A) 7
B) 10
C) 15
D) 27
Correct Answer: B
Solution :
Charge of electron \[=1.6\times {{10}^{-19}}C\] Dipole moment of \[HBr=1.6\times {{10}^{-30}}\] Inter-atomic spacing \[=1\,\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=1\times {{10}^{-10}}m\] % of ionic character in HBr \[=\frac{\text{dipole}\,\text{momentof}\,\text{HBr}\,\text{ }\!\!\times\!\!\text{ 100}}{\text{inter}\,\text{spacing}\,\text{distance}\,\text{ }\!\!\times\!\!\text{ }\,\text{q}}\,\] \[=\frac{1.6\times {{10}^{-30}}}{1.6\times {{10}^{-19}}\times {{10}^{-10}}}\times 100\] \[={{10}^{-30}}\times {{10}^{29}}\times 100\] \[={{10}^{-1}}\times 100\] \[=0.1\,\times \,100\] \[=10%\]
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