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\[=1{\AA}\] \[=1\times {{10}^{-10}}m\] \[%\] of ionic character in\[HBr\] \[=\frac{\text{dipole}\,\,\text{moment}\,\,\text{of}\,\,\text{HBr }\!\!\times\!\!\text{ 100}}{\text{inter}\,\,\text{spacing}\,\,\text{distance }\!\!\times\!\!\text{ q}}\] \[=\frac{1.6\times {{10}^{-30}}}{1.6\times {{10}^{-19}}\times {{10}^{-10}}}\times 100\] \[={{10}^{-30}}\times {{10}^{29}}\times 100\] \[={{10}^{-1}}\times 1000\] \[=0.1\times 100\] \[=10%\]
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