Question
Calculate the percentage ionic character of bond in CsF. If electronegativity of Cs and F is 0.7 and 4.0 respectively.
✓
Answer
$\begin{array}{l} \text { % ionic character of bond in CsF } \\
=16 \Delta+3.5 \Delta^2 \\ \Delta=\text { Difference in electronegativity } \\
=4.0-0.7=3.3 \\ \text { Hence % ionic of bond }=(16 \times 3.3)+\left[3.5(3.3)^2\right] \\
=52.8+3.5 \times 10.89 \\ \text { Character of bond }=52.8+38.1 \\
=90.9 \%\end{array}$
=16 \Delta+3.5 \Delta^2 \\ \Delta=\text { Difference in electronegativity } \\
=4.0-0.7=3.3 \\ \text { Hence % ionic of bond }=(16 \times 3.3)+\left[3.5(3.3)^2\right] \\
=52.8+3.5 \times 10.89 \\ \text { Character of bond }=52.8+38.1 \\
=90.9 \%\end{array}$
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