calculate the spin only magnetic and total angular moment for

1-Na2(Ni CL4)

2-CL2(Cr(H2O))

3-K4(CO(CN)4)

To calculate the spin-only magnetic moment and total angular momentum for the given coordination complexes, we need to consider the electronic configurations of the metal ions involved, their oxidation states, and the geometry of the complexes. Let's break this down for each compound step by step.

1. Na2(NiCl4)

In this complex, nickel (Ni) is in the +2 oxidation state. The electron configuration for Ni is [Ar] 3d⁸ 4s². When it loses two electrons to form Ni²⁺, the configuration becomes 3d⁸.

Determining the Spin State

Nickel(II) typically has a square planar geometry in complexes like NiCl₄^2-. In this arrangement, the 3d orbitals split into two sets: the lower energy (d_{x²-y² and dz²) and the higher energy (dxy, dyz, and dzx). For Ni²⁺, the 3d⁸ configuration will have two unpaired electrons, leading to a high-spin state.}

Calculating the Spin-Only Magnetic Moment

The formula for the spin-only magnetic moment (μ_s) is:

μ_s = √(n(n + 2))

where n is the number of unpaired electrons. For Ni²⁺, n = 2:

μ_s = √(2(2 + 2)) = √(8) = 2.83 BM (Bohr Magneton)

Total Angular Momentum

The total angular momentum (J) can be calculated using the formula:

J = √[j(j + 1)]

For two unpaired electrons, j = 1 (since j = s + l, where s = 1/2 and l = 0 for d electrons). Thus:

J = √[1(1 + 1)] = √2 = 1.41

2. Cr(H2O)Cl2

In this complex, chromium (Cr) is in the +3 oxidation state. The electron configuration for Cr is [Ar] 3d⁵ 4s¹. For Cr³⁺, the configuration becomes 3d³.

Analyzing the Spin State

Chromium(III) typically adopts an octahedral geometry. In this case, the 3d orbitals will split into two sets (t_2g and e_g). With three electrons in the 3d orbitals, all three will occupy the lower energy t_2g orbitals, resulting in three unpaired electrons.

Spin-Only Magnetic Moment Calculation

Using the same formula:

μ_s = √(n(n + 2))

For Cr³⁺, n = 3:

μ_s = √(3(3 + 2)) = √(15) = 3.87 BM

Finding Total Angular Momentum

For three unpaired electrons, j = 3/2:

J = √[3/2(3/2 + 1)] = √(15/4) = 1.94

3. K4(CO(CN)4)

In this complex, cobalt (Co) is in the +2 oxidation state. The electron configuration for Co is [Ar] 3d⁷ 4s². For Co²⁺, the configuration becomes 3d⁷.

Understanding the Spin State

Cobalt(II) typically forms a tetrahedral complex in this case. In a tetrahedral field, the 3d orbitals split into two sets (e and t₂). For Co²⁺ with 3d⁷, there will be three unpaired electrons in the higher energy t₂ orbitals, resulting in a high-spin state.

Calculating the Spin-Only Magnetic Moment

For Co²⁺, n = 3:

μ_s = √(3(3 + 2)) = √(15) = 3.87 BM

Total Angular Momentum Calculation

For three unpaired electrons, j = 3/2:

J = √[3/2(3/2 + 1)] = √(15/4) = 1.94

Summary of Results

• Na2(NiCl4): μ_s = 2.83 BM, J = 1.41
• Cr(H2O)Cl2: μ_s = 3.87 BM, J = 1.94
• K4(CO(CN)4): μ_s = 3.87 BM, J = 1.94

These calculations provide insight into the magnetic properties and angular momentum of the complexes based on their electronic configurations and geometries. Understanding these concepts is crucial for predicting the behavior of transition metal complexes in various chemical environments.