Read the passage carefully and answer.
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. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. 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 (either anions or the negative ends of dipolar molecules like \( NH_3 \) and \( H_2O \)) 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.
Which of the following pair exists as high spin, tetrahedral complexes?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \([NiCl_4]^{2-}\) and \([MnBr_4]^{2-}\)
In a tetrahedral complex the crystal field splitting is small, \( \Delta_t = \frac{4}{9}\Delta_o \), so it is never large enough to force electrons to pair. Four-coordinate complexes of weak field ligands such as \( \ce{Cl-} \), \( \ce{Br-} \) and \( \ce{F-} \) are therefore tetrahedral and high spin. A strong field ligand such as \( \ce{CN-} \) behaves differently: it pairs the d electrons, and with \( \ce{Ni^2+} \) it gives a square planar complex.
\( \ce{[NiCl4]^2-} \): Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \), so \( \ce{Ni^2+} \) is \( \mathrm{3d^{8}} \). \( \ce{Cl-} \) is a weak field ligand: \( sp^3 \) hybridisation, tetrahedral, 2 unpaired electrons, high spin.
\( \ce{[MnBr4]^2-} \): Mn (Z = 25) is \( \mathrm{[Ar]\,3d^{5}\,4s^{2}} \), so \( \ce{Mn^2+} \) is \( \mathrm{3d^{5}} \). \( \ce{Br-} \) is a weak field ligand: tetrahedral, 5 unpaired electrons, high spin.
So both complexes of Option (3) are high spin tetrahedral complexes.
Why the other options are incorrect
Option (1) \([Ni(SCN)_4]^{2-}\) and \([Mn(CN)_4]^{2-}\) — Incorrect. The cyanide complex has the strong field ligand \( \ce{CN-} \), which pairs the electrons, so the pair is not high spin.
Option (2) \([Ni(CN)_4]^{2-}\) and \([MnBr_4]^{2-}\) — Incorrect. \( \ce{[Ni(CN)4]^2-} \) is square planar (\( dsp^2 \)) and diamagnetic: the strong \( \ce{CN-} \) pairs the eight 3d electrons of \( \ce{Ni^2+} \) into four orbitals.
Option (4) \([NiF_4]^{2-}\) and \([Mn(CN)_4]^{2-}\) — Incorrect. \( \ce{[NiF4]^2-} \) is high spin tetrahedral, but its partner has the strong field \( \ce{CN-} \) ligand.
NCERT (2026-27 edition):
Valence Bond Theory – “… In tetrahedral complexes one s and three p orbitals are hybridised to form four equivalent orbitals oriented tetrahedrally. This is illustrated below for [NiCl4]2–. Here nickel is in +2 oxidation state and the ion has the electronic configuration 3d8. … Each Cl– ion donates a pair of electrons. The compound is paramagnetic since it contains two unpaired electrons. … In the square planar complexes, the hybridisation involved is dsp2. An example is [Ni(CN)4]2–. Here nickel is in +2 oxidation state and has the electronic configuration 3d8.”