The dipole moment of HCl is 1.03 D, if H—Cl bond distance is 1.26 Å, what is the percentage of ionic character in the H—Cl bond?Correct answer is between '16.7,17.4'. Can you explain this answer?

Question

The dipole moment of HCl is 1.03 D, if H&mdash;Cl bond distance is 1.26 &Aring;, what is the percentage of ionic character in the H&mdash;Cl bond?Correct answer is between '16.7,17.4'. Can you explain this answer?

Answer

The dipole moment of a molecule is given by the product of the magnitude of the charge separation (or the bond length) and the magnitude of the charge on each end of the bond. Therefore, the dipole moment of HCl is given by:

μ = q × d

where μ is the dipole moment, q is the magnitude of the charge separation, and d is the bond length.

Given that the dipole moment of HCl is 1.03 D, we can use this equation to solve for either the magnitude of the charge separation or the bond length. Let's assume that the bond length is known (we will use the value of 127 pm, which is the experimental bond length of HCl).

μ = q × d
1.03 D = q × 127 pm

To convert the bond length from picometers to meters, we need to divide by 10^12:

1.03 D = q × 1.27 × 10^-10 m

Solving for q, we get:

q = 1.03 D / (1.27 × 10^-10 m) = 8.11 × 10^-19 C

Therefore, the magnitude of the partial charge on the hydrogen atom in HCl is:

qH = q / 2 = 4.06 × 10^-19 C

The magnitude of the partial charge on the chlorine atom is the same but opposite in sign:

qCl = -4.06 × 10^-19 C

Note that the dipole moment points from the positive end (hydrogen) to the negative end (chlorine).

Answer

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