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?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → tetrahedral
The spin only magnetic moment tells how many unpaired electrons the ion has:
\( \displaystyle \mu = \sqrt{n(n+2)}\ \mathrm{BM} \)
\( \displaystyle n = 5:\quad \mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.9\ \mathrm{BM} \)
So the complex has five unpaired electrons. This agrees with the metal ion: x + 4(−1) = −2 gives Mn as +2, and Mn (Z = 25), \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \), loses its two 4s electrons to give \( \ce{Mn^2+} \): \( \mathrm{3d^{5}} \), one electron in each 3d orbital.
With four ligands the complex can be square planar (\( \mathrm{dsp^{2}} \)) or tetrahedral (\( \mathrm{sp^{3}} \)). \( \mathrm{dsp^{2}} \) needs one empty 3d orbital, which would force two electrons to pair and leave only three unpaired. Since all five stay unpaired, no 3d orbital is free, the 4s and three 4p orbitals are used, and the complex is tetrahedral.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Magnetic Properties of Coordination Compounds – …
