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

Hint:
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)
\[ {{h = \text{planck's constant} = 6}}{{.63 \times 1}}{{{0}}^{{{ - 34}}}}{{ Js}} \\
  {{C = \text{Velocity of light} = 3}}{{.0 \times 1}}{{{0}}^{{8}}}{{m/s}} \\
  {{v = \text{Wave number } = 20,300c}}{{{m}}^{{{ - 1}}}} \\ \]
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
\[2(0.6{\Delta _o}) + 3( - 0.4{\Delta _o}) = 0\]