Question

${[Ti{({H_2}O)_6}]^{3 + }}$ absorbs light of wavelength $498nm$ during d-d transition. The octahedral splitting energy for the above complex is $\_\_\_\_\_ \times {10^{ - 19}}J$.
(Round off to the nearest integer)
Given: $h = 6.626 \times {10^{ - 34}}Js$ and $c = 3 \times {10^8}m{s^{ - 1}}$

Answer 1

Hint: The d orbitals of central metal ions in octahedral complexes splits into two different sets consisting of different energies. This separation in energy is termed as splitting energy and the electrons tend to absorb light when the electrons get excited performing d-d transitions and the related energy can be calculated on applying values in the formula.

Formula used: Octahedral splitting energy ${\Delta _o} = \frac{{hc}}{{{\lambda _{abs}}}}$

Complete Step by Step Solution:
The transition metal absorbs light of a certain wavelength in both solid state and aqueous solution because of the excitation of the electrons from lower energy degenerated d-orbitals to the orbitals with higher energy.

The d orbitals of the transition metal are of different energy levels and split into two sets to form degenerate d orbitals. In octahedral complexes, ${t_2}g$ orbitals consisting of ${d_{xy}},{d_{yz}},{d_{xz}}$ have comparatively lower energy as compared to ${e_g}$ orbitals consisting of ${d_{{x^2} - {y^2}}},{d_{{z^2}}}$. When white light falls on these compounds, some wavelength is absorbed and the electrons are excited from one set of lower energy orbitals to higher energy orbitals. When the electron gets back to the ground state, an energy is released known as octahedral splitting energy and the process is known as d-d transition.

As per question, the wavelength of absorption ${\lambda _{abs}} = 498nm$. Substituting values in the given formula of energy:
${\Delta _o} = \frac{{hc}}{{{\lambda _{abs}}}}$
$ \Rightarrow {\Delta _o} = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{498 \times {{10}^{ - 9}}}}$
$ \Rightarrow {\Delta _o} = 0.0399 \times {10^{ - 17}}J$
$ \Rightarrow {\Delta _o} = 3.99 \times {10^{ - 19}}J$
On rounding off, the value of octahedral splitting energy will be ${\Delta _o} = 4 \times {10^{ - 19}}J$.

Note: It is important to note that the octahedral splitting energy depends on the type of ligand bonded to the central metal transition metal. If the ligands bonded are strong ligands, the splitting energy will be higher and the relative frequency will be lower whereas if the ligands bonded are weak field ligands, then the splitting energy will be lower.