A) 54%
B) 46%
C) 23%
D) 72%
Correct Answer: B
Solution :
[b] The percentage ionic character is related to electro negativity difference between the two atoms. It is given by Hannay and Smith relationship % age ionic character\[=16\,({{X}_{A}}-{{X}_{B}})+3.5\,{{({{X}_{A}}-{{X}_{B}})}^{2}}\] Here \[{{X}_{A}}\] and \[{{X}_{B}}\] are electro negativities of the two atoms A and B respectively. \[\therefore \] % ionic character \[=16(2)+3.5\,{{(2)}^{2}}=46%\]You need to login to perform this action.
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