Coordination Complexes Magnetic Properties

The question asks for the correct set of information regarding the magnetic properties and magnetic moment of two coordination complexes: [Mn(H2O)6]2+ and [Co(H2O)6]3+.

To determine the correct information, we need to analyze each complex separately based on its electronic configuration, the nature of the ligand (H2O), and its effect on crystal field splitting.

Analyzing [Mn(H₂O)₆]²⁺

In the complex [Mn(H2O)6]2+, manganese is in the +2 oxidation state. The electronic configuration of neutral manganese (Mn) is [Ar] 3d5 4s2. Therefore, the electronic configuration of Mn2+ is [Ar] 3d5.

H2O is generally considered a weak field ligand. In an octahedral complex with a weak field ligand, the crystal field splitting energy (<tex>\Delta_o</tex>) is less than the pairing energy (P). This leads to a high-spin complex where electrons occupy orbitals singly as much as possible before pairing up.

The 5 d electrons in Mn2+ (d5) will occupy the t2g and eg orbitals in a high-spin configuration:

• t2g orbitals: 3 electrons (one in each orbital)
• eg orbitals: 2 electrons (one in each orbital)

Configuration: t2g3 eg2.

This configuration results in 5 unpaired electrons (n = 5).

The spin-only magnetic moment (<tex>\mu</tex>) is calculated using the formula:

<tex>\mu = \sqrt{n(n+2)}</tex> Bohr Magnetons (BM)

For n = 5:

<tex>\mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35}</tex> BM

<tex>\sqrt{35}</tex> is approximately 5.92 BM.

For d5 high-spin octahedral complexes like [Mn(H2O)6]2+, there is no orbital contribution to the magnetic moment because the t2g orbitals are half-filled and the eg orbitals are also half-filled, and there is no degeneracy in the ground state allowing for orbital angular momentum. Therefore, the observed magnetic moment (<tex>\mu_{observed}</tex>) is very close to the spin-only magnetic moment (<tex>\mu</tex>). The options suggest <tex>\mu_{observed} = \mu</tex>, which is a reasonable approximation for this case.

Analyzing [Co(H₂O)₆]³⁺

In the complex [Co(H2O)6]3+, cobalt is in the +3 oxidation state. The electronic configuration of neutral cobalt (Co) is [Ar] 3d7 4s2. Therefore, the electronic configuration of Co3+ is [Ar] 3d6.

While H2O is typically considered a weak field ligand, its position in the spectrochemical series can cause Co3+ complexes to be low-spin. For Co3+ (d6), the crystal field splitting energy (<tex>\Delta_o</tex>) due to H2O is large enough to overcome the pairing energy (P). This leads to a low-spin complex.

The 6 d electrons in Co3+ (d6) will occupy the t2g and eg orbitals in a low-spin configuration:

• t2g orbitals: 6 electrons (all paired up)
• eg orbitals: 0 electrons

Configuration: t2g6 eg0.

This configuration results in 0 unpaired electrons (n = 0).

A substance with 0 unpaired electrons is diamagnetic. It is repelled by an external magnetic field.

Evaluating the Options

Based on our analysis:

• For [Mn(H2O)6]2+: There are 5 unpaired electrons, leading to a paramagnetic complex with <tex>\mu_{observed} \approx \mu</tex>.
• For [Co(H2O)6]3+: There are 0 unpaired electrons, leading to a diamagnetic complex.

Let's check the options:

Comparing our findings with the options:

• Option 1: [Mn(H2O)6]2+ part is consistent, but [Co(H2O)6]3+ is listed as Paramagnetic, which is incorrect.
• Option 2: [Mn(H2O)6]2+ part is inconsistent (<tex>\mu_{observed} > \mu</tex> is generally not true for d5 high spin where orbital contribution is quenched), though [Co(H2O)6]3+ part is consistent.
• Option 3: [Mn(H2O)6]2+ part is consistent (<tex>\mu_{observed} = \mu</tex>), and [Co(H2O)6]3+ is listed as Diamagnetic, which is also consistent.
• Option 4: Both parts are inconsistent with our analysis.

Therefore, the set of information in Option 3 correctly describes the magnetic properties of the two complexes.