Question
Question: Bond length of A - A bond is 124 pm and bond length of B - B bond is 174 pm. The bond length (in pm)...
Bond length of A - A bond is 124 pm and bond length of B - B bond is 174 pm. The bond length (in pm) of A - B molecule if percent of ionic character of A - B bond is 19.5% is ________.

148.9
Solution
The problem asks us to calculate the bond length of an A-B molecule, given the bond lengths of A-A and B-B, and the percentage of ionic character of the A-B bond.
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Calculate the covalent radii of A and B: The covalent radius of an atom is half of its bond length in a homonuclear diatomic molecule. Covalent radius of A, rA=2Bond length of A-A=2124 pm=62 pm Covalent radius of B, rB=2Bond length of B-B=2174 pm=87 pm
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Calculate the theoretical bond length of a purely covalent A-B bond: For a purely covalent A-B bond, the bond length would be the sum of the covalent radii. dcovalent(A−B)=rA+rB=62 pm+87 pm=149 pm
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Determine the electronegativity difference (ΔX) using the percentage ionic character: The percentage ionic character (P) is related to the electronegativity difference (ΔX=∣XA−XB∣) by Pauling's formula: P=(1−e−0.25(ΔX)2)×100% Given P=19.5%=0.195 (as a fraction). 0.195=1−e−0.25(ΔX)2 e−0.25(ΔX)2=1−0.195=0.805 Take the natural logarithm of both sides: −0.25(ΔX)2=ln(0.805) −0.25(ΔX)2≈−0.2169 (ΔX)2=−0.25−0.2169=0.8676 ΔX=0.8676≈0.9314
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Calculate the actual bond length of A-B using the Schomaker-Stevenson equation: The Schomaker-Stevenson equation accounts for the shortening of bond length due to electronegativity difference (ionic character): dA−B=rA+rB−0.09∣ΔX∣ Substitute the values: dA−B=149 pm−0.09×0.9314 dA−B=149 pm−0.083826 pm dA−B=148.916174 pm
Rounding to a reasonable number of significant figures (e.g., one decimal place as inputs are integers), the bond length is approximately 148.9 pm.