Question

The electronegativity of f and h are 4.0 and 2.1 respectively. the percentage ionic character in h and f bond is

Answer 1

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Answer 2

Answer:

The correct answer to this question is $43.035$

Explanation:

Given - The electronegativity of f and h are 4.0 and 2.1

To Find - Write the percentage ionic character in the h and f bond.

Electronegativity is a chemical property that describes an atom's or functional group's capacity to attract electrons to itself. An atom's electronegativity is influenced by its atomic number as well as the distance between its valence electrons and the charged nuclei.

The percentage of ionic character -

= $16(XA - XB ) + 3.5 (XA - XB )^{2} \\16(4 -2.1) + 3.5 (4 - 2.1 )^{2}$

$16(1.9) + 3.5 ( 1.9) ^{2} \\30.4 + 12.635\\43.035$

$43.035$ is the percentage ionic character in the h and f bond.

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