Answer the question on the basis of passage given below:

All transition elements display typical metallic properties like high melting point, boiling point, tensile strength, high malleability and ductility, luster, etc. Their metallic character increases down the group and elements of 4d and 5d series exhibit higher metallic characters. They show variable oxidation states and forms a number of coloured compounds due to the presence of unpaired electrons which shows d-d transitions and makes them paramagnetic.

Which out of the following coordination compounds are colourless in nature?

A. $[Cr(NH_3)_6]Cl_3$
B. $[Co(NH_3)_6]Br_3$
C. $K_3[Fe(CN)_6]$
D. $[MnCl_2(en)_2]^+$
E. $[Ni(CO)_4]$

Choose the correct answer from the options given below:

B, E

The correct answer is Option (3) → B, E.

To determine whether a coordination compound is colorless or colored, we must consider its electronic structure, the metal ion's oxidation state, and the type of ligands involved. The color of a coordination compound arises due to d-d transitions (where electrons move between split d-orbitals) and ligand-to-metal charge transfers (LMCT). Compounds without unpaired electrons or with fully filled d-orbitals tend to be colorless.

Let us examine each compound in detail:

A. \([Cr(NH_3)_6]Cl_3\)

Oxidation State of Chromium: In \([Cr(NH_3)_6]^{3+}\), chromium is in the +3 oxidation state.

Electronic Configuration of \(Cr^{3+}\):

Chromium's ground-state configuration is \( [Ar] 3d^5 4s^1 \).

In the +3 state, chromium loses 3 electrons: 2 from the 4s and 1 from the 3d orbital, resulting in \(3d^3\).

Unpaired Electrons: \(Cr^{3+}\) has 3 unpaired electrons in its 3d orbitals.

Color: Since the presence of unpaired electrons allows for d-d transitions, the compound will absorb visible light, and thus, it will be colored.

Conclusion: Not colorless (the compound is colored).

B. \([Co(NH_3)_6]Br_3\)

Oxidation State of Cobalt: In \([Co(NH_3)_6]^{3+}\), cobalt is in the +3 oxidation state.

Electronic Configuration of \(Co^{3+}\):

Cobalt’s ground-state configuration is \( [Ar] 3d^7 4s^2 \).

In the +3 state, cobalt loses 3 electrons: 2 from the 4s orbital and 1 from the 3d orbital, resulting in \(3d^6\).

Unpaired Electrons: In an octahedral field (due to the \(NH_3\) ligands), the \(3d^6\) configuration of \(Co^{3+}\) leads to the splitting of the d-orbitals. Depending on the field strength of the ligand (which in this case is \(NH_3\), a weak field ligand), this configuration has unpaired electrons.

Color: The presence of unpaired electrons allows for d-d transitions, leading to visible light absorption. Therefore, this compound will be colored.

Conclusion: Not colorless (the compound is colored).

C. \(K_3[Fe(CN)_6]\)

Oxidation State of Iron: In \([Fe(CN)_6]^{3-}\), iron is in the +3 oxidation state.

Electronic Configuration of \(Fe^{3+}\):

Iron’s ground-state configuration is \( [Ar] 3d^6 4s^2 \).

In the +3 state, iron loses 3 electrons: 2 from the 4s orbital and 1 from the 3d orbital, resulting in \(3d^5\).

Unpaired Electrons: The \(3d^5\) configuration in the +3 oxidation state leaves 5 unpaired electrons. Although \(CN^-\) is a strong field ligand, in \(Fe^{3+}\), the pairing energy is quite high, so some unpaired electrons remain even in the octahedral field.

Color: The presence of unpaired electrons in \(Fe^{3+}\) allows for d-d transitions, meaning the compound absorbs visible light and is colored.

Conclusion: Not colorless (the compound is colored).

D. \([MnCl_2(en)_2]^+\)

Oxidation State of Manganese: In \([MnCl_2(en)_2]^+\), manganese is likely in the +2 oxidation state.

Electronic Configuration of \(Mn^{2+}\):

Manganese’s ground-state configuration is \( [Ar] 3d^5 4s^2 \).

In the +2 state, manganese loses 2 electrons from the 4s orbital, resulting in \(3d^5\).

Unpaired Electrons: The \(3d^5\) configuration of \(Mn^{2+}\) results in 5 unpaired electrons.

Color: The presence of these unpaired electrons leads to d-d transitions, making the compound colored.

Conclusion: Not colorless (the compound is colored).

E. \([Ni(CO)_4]\)

Oxidation State of Nickel: In \([Ni(CO)_4]\), nickel is in the 0 oxidation state.

Electronic Configuration of \(Ni^0\):

Nickel’s ground-state configuration is \( [Ar] 3d^8 4s^2 \).

In the neutral state, nickel retains its full electron configuration, resulting in \(3d^{10}\).

Unpaired Electrons: In the neutral state, all 10 electrons occupy the d-orbitals, meaning there are no unpaired electrons.

Color: Since there are no unpaired electrons, d-d transitions cannot occur, and the compound does not absorb visible light. Thus, it is colorless.

Conclusion: Colorless.

Summary:

Colored Compounds: A (due to unpaired electrons in \(Cr^{3+}\)), B (unpaired electrons in \(Co^{3+}\)), C (unpaired electrons in \(Fe^{3+}\)), and D (unpaired electrons in \(Mn^{2+}\)).

Colorless Compounds: E (because \(Ni^0\) has no unpaired electrons and no d-d transitions).