Identifying Paramagnetic Coordination Compounds

A substance is considered paramagnetic if it is attracted to a magnetic field. This property arises from the presence of one or more unpaired electrons within the atoms or ions of the substance. Conversely, a substance with no unpaired electrons is repelled by a magnetic field and is called diamagnetic.

To determine if a coordination compound is paramagnetic, we need to analyze the electronic configuration of the central metal ion and how the ligands interact with the metal's d-orbitals (crystal field theory). The key steps involve:

1. Determine the oxidation state of the central metal atom.
2. Write the electronic configuration of the central metal ion in that oxidation state.
3. Consider the nature of the ligands (strong field or weak field). Strong field ligands cause a larger splitting of d-orbitals and favor pairing of electrons, while weak field ligands cause smaller splitting and favor filling orbitals singly first (Hund's rule).
4. Determine the number of unpaired electrons based on the electronic configuration and ligand field splitting.

Analyzing Each Coordination Compound Option

1. Hexa amine chromium (III) ion, \([Cr(NH_3)_6]^{3+}\)

• Central metal: Chromium (Cr)
• Oxidation state: Let the oxidation state of Cr be x. The charge of NH₃ is 0. The total charge of the complex is +3. So, \(x + 6(0) = +3\), which gives \(x = +3\). Chromium is in the +3 oxidation state.
• Electronic configuration of neutral Cr: \([Ar] 3d^5 4s^1\)
• Electronic configuration of Cr³⁺: Remove 3 electrons (1 from 4s and 2 from 3d) = \([Ar] 3d^3\).
• Ligand: NH₃ is generally considered a strong field ligand, especially for 3d metal ions.
• Crystal field splitting (Octahedral complex): In an octahedral field, the five d-orbitals split into a lower energy set of three orbitals (\(t_{2g}\)) and a higher energy set of two orbitals (\(e_g\)).
• Electron filling: We have 3 electrons in the 3d orbitals of Cr³⁺. According to Hund's rule, these 3 electrons will occupy the three \(t_{2g}\) orbitals singly before any pairing occurs.

• d-orbital level	Occupancy
  \(e_g\)	\(\_ \quad\_\)
  \(t_{2g}\)	\(\uparrow \quad\uparrow \quad\uparrow\)

  
  
• Number of unpaired electrons: 3.
• Conclusion: Since there are 3 unpaired electrons, \([Cr(NH_3)_6]^{3+}\) is paramagnetic. This finding confirms it as a paramagnetic substance.

2. Tetraamminezinc (II) ion, \([Zn(NH_3)_4]^{2+}\)

• Central metal: Zinc (Zn)
• Oxidation state: Let the oxidation state of Zn be x. The charge of NH₃ is 0. The total charge is +2. So, \(x + 4(0) = +2\), which gives \(x = +2\). Zinc is in the +2 oxidation state.
• Electronic configuration of neutral Zn: \([Ar] 3d^{10} 4s^2\)
• Electronic configuration of Zn²⁺: Remove 2 electrons from 4s = \([Ar] 3d^{10}\).
• Electron filling: The 3d orbitals are completely filled with 10 electrons.
• Number of unpaired electrons: 0.
• Conclusion: With no unpaired electrons, \([Zn(NH_3)_4]^{2+}\) is diamagnetic.

3. Tetracyanonickelate (II) ion, \([Ni(CN)_4]^{2-}\)

• Central metal: Nickel (Ni)
• Oxidation state: Let the oxidation state of Ni be x. The charge of CN⁻ is -1. The total charge is -2. So, \(x + 4(-1) = -2\), which gives \(x - 4 = -2\), so \(x = +2\). Nickel is in the +2 oxidation state.
• Electronic configuration of neutral Ni: \([Ar] 3d^8 4s^2\)
• Electronic configuration of Ni²⁺: Remove 2 electrons from 4s = \([Ar] 3d^8\).
• Ligand: CN⁻ is a very strong field ligand.
• Crystal field splitting (Square Planar complex): For d⁸ complexes with strong field ligands, the geometry is often square planar. The d-orbital splitting in square planar complexes is significant. The order of energy levels is typically \(d_{x^2-y^2} > d_{xy} > d_{z^2} > d_{xz}, d_{yz}\).
• Electron filling: We have 8 electrons in the 3d orbitals of Ni²⁺. With a strong field ligand, electrons pair up in the lower energy orbitals before occupying higher energy ones. The 8 electrons will fill the \(d_{xz}, d_{yz}\) (4 electrons), \(d_{z^2}\) (2 electrons), and \(d_{xy}\) (2 electrons) orbitals. The highest energy orbital, \(d_{x^2-y^2}\), remains empty.
• Number of unpaired electrons: 0. All 8 electrons are paired.
• Conclusion: With no unpaired electrons, \([Ni(CN)_4]^{2-}\) is diamagnetic.

4. Diammine silver (I) ion, \([Ag(NH_3)_2]^{+}\)

• Central metal: Silver (Ag)
• Oxidation state: Let the oxidation state of Ag be x. The charge of NH₃ is 0. The total charge is +1. So, \(x + 2(0) = +1\), which gives \(x = +1\). Silver is in the +1 oxidation state.
• Electronic configuration of neutral Ag: \([Kr] 4d^{10} 5s^1\)
• Electronic configuration of Ag⁺: Remove 1 electron from 5s = \([Kr] 4d^{10}\).
• Electron filling: The 4d orbitals are completely filled with 10 electrons.
• Number of unpaired electrons: 0.
• Conclusion: With no unpaired electrons, \([Ag(NH_3)_2]^{+}\) is diamagnetic.

Summary of Unpaired Electrons

Let's summarize the number of unpaired electrons for each coordination compound:

Based on the analysis of unpaired electrons, only the hexa amine chromium (III) ion, \([Cr(NH_3)_6]^{3+}\), has unpaired electrons and is therefore paramagnetic. Understanding the electronic configuration and the effect of ligands is crucial for determining the magnetic properties of coordination compounds.