Question

In an octahedral crystal field, the energies of which d-orbitals will be raised when ligands approach the central metal atom/ion? Give a reason in support of your answer.

Answer 1

Step 1: Picture the octahedral setup.
In an octahedral complex the six ligands come in along the three axes, that is along \(+x, -x, +y, -y, +z, -z\). We treat each ligand as a tiny negative charge approaching the metal.

Step 2: Split the d-orbitals into two groups.
Two d-orbitals, \(d_{x^2-y^2}\) and \(d_{z^2}\), have their lobes pointing straight along the axes. These form the \(e_g\) set. The other three, \(d_{xy}, d_{yz}\) and \(d_{zx}\), point in between the axes. These form the \(t_{2g}\) set.

Step 3: See which group meets the ligands head on.
The \(e_g\) orbitals point right at the incoming ligands, so the electrons in them face strong repulsion from the ligand charges.

Step 4: Compare the repulsion.
The \(t_{2g}\) orbitals lie between the ligands, so they feel less repulsion. More repulsion means higher energy.

Step 5: State the result.
So the \(e_g\) orbitals, \(d_{x^2-y^2}\) and \(d_{z^2}\), are pushed up in energy.

Answer: The \(e_g\) orbitals \((d_{x^2-y^2}\) and \(d_{z^2})\) are raised in energy, because they point directly towards the approaching ligands and so experience greater electrostatic repulsion than the \(t_{2g}\) orbitals, which lie between the axes.

Answer 2

Step 1: Find the oxidation state of cobalt.
In both \([CoF_6]^{3-}\) and \([Co(NH_3)_6]^{3+}\), the cobalt is in the \(+3\) state.

Step 2: Get the d-electron count.
Co has atomic number 27, so its configuration is \([Ar]3d^7 4s^2\). Removing three electrons for \(Co^{3+}\) leaves \(3d^6\). So we place six electrons in the \(t_{2g}\) and \(e_g\) levels.

Step 3: Look at the ligand for (i) \([CoF_6]^{3-}\).
\(F^-\) is a weak-field ligand. It gives a small splitting, so electrons stay unpaired as far as possible (high spin). The six electrons fill as \(t_{2g}^4 e_g^2\).

Step 4: Look at the ligand for (ii) \([Co(NH_3)_6]^{3+}\).
\(NH_3\) is a strong-field ligand. It gives a large splitting, so electrons pair up in the lower set first (low spin). The six electrons fill as \(t_{2g}^6 e_g^0\).

Step 5: Write the final configurations.
Both are \(d^6\) ions but they differ because of the ligand strength.

Answer: (i) \([CoF_6]^{3-}\): \(t_{2g}^4 e_g^2\) (high spin, weak field \(F^-\)); (ii) \([Co(NH_3)_6]^{3+}\): \(t_{2g}^6 e_g^0\) (low spin, strong field \(NH_3\)).

Answer 3

Step 1: Find the oxidation state in each complex.
In \([NiCl_4]^{2-}\), chloride is \(-1\) each, so nickel is \(+2\). In \([Ni(CO)_4]\), CO is neutral, so nickel is \(0\).

Step 2: Write the d-electron count.
Ni has atomic number 28, configuration \([Ar]3d^8 4s^2\). For \(Ni^{2+}\) we remove two electrons to get \(3d^8\). For \(Ni(0)\) the metal keeps \(3d^8 4s^2\), and on bonding it becomes effectively \(3d^{10}\).

Step 3: Check the ligand field strength.
\(Cl^-\) is a weak-field ligand, so it cannot force the electrons of \(Ni^{2+}\) to pair. The \(d^8\) ion keeps two unpaired electrons.

Step 4: Look at the CO complex.
CO is a strong-field ligand. It pushes all the electrons in \(Ni(0)\) to pair up, giving a fully filled \(3d^{10}\) set with no unpaired electrons.

Step 5: Link unpaired electrons to magnetism.
Unpaired electrons make a complex paramagnetic; all-paired electrons make it diamagnetic.

Answer: \([NiCl_4]^{2-}\) has \(Ni^{2+}\) (\(d^8\)) with weak-field \(Cl^-\), so two electrons stay unpaired and it is paramagnetic. \([Ni(CO)_4]\) has \(Ni(0)\) with strong-field CO, so all electrons pair up to give \(d^{10}\) with no unpaired electrons, making it diamagnetic.

Answer 4

Step 1: Find the oxidation state of iron.
In \([Fe(CN)_6]^{3-}\), each \(CN^-\) is \(-1\), six of them give \(-6\), and the overall charge is \(-3\), so iron is \(+3\).

Step 2: Get the d-electron count.
Fe has atomic number 26, configuration \([Ar]3d^6 4s^2\). For \(Fe^{3+}\) we remove three electrons to get \(3d^5\).

Step 3: Check the ligand strength.
\(CN^-\) is a strong-field ligand. It causes the d-electrons to pair up.

Step 4: Arrange the electrons.
The five electrons go into the lower set: \(t_{2g}^5 e_g^0\). This leaves two inner d-orbitals empty.

Step 5: Decide the hybridisation and magnetism.
With two inner \(3d\) orbitals free, the complex uses \(d^2sp^3\) hybridisation (inner orbital, octahedral). One unpaired electron is left, so the complex is paramagnetic.

Answer: \([Fe(CN)_6]^{3-}\) has \(d^2sp^3\) hybridisation (inner orbital complex) and is paramagnetic due to one unpaired electron.