The percentage ionic character for C-F bond (electronegativity for C is 2.5; for F is 4.0) in CF4(subscript)is

Question

a. 9%

b.15%

c.38%

d.62%

Answer 1

http://www.chemistry.mcmaster.ca/esam/Ch...

% = [1 - e ^-(0.25)(xa -xb)] times 100

% = [1 - e ^-(0.25)(4.0 - 2.5)] times 100

% = [1 - e ^-(0.25)(1.5)] times 100

% = [1 - e ^-(0.375)] times 100

% = [1 - 0.69]times 100

% = 31%

closest answer: c.38%

=================

&

http://web.mit.edu/3.091/www/Fall07/home...

states thaat for C-F it is roughly 40%