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The electronegativity of hydrogen is $2.1$ and that of chlorine is $3.2$. The percent ionic character of the bond in $HCl$ molecule is approximatelyA) $50\%$ B) $30\%$ C)$70\%$ D)$20\%$

Last updated date: 02nd Aug 2024
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Hint: To answer this question we need to have knowledge about the formula to calculate the % ionic character will be necessary. $HCl$ Is a polar molecule with a big electronegativity difference.

Formula used: $16({x_B} - {x_A}) + 3.5{({x_B} - {x_A})^2}$

Electronegativity is a tendency of an atom to attract a shared pair of electrons towards itself. Halogen atoms have the highest electronegativity as they are just before the inert gas family. So they always try to get an electron to stabilise themselves.

As chlorine is a halogen atom it has higher electronegativity value, hence it has a tendency to pull shared pairs of electrons towards itself.

Given: electronegativity of $H = 2.1;Cl = 3.2$
The bond of $HCl$is polar due to the high electronegativity of chlorine atoms. Thus $HCl$ is also a very strong acid as it can donate ${H^ + }$ ions very easily in a solution.By using the formula, $16({x_B} - {x_A}) + 3.5{({x_B} - {x_A})^2}$ where ${x_A}$ and ${x_B}$ are the electronegativity values of hydrogen and chlorine atom and ${x_B} - {x_A}$ is the electronegativity difference between both the atoms, the % ionic character in bond can be calculated.

% ionic character in $HCl = 16({x_B} - {x_A}) + 3.5{({x_B} - {x_A})^2}$

By substituting the values ${x_A} = 2.1,{x_B} = 3.2$ in the above equation we get,
$\Rightarrow 16(3.2 - 2.1) + 3.5{(3.2 - 2.1)^2} \\ \Rightarrow 16 \times 1.1 + 3.5 \times {(1.1)^2} \\ \Rightarrow 17.6 + 3.5 \times 1.21 \\ \Rightarrow 17.6 + 4.235 \\ \Rightarrow 21.835\% \\$
Hence the percent ionic character of the bond in $HCl$ molecule is $21.835\%$ which is approximately equal to$20\%$.

Additional information: The electronegativity of halogen atoms are very high , as they are in $Group17$ and wanted to have stable , inert gas like configuration.

Hence the correct answer is option ‘D’.

Note: The substitution of values must be correct. Squaring the next term is important, as it can be easily skipped by mistake during the numerical process. If the % ionic character is $20\%$ then % covalent character in the bond will be $80\%$.