Verified Solution • 2025–26 Session moderate

QJEEM23Ch1935: JEE Main 2023 Chemistry - Class 12 JEEM23Ch1935 Solution: If the CFSE of is , this complex will absorb maximum at…

If the CFSE of [Ti(H2O)6]3+\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+} is −96.0 kJ/mol-96.0 \mathrm{~kJ} / \mathrm{mol} , this complex will absorb maximum at wavelength ___________ nm\mathrm{nm} . (nearest integer)

Assume Planck's constant (h) =6.4×10−34Js=6.4 \times 10^{-34} \mathrm{Js} , Speed of light (c)=3.0×108 m/s(\mathrm{c})=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s} and Avogadro's

Constant (NA)=6×1023/mol\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} / \mathrm{mol}

Step-by-Step Verified Solution

amp;[Ti(H2O)6]3+,amp;CFSE=−0.4Δ0=−96.0 kJ mol−1amp;∴Δ0=−96.0−0.4amp;∴Δ0=240 kJ mol−1\begin{aligned}& {\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},} \\\\& \mathrm{CFSE}=-0.4 \Delta_0 =-96.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\& \therefore \Delta_0=\frac{-96.0}{-0.4} \\\\& \therefore \Delta_0=240 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\\end{aligned}

hcλamp;=96×1030.4×6×1023λamp;=0.4×6×1023×6.4×10−34×3×10896×103amp;=0.48×10−6 mamp;=480×10−9 mamp;=480 nm\begin{aligned}\frac{\mathrm{hc}}{\lambda} & =\frac{96 \times 10^3}{0.4 \times 6 \times 10^{23}} \\\\\lambda & =\frac{0.4 \times 6 \times 10^{23} \times 6.4 \times 10^{-34} \times 3 \times 10^8}{96 \times 10^3} \\\\& =0.48 \times 10^{-6} \mathrm{~m} \\\\& =480 \times 10^{-9} \mathrm{~m} \\\\& =480 \mathrm{~nm}\end{aligned}

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