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.

How many unpaired electrons are present in [Fe(H₂O)₆]²⁺?

4

The correct answer is option 2. 4.

To determine the number of unpaired electrons in \([Fe(H_2O)_6]^{2+}\), we need to examine the electronic configuration of the iron ion and its interaction with the ligands in the complex.

Oxidation State of Iron:

In \([Fe(H_2O)_6]^{2+}\), the oxidation state of iron (Fe) is \(+2\).

Electronic Configuration of Fe:

The atomic number of iron (Fe) is 26. The ground-state electronic configuration of Fe is: \([Ar] 3d^6 4s^2\).

Electronic Configuration of \(Fe^{2+}\):

When iron loses two electrons to form Fe\(^{2+}\), the two electrons are removed from the 4s orbital first:

Fe: \([Ar] 3d^6 4s^2\)

Fe\(^{2+}\): \([Ar] 3d^6\)

Complex Formation:

In \([Fe(H_2O)_6]^{2+}\), water (H\(_2\)O) is a weak field ligand. Weak field ligands do not cause significant splitting of the \(d\)-orbitals in the crystal field.

d-Orbital Splitting and Electron Distribution:

In an octahedral crystal field, the \(3d\) orbitals split into two sets: \(t_{2g}\) (lower energy) and \(e_g\) (higher energy). Because water is a weak field ligand, the \(d\)-orbital splitting energy (\(\Delta_0\)) is small, and electrons will occupy the orbitals according to Hund's rule, maximizing the number of unpaired electrons.

Electron Configuration in the Octahedral Field:

For Fe\(^{2+}\) (\([Ar] 3d^6\)):

The six \(d\) electrons will be distributed among the \(t_{2g}\) and \(e_g\) orbitals. According to Hund's rule, the electrons will fill the \(t_{2g}\) set first (each orbital gets one electron before any gets two) and then the \(e_g\) set.

Arrangement of Electrons in \(d\) Orbitals:

\(t_{2g}\) orbitals: 3 electrons (one in each)

\(e_g\) orbitals: 3 electrons (one in each)

This distribution results in 4 unpaired electrons in the \(d\) orbitals.