Read the passage carefully and answer the Questions.

According to Valence Bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field.

The spin only magnetic moment of $[MnBr_4]^{2-}$ is 5.9 BM. Predict the geometry of the complex ion?

tetrahedral

The correct answer is Option (1) → tetrahedral **

Statement: The complex ion $[MnBr_4]^{2-}$ has tetrahedral geometry.

Reason: Its magnetic moment ($5.9$ BM) indicates five unpaired electrons, which is characteristic of a high-spin tetrahedral complex.

Step-by-step reasoning

1. Find oxidation state of Mn

In $[MnBr_4]^{2-}$:

• Let oxidation state of $Mn = x$
• $x + 4(-1) = -2$
• $x - 4 = -2$
• $x = +2$

So $Mn^{2+} \rightarrow$ electronic configuration $= 3d^5$.

2. Use magnetic moment formula

Spin-only formula:

$\mu = \sqrt{n(n + 2)} \text{ BM}$

Given $\mu \approx 5.9 \text{ BM}$

This corresponds to $n = 5$ unpaired electrons.

3. Interpretation

• $d^5$ with $5$ unpaired electrons means no pairing occurred.
• $Br^-$ is a weak field ligand, so it cannot cause pairing.
• Four-coordinate complexes with weak field ligands usually form tetrahedral ($sp^3$) structures.
• Tetrahedral fields are small splitting fields $\rightarrow$ always high spin.

4. Reject other options

• Square planar $\rightarrow$ usually low spin and needs strong field ligands.
• Trigonal pyramidal $\rightarrow$ uncommon for such complexes.
• Distorted octahedral $\rightarrow$ requires coordination number $6$ (not $4$).

Conclusion:

Because $Mn^{2+}$ is $d^5$ high spin with weak-field $Br^-$ ligands and coordination number $4$, the complex $[MnBr_4]^{2-}$ is tetrahedral.