Problem 176 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following facts about the complex \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is wrong? [2011] (a) The complex gives white precipitate with silver nitrate solution. (b) The complex involves \(\mathrm{d}^{2} \mathrm{sp}^{3}\) hybridisation, and is octahedral in shape. (c) The complex is an outer orbital complex. (d) The complex is paramagnetic.

Step-by-Step Solution

Answer

Statement (c) is wrong; the complex is an inner orbital complex, not outer.

Step 1: Identify the Coordination Complex Structure

The given coordination compound is \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\). It consists of the central metal ion \(\mathrm{Cr}^{3+}\) and ammonia ligands \(\left(\mathrm{NH}_3\right)\), which are neutral, and three chloride ions that are outside of the coordination sphere.

Step 2: Determine the Precipitation by Silver Nitrate

When \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}\) is treated with silver nitrate, the free chloride ions will react to form a white precipitate of silver chloride (\(\mathrm{AgCl}\)). Therefore, statement (a) is correct.

Step 3: Assess the Geometry and Hybridisation

The complex \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is an octahedral complex since it contains six ligands. The correct hybridisation for \(\mathrm{Cr}^{3+}\) with ammonia, a strong field ligand, is \(\mathrm{d^2sp^3}\). Therefore, statement (b) is correct.

Step 4: Determine Whether it is Inner or Outer Orbital Complex

The hybridisation \(\mathrm{d}^{2}\mathrm{sp}^{3}\) indicates it is an inner orbital or low spin complex because the electrons fill the inner \(d\) orbitals in the presence of strong field ligands like ammonia. Statement (c) suggests it is an outer orbital complex, which is wrong because it should be inner orbital due to \(\mathrm{d}^2\mathrm{sp}^3\) hybridisation.

Step 5: Check the Magnetic Properties

\(\mathrm{Cr}^{3+}\) has a configuration of \([\mathrm{Ar}]3\mathrm{d}^{3}\) with three unpaired electrons, hence the complex \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right]\) is paramagnetic. Thus, statement (d) is correct.

Key Concepts

Hybridisation in Coordination Chemistry

In coordination chemistry, hybridisation describes the mixing of atomic orbitals to form new, hybrid orbitals. These form bonds with ligands surrounding a central metal ion. Hybridisation dictates the geometry of the complex, influencing its structure and properties.
The complex \([\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}\) has \(d^2sp^3\) hybridisation. This involves the mixing of two inner \(d\) orbitals with one \(s\) and three \(p\) orbitals, forming six equivalent orbitals.
This hybridisation results in an octahedral geometry, which is characteristic when six ligands coordinate around a central metal ion.

• Octahedral structures are commonly found in many transition metal complexes.
• Hybridisation determines not only the shape but also the electronic and magnetic properties of a compound.

Understanding hybridisation helps predict how a complex will interact with other substances and its behavior in chemical reactions.

Magnetic Properties

Magnetic properties of coordination compounds depend on the arrangement of electrons, particularly unpaired electrons in their d-orbitals. The presence of unpaired electrons typically indicates a compound is paramagnetic.
The electron configuration of \(\mathrm{Cr}^{3+}\) is \([\mathrm{Ar}]3\mathrm{d}^3\), suggesting three unpaired electrons. Thus, \([\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}\) is paramagnetic.

• Paramagnetic materials are attracted to magnetic fields due to unpaired electrons.
• The magnitude of paramagnetism increases with the number of unpaired electrons.
• Complexes with strong field ligands can exhibit low or no unpaired electrons, becoming diamagnetic, but this is not the case here.

The presence of unpaired electrons in the \(d\) orbitals is crucial in defining the magnetic and spectral properties of a coordination compound.

Inner vs Outer Orbital Complexes

The distinction between inner and outer orbital complexes is based on which of the metal's d orbitals participate in hybridisation. This affects the spin state and magnetic properties of the complex.
Inner orbital complexes use their inner 3d orbitals in hybridisation, forming low-spin configurations due to strong field ligands. This is observed in \(d^2sp^3\) hybridisation, such as in the complex \([\mathrm{Cr}(\mathrm{NH}_3)_6]^{3+}\).

• Inner orbital complexes are usually low spin, often with fewer unpaired electrons.
• Outer orbital complexes use 4d or 5d orbitals and are typically high-spin, allowing more unpaired electrons.
• The ligand's field strength determines whether the complex is high or low spin.

In this exercise, because ammonia is a strong field ligand, the complex is inner orbital with \(d^2sp^3\) hybridisation. Contrary to outer orbital complexes, these have specific geometries and magnetic characteristics, usually making them more stable.