dipole moment of HBr ×100inter spacing distance × q\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{dipole\, moment\, of\, HBr\, \times100}{inter\, spacing\, distance\, \times\, q}                  =1.6×10−301.6×10−19× 10−10×100\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\frac{1.6 \times10^{-30}}{1.6 \times 10^{-19} \times\, 10^{-10}}\times100                  =10−30×1029×100\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10^{-30}\times10^{29} \times100                  =0.1×100\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =0.1\times 100                  =10%\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =10\%