Question

According to CFT, wave number obtained is ${{20,000c}}{{{m}}^{{{ - 1}}}}$for transition ${{{e}}_{{g}}} \leftarrow {{{t}}_{{{2g}}}}$. The CFSE (in ${{KJ/mole}}$) for complex ${{{[Ti(}}{{{H}}_{{2}}}{{O}}{{{)}}_{{6}}}{{]}}^{{{3 + }}}}$ is;
1. ${{ - 243 KJ/mol}}$
2. ${{ - 92}}{{.7 KJ/mol}}$
3. ${{ - 194 KJ/mol}}$
4. ${{ - 143 KJ/mol}}$

Answer 1

Wave number is the number of waves in unit length. In CFT, the five degenerate orbitals are split into two sets of ${{d}}$- orbitals. And these two sets are indicated by the $t_{2g}$ and $e_g$ respectively which have different energies.

Complete step by step answer:
The given complex – ${{{[Ti(}}{{{H}}_{{2}}}{{O}}{{{)}}_{{6}}}{{]}}^{{{3 + }}}}$
The central metal atom - ${{Ti}}$
Electronic configuration - ${{[Ar] 3}}{{{d}}^{{2}}}{{4}}{{{s}}^{{2}}}$
In the given complex ${{Ti}}$ is present in the ${{ + 3}}$state.
The electronic configuration of ${{T}}{{{i}}^{{{ + 3}}}}$will be ${{[Ar]3}}{{{d}}^{{1}}}$.
The ${{d}}$- orbital has only one electron.
This single electron will occupy one of the more stable (lower energy) ${{{t}}_{{{2g}}}}$orbital and the complex is stabilized by an amount of ${{0}}{{.4}}{{{\Delta }}_{{o}}}$.
So, the energy will be calculated by the following formula.
${{\Delta E = hc}}\overline {{\nu }}$……………………………….(1)

{{C = \text{Velocity of light} = 3}}{{.0 \times 1}}{{{0}}^{{8}}}{{m/s}} \\\ {{v = \text{Wave number } = 20,300c}}{{{m}}^{{{ - 1}}}} \\\ $$ Substitute the values in the equation (1) $${{\Delta E = 6}}{{.63 \times 1}}{{{0}}^{ - 34}}{{Js \times 3}}{{.0 \times 1}}{{{0}}^{{8}}}{{ m/s \times 20,300c}}{{{m}}^{{{ - 1}}}} \\\ {{ = 4}}{{.037 \times 1}}{{{0}}^{{{ - 19}}}}{{J}} \\\ $$ This energy is just for one ion. When it is calculating for one mole then, it will be One mole contains $${{6}}{{.022 }} \times {10^{23}}$$ atoms. $${{\Delta E = 4}}{{.037 \times 1}}{{{0}}^{{{ - 19}}}}{{J }} \times {{ 6}}{{.022 }} \times {10^{23}}$$ $${{ = 243067 J/mol}}$$ Let’s convert the “joules” into “Kilojoules”. $${{1 Kj = 1000 j}}$$ $${{ = 243 J/mol}}$$ Hence, the energy is $${{243 J/mol}}$$ As CFSE = $${{0}}{{.4}}{{{\Delta }}_{{o}}}$$ Then, $$ {{CFSE = 0}}{{.4 \times 243 KJ/mol}} \\\ {{CFSE = 97}}{{.1 KJ/mol}} \\\ $$ **Hence, the correct option is $${{2}}$$.** **Additional information:** The energy difference between two sets of $${{d}}$$ orbitals is called the crystal field splitting energy $${{{\Delta }}_{{o}}}$$. Here, $${{o}}$$ stands for octahedral. The magnitude of splitting depends on the charge on the metal ion, the position of metal in periodic table and nature of ligands. **Note:** It is important to note that the splitting of the d orbitals in a crystal field does not change the total energy of the five $${{d}}$$ orbitals: the two $${{{e}}_{{g}}}$$ orbitals increase in energy by $${{0}}{{.6}}{{{\Delta }}_{{o}}}$$ whereas the three $${{{t}}_{{{2g}}}}$$ orbitals decrease in energy by $${{0}}{{.4}}{{{\Delta }}_{{o}}}$$. Thus, the total change in energy is $$2(0.6{\Delta _o}) + 3( - 0.4{\Delta _o}) = 0$$