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.

What is the magnetic moment of compound [Cr(NH₃)₄Cl₂]Br?

3.87 B.M.

The correct answer is option 1. 3.87 B.M.

To determine the magnetic moment of the compound \([Cr(NH_3)_4Cl_2]Br\), we first need to determine the oxidation state of the chromium ion and the number of unpaired electrons.

Determining the Oxidation State of Chromium:

The compound is \([Cr(NH_3)_4Cl_2]Br\).

\(NH_3\) (ammonia) is a neutral ligand.

\(Cl^-\) (chloride) is a -1 ligand.

\(Br^-\) (bromide) is an ion outside the coordination sphere.

Let the oxidation state of Cr be \(x\):

\(x + 4(0) + 2(-1) + (-1) = 0 \)

\(x - 2 - 1 = 0 \)

\(x = +3 \)

So, the oxidation state of Cr in \([Cr(NH_3)_4Cl_2]Br\) is +3.

Electronic Configuration of Cr\(^3+\):

Chromium (Cr) has an atomic number of 24. The electronic configuration of \(Cr\) is: \([Ar] 3d^5 4s^1\)

For Cr\(^3+\):

It loses 3 electrons: 2 from the 4s orbital and 1 from the 3d orbital.

The electronic configuration of \(Cr^3+\) is: \([Ar] 3d^3 \)

The \(3d\) orbital has three electrons. The three \(d\) electrons will occupy three different \(d\) orbitals according to Hund's rule, resulting in three unpaired electrons.

The magnetic moment (\(\mu\)) can be calculated using the formula:

\(\mu = \sqrt{n(n + 2)}\)

where \(n\) is the number of unpaired electrons.

For Cr\(^3+\), \(n = 3\):

\(\mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ B.M.}\)

Conclusion: The magnetic moment of \([Cr(NH_3)_4Cl_2]Br\) is approximately 3.87 Bohr Magnetons (B.M.).