Question

The ${\left[Ti\right(NCS{)}_6]}^{3-}$ ion exhibits a single absorption band at 544 nm. What will be the crystal field splitting energy $\left({\text{in kJmol}}^{-1}\right)$ of the complex? $(h=6.626\times {10}^{-34}Js;c=3.0\times {10}^{8}m/s;N_A=6.02\times {10}^{23}\text{ions/mole})$

• A. 240
• B. 220
• C. 270
• D. 250

Answer 1

The correct option is B 220

Given:
$\lambda =544nm,h=6.626\times {10}^{-34}Js;c=3.0\times {10}^{8}m/s;N_A=6.02\times {10}^{23}\text{ions/mole}$
We know that, $E=hc/\lambda$
Upon substituting values, we get: $E=\frac{6.626\times {10}^{-34}Js\times 3.0\times {10}^{8}m/s}{544\times {10}^{-9}m}$
On simplifying, we get: $E=0.036\times {10}^{-17}J/atom$
To convert this to one mole, we can write $E=0.036\times {10}^{-17}\times 6.02\times {10}^{23}J/mole$
$E=0.21997\times {10}^{6}J/mole$
$E=219.97kJ/mole$
$E=220kJ/mole$