Question

Calculate the %age ionic character in HF molecule. Electronegative of H and F are 2.1 and 4.0 respectively

Answer 1

Answer:

i

Explanation:

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Answer 2

Answer:

43%

Explanation:

Percentage of ionic character

= [16 |(Ea - Eb)|+ 3.5 |(Ea - Eb)|^2 ]

Where Ea = Electronegativity of atom A

Eb = Electronegativity of atom B

| (Ea - Eb) | = The subtraction will give a positive value always since modulus (|| = modulus) sign is used !

Example :

Calculate the percentage ionic character of HF bond where i) Electronegativity of H = 2.1 ii) Electronegativity of F = 4.0

Solution:

% = 16 |(H - F)| + 3.5 |(H - F)|^2

= 16 |(2.1–4.0)| + 3.5 |(2.1–4.0)|^2

= 16 * 1.9 + 3.5 (1.9)^2

= 16 * 1.9 + 3.5 * 3.61

= 30.4 + 12.635

= 43.035 %