A) 54%
B) 46%
C) 23%
D) 72%
Correct Answer: B
Solution :
According to Hannay and Smith equation \[\therefore \]% ionic character \[=16({{x}_{A}}-{{x}_{B}})+3.5{{({{x}_{A}}-{{x}_{B}})}^{2}}\] where,\[{{x}_{A}}\]and\[{{x}_{B}}\]are the electronegativites of the atoms A and B respectively. \[\therefore \]% ionic character \[=16(2)+3.5{{(2)}^{2}}\] \[=32+14=46%\]You need to login to perform this action.
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