Question

Given, \[{{\text{ }\!\![\!\!\text{ Cr(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}\] complex typically absorbs at around \[\text{574nm}\]. It is allowed to react with ammonia to form a new complex ${{\text{ }\!\![\!\!\text{ Cr(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$that should have absorption at:
(A)$\text{800nm}$
(B)\[\text{580nm}\]
(C)$\text{620nm}$
(D)$\text{320nm}$

Answer 1

Hint: When d-electron of a complex absorbs energy from a visible region they will get excited and when electrons fall into a lower energy level it will emit light whose wavelength is complementary to the absorbed wavelength.
\[\vartriangle {{\text{E}}_{\text{absorbed}}}=\dfrac{\text{he}}{{{\text{ }\!\!\lambda\!\!\text{ }}_{\text{absorbed}}}}\]
Compound ${{\text{ }\!\![\!\!\text{ Ni(CN}{{\text{)}}_{\text{4}}}\text{ }\!\!]\!\!\text{ }}^{\text{2-}}}$is colourless due to high splitting energy of the compound.

Complete answer:
In a coordination compound colour emitted by complexes depends on the nature of the ligands.
Wavelength of absorbed light is inversely proportional to the energy of the absorbed photon. This colour emitted by complex compounds depends on the magnitude of the splitting energy.
High spin complex or the complex which have weak field ligand \[{{\text{ }\!\![\!\!\text{ Cr(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3-}}}\]absorb photon of lower energy corresponding to high wavelength (\[\text{574nm}\]) for the excitation of electron. In low spin complex ${{\text{ }\!\![\!\!\text{ Cr(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$have a strong field ligand and have large splitting energy. As a result this compound will absorb high energy photons for d-d transition.
Since new complex ${{\text{ }\!\![\!\!\text{ Cr(N}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{6}}}\text{ }\!\!]\!\!\text{ }}^{\text{3+}}}$will absorb high energy photons, so its absorption wavelength will be smaller than \[\text{574nm}\].

So, option (D) will be the correct option.

Additional information
When a strong field ligand approaches a metal ion which causes splitting greater than pairing energy. So it will not be able to put electrons into higher energy orbital so it will absorb high energy electrons for the excitation of electrons.
However a weak field ligand causes a small splitting of the d- orbital where splitting energy is less than pairing energy. It will be easier to put electrons into the higher energy orbital, so it will absorb low energy photons for d-d transition.

Note
Coordination compounds show colour property due to d-d transition.
${{\text{H}}_{\text{2}}}\text{O}$ Acts as a weak field or high spin ligand while $\text{N}{{\text{H}}_{\text{3}}}$acts as strong field or low spin ligand.