Colour of Transition Elements | Transition Metal Properties

Transition metal ions form colours in an aqueous solution; this is due to the adsorption of visible light radiation, which promotes an electron from one d-orbital to another. The ions of d-orbital transition elements absorb a specific wavelength of radiation and reflect the rest, giving the solution colour. Transition metal complexes have a wide range of hues in different solvents. The ligand determines the hue of the complex. In water, Fe2+ is pale green, but in a concentrated hydroxide base solution, carbonate solution, or ammonia, it forms a dark green precipitate. In water Co2+ forms a pink solution, ammonia, and carbonate solution, it forms a blue-green precipitate, a straw-coloured solution, and a pink precipitate.

Coloured complexes are also formed by elements in a lanthanide class. Lanthanides are sometimes known as transition metals or simply as transition metals subclass. The colourful complexes, and on the other hand, are caused by 4f electron transition. When transitioning to a transition metal complex, the hues of lanthanide complexes are less affected by the nature of their ligands. A coloured solution is produced when visible light of a certain energy level is not absorbed. When transition metal compounds are dissolved in water, they take on a wide range of vibrant colours.

Colour Of Transition Elements

When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. In aqueous solutions where water molecules are the ligands, the colours of the ions are observed.

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Some Solved Examples

Q.1 Which one of the following when dissolved in water gives the coloured solution in nitrogen atmosphere?

(1) Cu₂Cl₂

(2) ZnCl₂

(3) CuCl₂

(4)AgCl

Solution:

As we learned -

When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand.

The given metal cations are

$\mathrm{Cu}^{+}, \mathrm{Zn}^{2+}, \mathrm{Cu}^{2+}$ and $\mathrm{Ag}^{+}$

Out of these, only $\mathrm{Cu}^{2+}$ contains unpaired elections and can show colour.

The other ions $\left(\mathrm{Cu}^{+}, \mathrm{Ag}^{+}\right.$and $\left.\mathrm{Zn}^{2+}\right)$ have $\mathrm{d}^{10}$ configuration and do not show color.

Hence, the answer is the option (3).

Q.2 The set having ions which are coloured and paramagnetic both is -

(1) $\mathrm{Cu}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Sc}^{+}$

(2) $\mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}, \mathrm{Mn}^{4+}$

(3) $\mathrm{Sc}^{3+}, \mathrm{V}^{5+}, \mathrm{Ti}^{4+}$

(4) $\mathrm{Ni}^{2+}, \mathrm{Mn}^{7+}, \mathrm{Hg}^{2+}$

Solution:

As we learned -

The set of ions which are coloured and paramagnetic is

$\mathrm{Cu}^{2+}\left(3 \mathrm{~d}^9\right), \mathrm{Cr}^{3+}\left(3 \mathrm{~d}^3\right)$ and $\mathrm{Sc}^{+}\left(4 \mathrm{~s}^1 3 \mathrm{~d}^1\right)$

Species having d⁰or d¹⁰ configuration are diamagnetic such as

$\mathrm{Zn}^{2+}\left(3 \mathrm{~d}^{10}\right), \mathrm{Sc}^{3+}\left(3 \mathrm{~d}^0\right), \mathrm{Ti}^{4+}\left(3 \mathrm{~d}^0\right), \mathrm{V}^{5+}\left(3 \mathrm{~d}^0\right)$ and $\mathrm{Mn}^{+7}\left(3 \mathrm{~d}^0\right)$

Hence, the correct answer is Option (1)

Q.3 When $\mathrm{XO}_2$ is fused with an alkali metal hydroxide in the presence of an oxidizing agent such as $\mathrm{KNO}_3$ ; a dark green product is formed which disproportionates in an acidic solution to afford a dark purple solution. X is :

1)Ti

2)V

3)Cr

4) (correct)Mn

Solution

As we have learnt,

The given reaction sequence is a method of preparation of $\mathrm{KMnO}_4$

The reactions mentioned are given below

$\mathrm{MnO}_2+\mathrm{KOH} \rightarrow \underset{\text { dark green }}{\mathrm{K}_2 \mathrm{MnO}_4} \xrightarrow[\text { solution }]{\text { Acidic }} \underset{\text { dark purple }}{\mathrm{KMnO}_4}$

Hence, the answer is an option (4).

Conclusion

Transition metals are elements with half-filled d-orbitals. The periodic table places this group of elements near the centre (between s-block and p-block elements). These elements have a proclivity for various oxidation states, the formation of complex compounds, toughness, and high density, and lanthanoid contraction can be seen in them. Colourful ions, complexes, and compounds are formed by transition metals. Light wavelengths that have not been absorbed flow through a complex. Transition metal metal complexes have a wide range of hues in different solvents. Transition metal ions form colours in an aqueous solution this is due to the absorption of visible light. A colourless solution is formed by a transition metal ion with zero or ten electrons.