Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: Coordination Compounds

Question:

The magnetic moment of coordination compounds can be measured by the magnetic susceptibility experiments. The results can be used to obtain information about the number of unpaired electrons and hence structures adopted by metal complexes. A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with up to three electrons in the d orbitals, like Ti3+(d1); V3+(d2); Cr3+(d3); two vacant d orbitals are available for octahedral hybridization with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridization is not directly available (as a consequence of Hund’s rule). Thus, for d4 (Cr2+, Mn3+), d5 (Mn2+, Fe3+), d6 (Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively. The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d 6 ions. However, with species containing d4 and d5 ions there are complications. [Mn(CN)6]3– has magnetic moment of two unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic. This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. [Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridization, the former two complexes are paramagnetic and the latter diamagnetic. On the other hand, [MnCl6]3–, [FeF6]3– and [CoF6]3– are outer orbital complexes involving sp3d2 hybridization and are paramagnetic corresponding to four, five and four unpaired electrons.

[Ni(CN)4]2- and [NiCl4]2- differs in which of the following?

Options:

Oxidation number

Co-ordination number

Structure

All of these

Correct Answer:

Structure

Explanation:

The correct answer is option 3. Structure.

To determine how \([Ni(CN)_4]^{2-}\) and \([NiCl_4]^{2-}\) differ, we need to analyze their oxidation states, coordination numbers, and structures.

1. Oxidation number

For both complexes, the oxidation state of nickel (Ni) is determined as follows:

\([Ni(CN)_4]^{2-}\)

\(CN^-\) is a neutral ligand.

The overall charge on the complex is -2.

Let the oxidation state of Ni be \(x\):

\(x + 4(-1) = -2 \)

\(x - 4 = -2 \)

\(x = +2\)

\([NiCl_4]^{2-}\)

\(Cl^-\) is a -1 ligand.

The overall charge on the complex is -2.

Let the oxidation state of Ni be \(x\):

\(x + 4(-1) = -2 \)

\(x - 4 = -2 \)

\(x = +2 \)

Both complexes have nickel in the +2 oxidation state.

2. Co-ordination number

The coordination number is the number of ligands directly bonded to the central metal ion.

\([Ni(CN)_4]^{2-}\)

There are 4 cyanide (CN\(^-\)) ligands.

Coordination number is 4.

\([NiCl_4]^{2-}\)

There are 4 chloride (Cl\(^-\)) ligands. Coordination number is 4.

Both complexes have a coordination number of 4.

3. Structure

The structure of the complexes can differ based on the nature of the ligands and the geometry they adopt

\([Ni(CN)_4]^{2-}\)

Cyanide is a strong field ligand. With a coordination number of 4 and strong field ligands, the complex is typically square planar. \([Ni(CN)_4]^{2-}\) is expected to be square planar.

\([NiCl_4]^{2-}\)

Chloride is a weak field ligand. With a coordination number of 4 and weak field ligands, the complex is typically tetrahedral. \([NiCl_4]^{2-}\) is expected to be tetrahedral.

Conclusion:

The differences between \([Ni(CN)_4]^{2-}\) and \([NiCl_4]^{2-}\) are primarily in their structures due to the different geometries they adopt:

\([Ni(CN)_4]^{2-}\) is square planar.

\([NiCl_4]^{2-}\) is tetrahedral.

Thus, they differ in structure but not in oxidation number or coordination number.

Therefore, the correct answer is: Structure.