Question

Identify the compound with the smallest percent ionic character. identify the compound with the smallest percent ionic character. hf lif ibr hcl

Answer 1

Answer: IBr has the smallest percentage ionic character amongst the given compounds.

Explanation: Percentage ionic character is calculated by using the formula:

$\%\text{ Ionic character}=[16(\Delta E.N.)+3.5(\Delta E.N.)^2]\%$    ....(1)

where, $\Delta E.N.$ = difference in the electronegativies of the element forming a compound.

Electronegativity of H = 2.1

Electronegativity of F = 3.98

Electronegativity of Li = 0.98

Electronegativity of I = 2.66

Electronegativity of Br = 2.96

Electronegativity of Cl = 3.16

• For HF

$\Delta E.N.=3.98-2.1=1.88$

Putting this in equation 1, we get:

$\%\text{ Ionic character of HF}=[16(0.188)+3.5(1.88)^2]=42.45\%$

• For LiF

$\Delta E.N.=3.98-0.98=3.00$

Putting this in equation 1, we get:

$\%\text{ Ionic character of HF}=[16(3)+3.5(3)^2]=79.5\%$

• For IBr

$\Delta E.N.=2.96-2.66=0.30$

Putting this in equation 1, we get:

$\%\text{ Ionic character of HF}=[16(0.30)+3.5(0.30)^2]=5.115\%$

• For HCl

$\Delta E.N.=3.16-2.1=1.06$

Putting this in equation 1, we get:

$\%\text{ Ionic character of HF}=[16(1.06)+3.5(1.06)^2]=20.89\%$

From the above calculations, we see that IBr has the smallest percent ionic character.