Question

Which of the following compound is both paramagnetic and coloured? (a) \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{TiCl}_{6}\right]\) (b) \(\mathrm{VOSO}_{4}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{K}_{3}\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]\)

Short answer

The compound \( \mathrm{VOSO}_{4} \) is both paramagnetic and colored.

Step-by-step solution

Understanding Paramagnetism

A compound is said to be paramagnetic if it has unpaired electrons. This can be determined by looking at the electron configuration of the central metal ion in the compound.

Checking Coloration

A compound is typically colored due to electronic transitions (d-d transitions) within the metal ion. These usually occur when the metal ion has a partially filled d-orbital.

Analyze Option (a)

- Compound: \( (\mathrm{NH}_{4})_{2}[\mathrm{TiCl}_{6}] \)- Titanium (Ti) is in the +4 oxidation state- Electronic configuration: [Ar] 3d^0- No unpaired electrons; it is diamagnetic- No d-d transitions; colorless

Analyze Option (b)

- Compound: \( \mathrm{VOSO}_{4} \)- Vanadium (V) is in the +4 oxidation state- Electronic configuration: [Ar] 3d^1- One unpaired electron; paramagnetic- d-d transitions are possible; colored

Analyze Option (c)

- Compound: \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \)- Chromium (Cr) is in the +6 oxidation state- Electronic configuration: [Ar] 3d^0- No unpaired electrons; diamagnetic- Chromate colors are usually due to ligand-to-metal charge transfer, not d-d transitions

Analyze Option (d)

- Compound: \( \mathrm{K}_{3}[\mathrm{Cu}(\mathrm{CN})_{4}] \)- Copper (Cu) is in the +1 oxidation state- Electronic configuration: [Ar] 3d^{10}- No unpaired electrons; diamagnetic- Color may arise from other sources, but typically d^10 complexes are not colored.

Conclusion

From the analysis, \( \mathrm{VOSO}_{4} \) is the only compound that is both paramagnetic, due to one unpaired electron, and colored, due to possible d-d transitions.

Key concepts

Electron Configuration

Electron configuration is like the roadmap of an atom's electrons. It shows us the distribution of electrons in atomic orbitals. These orbitals fill up according to rules: first the lower energy levels, then the higher ones. For a chemistry student, identifying the electron configuration helps in predicting the chemical and physical properties of elements, such as their ability to form bonds, their reactivity, and their magnetic behavior.

For instance, titanium (Ti) with oxidation state +4 has an electron configuration of [Ar] 3d⁰. This means all its 3d orbitals are empty, hence no unpaired electrons result, indicating diamagnetism. Contrast this with vanadium (V) in its +4 oxidation state. Here, the electron configuration is [Ar] 3d¹ showing a single unpaired electron, which results in paramagnetism. So, the presence or absence of unpaired electrons in the electron configuration can tell us a lot about the compound's magnetic properties.

d-d Transitions

d-d transitions involve the movement of electrons between d orbitals of a transition metal ion. This is a crucial concept in understanding why some compounds are colored. These transitions occur when the metal has partially filled d orbitals. When light hits the compound, electrons absorb specific wavelengths to jump from one d orbital to another. The wavelengths of light absorbed correspond to the energy difference between these orbitals, and this process creates visible color.

Consider the compound \( ext{VOSO}_{4}\). Here, vanadium has a 3d¹ electron configuration due to its +4 oxidation state. The presence of one electron means that d-d transitions can occur, leading to coloration. In contrast, compounds like \( \mathrm{K}_{3}[\mathrm{Cu}(\mathrm{CN})_{4}] \), in a fully filled 3d¹⁰ state, don't exhibit these transitions as effectively, since there are no available low-energy d orbitals for electrons to transition to, hence they're often colorless.

Oxidation State

The oxidation state of an element is a measure of its degree of oxidation in a compound. It tells us how many electrons an atom has gained, lost, or shared when forming bonds. Understanding the oxidation state is vital for grasping the electronic structure of compounds.

In transition metals, the oxidation state plays a significant role because it affects both electron configuration and the occurrence of d-d transitions. Vanadium in \( \text{VOSO}_{4} \) is in a +4 oxidation state, which means the metal ion has lost four electrons. Such configurations lead to complex properties like color and magnetism, due to partially filled d orbitals. Other compounds, such as \( (\mathrm{NH}_{4})_{2}[\mathrm{TiCl}_{6}] \), where titanium is in a +4 oxidation state, have electron configurations without unpaired electrons, rendering them diamagnetic and often colorless.

Thus, knowing the oxidation state helps predict the behavior of a compound, especially in regard to its electronic and magnetic properties, and whether it will be colored or not.