Question

The spin-only magnetic moment of ${[MnBr4]^2-}$ is 5.9 BM. The geometry of the complex and x respectively are

• A. tetrahedral, 1
• B. square planar, 1
• C. square planar, 2
• D. tetrahedral, 2

Hint:

Magnetic Moment and Geometry:

• Weak field ligands (like Br$^-$) lead to high-spin complexes.
• Tetrahedral complexes are generally high spin.
• A spin-only value near 5.9 BM confirms 5 unpaired electrons.
• For 3d$^5$ Mn$^2+$, the complex is high-spin tetrahedral.

Answer 1

The Correct Option is D

Manganese in ${[MnBr4]^2-}$ is in the +2 oxidation state (Mn$^{2+}$), and the electronic configuration of Mn$^{2+}$ is [Ar] 3d$^5$. Given spin-only magnetic moment ($\mu_s$) = 5.9 BM, we use the formula: \[ \mu_s = \sqrt{n(n+2)} \text{ BM} \] where $n$ = number of unpaired electrons. Solving: \[ 5.9 = \sqrt{n(n+2)} \Rightarrow n(n+2) = 34.81 \Rightarrow n = 5 \] Thus, Mn$^{2+}$ has 5 unpaired electrons, consistent with high spin 3d$^5$ configuration. Since bromide ({Br^-}) is a weak field ligand, it cannot pair the d-electrons. Hence, the geometry is tetrahedral (which favors high spin complexes). The number of unpaired electrons = 5, so total x = 2 (for octahedral: coordination number is 6; for tetrahedral: 4; here geometry confirms it as 2 based on the matching).

Answer 2

Step 1: Identify the metal and its oxidation state
The complex ion is \([MnBr_4]^{2-}\). Bromide (Br⁻) has a charge of -1, so for four Br⁻ ions the total charge is -4.
Given the overall charge of the complex ion is -2, the oxidation state of Mn must be +2.

Step 2: Determine the electronic configuration of Mn(II)
Manganese (Mn) atomic number = 25.
Electronic configuration of Mn: [Ar] 3d⁵ 4s².
For Mn(II), two electrons are lost from 4s orbital: configuration becomes [Ar] 3d⁵.
Hence, Mn(II) has 5 unpaired electrons in 3d orbitals.

Step 3: Calculate the spin-only magnetic moment
Spin-only magnetic moment, μ = \(\sqrt{n(n+2)}\) BM, where n = number of unpaired electrons.
Here, n = 5, so μ = \(\sqrt{5 \times 7} = \sqrt{35} \approx 5.92\) BM, close to the given value 5.9 BM.

Step 4: Deduce the geometry and coordination number
The complex \([MnBr_4]^{2-}\) with 4 ligands and magnetic moment indicating high spin corresponds to a tetrahedral geometry.
The coordination number (x) is 4 for tetrahedral complexes.

Step 5: Conclusion
The geometry of the complex is tetrahedral, and the coordination number x is 4.