Question

The electronegativity of h and cl are 2.1 and 3.0 respectively.then hcl is ionic by what %

Answer 1

Answer:- Percent ionic character for HCl molecule is 17.2%.

Solution:- Using electron negativity values the percentage ionic character could be calculated as:

percent ionic character = $16(\Delta E)+3.5(\Delta E)^2$

where, $\Delta E$ is the electron negativity difference of two atoms.

electron negativity of H is 2.1 and for Cl it is 3.0.

So, $\Delta E$ = 3.0 - 2.1 = 0.9

Let's plug in this value in the formula:

percent ionic character = $16(0.9)+3.5(0.9)^2$

percent ionic character = 14.4 + 2.835 = 17.2

Hence, the percent ionic character for the HCl molecule is 17.2%.