Question

The spin-only magnetic moment value of \([\mathrm{MnBr}_4]^{2-}\) is \(5.9\,\mathrm{BM}\). What will be the geometry of the complex ion?

Hint:

Use \(\mu_{\text{so}}=\sqrt{n(n+2)}\) to get \(n\). For four-coordinate complexes: weak-field ligands (e.g., \(\mathrm{Br^-}\), \(\mathrm{I^-}\)) \(\Rightarrow\) usually tetrahedral; strong-field cases and \(d^8\) metals \(\Rightarrow\) often square planar.

Answer 1

Step 1: Find number of unpaired electrons from \(\mu_{\text{so}}\).
Spin-only formula: \(\mu_{\text{so}}=\sqrt{n(n+2)}\,\mathrm{BM}\). Given \(5.9\,\mathrm{BM}\) \(\Rightarrow\) \(n(n+2)\approx(5.9)^2\approx34.8\). The nearest integer solution is \(n=5\) (since \(\sqrt{5(5+2)}=\sqrt{35}=5.92\,\mathrm{BM}\)).

Step 2: Determine metal oxidation state and \(d\)-count.
Let oxidation state of Mn be \(x\): \(x+4(-1)=-2 $\Rightarrow$ x=+2\). Thus \(\mathrm{Mn^{2+}}\) is \(d^5\).

Step 3: Decide geometry.
\(\mathrm{Br^-}\) is a weak-field ligand, so pairing is unfavorable; a \(d^5\) ion remains high spin with 5 unpaired electrons—consistent with the observed \(\mu\). Four-coordinate \(\mathrm{Mn^{2+}}\) with weak-field ligands prefers tetrahedral over square planar (which is typical for \(d^8\) ions and would require strong-field stabilization).
\[ \boxed{\text{Geometry of }[\mathrm{MnBr}_4]^{2-}\ \text{is tetrahedral.}} \]