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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 ________.

Answer

148.9

Explanation

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.

  1. 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=Bond length of A-A2=124 pm2=62 pmr_A = \frac{\text{Bond length of A-A}}{2} = \frac{124 \text{ pm}}{2} = 62 \text{ pm} Covalent radius of B, rB=Bond length of B-B2=174 pm2=87 pmr_B = \frac{\text{Bond length of B-B}}{2} = \frac{174 \text{ pm}}{2} = 87 \text{ pm}

  2. 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(AB)=rA+rB=62 pm+87 pm=149 pmd_{\text{covalent}}(A-B) = r_A + r_B = 62 \text{ pm} + 87 \text{ pm} = 149 \text{ pm}

  3. Determine the electronegativity difference (ΔX\Delta X) using the percentage ionic character: The percentage ionic character (PP) is related to the electronegativity difference (ΔX=XAXB\Delta X = |X_A - X_B|) by Pauling's formula: P=(1e0.25(ΔX)2)×100%P = (1 - e^{-0.25(\Delta X)^2}) \times 100\% Given P=19.5%=0.195P = 19.5\% = 0.195 (as a fraction). 0.195=1e0.25(ΔX)20.195 = 1 - e^{-0.25(\Delta X)^2} e0.25(ΔX)2=10.195=0.805e^{-0.25(\Delta X)^2} = 1 - 0.195 = 0.805 Take the natural logarithm of both sides: 0.25(ΔX)2=ln(0.805)-0.25(\Delta X)^2 = \ln(0.805) 0.25(ΔX)20.2169-0.25(\Delta X)^2 \approx -0.2169 (ΔX)2=0.21690.25=0.8676(\Delta X)^2 = \frac{-0.2169}{-0.25} = 0.8676 ΔX=0.86760.9314\Delta X = \sqrt{0.8676} \approx 0.9314

  4. 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): dAB=rA+rB0.09ΔXd_{A-B} = r_A + r_B - 0.09 |\Delta X| Substitute the values: dAB=149 pm0.09×0.9314d_{A-B} = 149 \text{ pm} - 0.09 \times 0.9314 dAB=149 pm0.083826 pmd_{A-B} = 149 \text{ pm} - 0.083826 \text{ pm} dAB=148.916174 pmd_{A-B} = 148.916174 \text{ 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.