Find separation energy of t 2 g and e g orbitals in a single [ Ti ( L ) 6 | TestHub

Passage / Comprehension

[Ti(L)₆]³⁺ absorbs 514 nm for a d-d transition (L = monodentate ligand). The 514 nm wavelength corresponds to an almost green coloration in the visible range.

( $h = 6.62 \times 1 0^{- 34} m^{2} k g / s$ ) and ( $c = 3 \times 1 0^{8} m / s$ ).

Find separation energy of $t_{2 g}$ and $e_{g}$ orbitals in a single $[Ti (L)_{6}]^{3 +}$ ion.

Options:(select one or more)

Answer:

A, C

Solution:

Energy of d–d transition is given by:

$Δ_{o} = \frac{h c}{λ}$

Substitute values: $h = 6.62 \times 1 0^{- 34}$ , $c = 3 \times 1 0^{8}$ , $λ = 514 \times 1 0^{- 9} m$ .

$Δ_{o} = \frac{6.62 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{514 \times 1 0 ^{- 9}} = 3.86 \times 1 0^{- 19} J$

Thus option A is correct.

Option C is also a correct form of the same expression. Option C is:

$\frac{h c}{514} \times 1 0^{9} J$

But wavelength must be in meters, so:

$514 nm = 514 \times 1 0^{- 9} m$

So the correct expression becomes:

$\frac{h c}{514 \times 1 0 ^{- 9}}$

Chemistry - Coordination Chemistry Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion