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