Coloured compound by d-d transition and charge transfer.

Formation of the colour compound.

Transition metal ions and their compounds are usually coloured formation of coloured compounds can be explained by either of two ways.

1. d-d transition.
2. Charge transfer.

1. d-d transition.

In isolated state the transition metal ion have all the five d-orbitals in a degenerate state. However, in bonded state or in solution the d-orbital split into two sets of orbitals having a different energy.

When transition metal ions or their compound are exposed to light, electrons are excited from lower energy d-orbital to higher energy d-orbital.This is called d-d transition.d-d the transition is accompanied by absorption of light of specific frequency from the visible region of the spectrum. The unabsorbed light is transmitted I,e reflected which give rise colour to the substance. It should be noted that we see the complementary colour of the absorbed light.

Transition metal ions having do (eg; Sc⁺³, Ti⁺⁴ etc) and d¹⁰ electronic configuration are typically colourless (d¹⁰=Zn⁺⁺, Cu⁺⁺ etc) . This is because such metal ions cannot exhibit d-transition.However, there are some transition metal ions having d⁰ and d¹⁰ electronic configuration and are coloured.Coloured of such metal ions can be explained on the basis of charge transfer.

2. Charge transfer.

MnO^4-(Mn=3d⁰), Cr₂O₇^--(Cr⁺⁶=3d^o), Cds(Cd⁺²=4d^1o) are deep coloured compound. Colour of such ions compound can be explained on the basis of charge transfer.

In order to exhibit coloured by charge transfer.One atom should act as strong oxidising agent and the other atom should act as a strong reducing agent.For example; In [MnO₄]^- Mn⁺⁷ act as oxidising agent and O^-2 act as reducing agent.Hence, charge transfer is possible between Mn⁺⁷ and-and O^-2.

When such substance are exposed to a light electron is transferred from one atom to another atom which is called charge (electron) transfer.Charge transfer is accompanied by absorption of light of specific frequency in the visible region.The unabsorbed light is transmitted which give rise coloured to the substance.

Explain why TiCl₄ is colourless but TiCl₃ is purple (violet).

In TiCl_^4, Ti⁺⁴ is in d^o state, there is no electron in d-orbital.Hence, it does not exhibit d-d transition and hence, it is colourless (white).

In TiCl₃, Ti⁺³ has d¹ electronic configuration.It has one electron in d-orbital and can exhibit d-d transition.d-dtransition is accompanied by absorption of light of specific frequency (yellow-green) in the visible region.The complementary colour of which is violet or purple.Therefore, TiCl₃ has purple violet.

Q. Transition metal compound are usually coloured but Ti(IV) compound is white.Explain with reason.

Transition metal usually forms a compound .It is due to incomplete filling of d-orbitals, where absorption of light could lead to d-d transition.Ti has valence shell configuration 3d²4s².So, Ti⁺⁴ has a d⁰ configuration and d-level is empty.Hence, a d-d electronic transition is not possible and Ti⁴⁺ compound be colourless or white.

1. Explain why TiO₂ is white while TiCl₃ is violet.

In TiO2, the oxidation state of Ti is ⁺⁴ and has a do configuration. The d-d electronic transition is not possible and hence TiO₂ should be colourless or white.But in TiCl₃, the oxidation state Ti is +3 and has 3d¹ configuration. Here, single- d electron transition occurs ie, a d-d transition is possible.The violet colour of TiCl₃ results by the absorption of light due to a promotion of one d electron from one level to higher energy level.

Cu²⁺ complexes are coloured while the Zn²⁺ are colourless.

Cu²⁺ has 3d₉ arrangement.In Cu²⁺ complexes, the d-d electronic transition is possible due to which it absorbs light and transmits certain light.But Zn²⁺ has 3d1o arrangement. Zn²⁺ does not exhibit coloured compound due to complete filling d-orbitals and hence, a d-d transition is not feasible. No light of any colour or wavelength is absorbed by Zn²⁺ complexes.

Q, Cupric salts are blue in colour while cuprous salts are colourless.

Outer electronic configuration of Cu²⁺ and Cu⁺ are

Cu²⁺=3d⁹, Cu⁺=3d^10., Cu⁺=3d^10.

There is a possibility od d-d transition in cupric (Cu²⁺) with the absorption of light from visible region.Hence, Cu²⁺ salt is blue in colour. But in with Cu⁺¹ salt, there is no possibility of d-transition due to complete filling of orbitals. Hence, no colour is shown by cuprous salts.

Explain why ? V₂O₅ is colourless but VO₂ is blue in colour.

V₂O₅ , the oxidation state of V is ⁺⁵ and has a 3d^o configuration. The d-d transition is not possible and hence V₂O₅ is colourless.Here, single -d electron transition occurs ie, a d-d transition is possible.The blue colour of VO₂ results by absorption of light blue to the promotion of one d-electron from low energy to high energy d-level.

Fe⁺³ is more paramagnetic than Fe⁺⁺ ion.

Here, the magnetic property of s substance depends on the number of unpaired electrons.In Fe⁺³ ion, the outermost configuration is 3d⁵. Thus there is five unpaired electron .In Fe⁺⁺ ion, the outermost configuration is 3d⁶.Thus there is four unpaired electron .Fe⁺³ has a number of a unpaired electron than Fe⁺⁺ ion. Hence, Fe⁺³ is more, paramagnetic than Fe⁺² ion.

Fe⁺³ ion =3d⁵

Fe⁺² ion= 3d^6.

Reference.

F.A.Cotton and Wilkinson G. Basic inorganic Chemistry. John,Wiley and Sons (Asia), 2007.

Lee., J.D. Concise Inorganic Chemistry. fifth edition. New Delhi: Oxford University Press., 2008.

Sharma, M.L and P.N Chaudhary. A textbook of B.S.C chemistry. Kathmandu Nepal: Ekta Books Thapathali Kathmandu, 2011.