Question #10d09

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Answer 1 · 2017-08-04T05:38:15

77.4%

The percentage of partial ionic character depends upon the difference between the electronegativity of two atoms.

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference Δχ:

percent ionic character #= 1−e^((−Δχ/2)^2)xx100#

A commonly used simplified version is:
% ionic character #= 1-exp^(-(0.25)(X_A - X_B)^2)xx100 #
http://sciencing.com/figure-ionic-percentage-once-difference-8578423.html

Electronegativity of Ca is 1.0
Electronegativity of O is 3.44
https://www.webelements.com/calcium/electronegativity.html

#(X_A – X_B)^2 = 2.44^2 = 5.95#

% ionic character #= 1-exp^(-(0.25)(5.95))xx100 #
% ionic character #= 1-exp^(-1.488)xx100 # = #0.774 xx 100 = 77.4%# ionic character.