Geometry of \( [Ni(CN)_4]^{2-} \) and \( [NiCl_4]^{2-} \) are.............
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Square planar, low spin and tetrahedral, high spin respectively
By valence bond theory, the geometry of a four-coordinate complex depends on the hybridisation of the metal ion, and this depends on the ligand. A strong field ligand such as \( \ce{CN-} \) forces the d electrons to pair up and frees an inner d orbital, giving dsp2 (square planar). A weak field ligand such as \( \ce{Cl-} \) leaves the d electrons unpaired, so the outer s and p orbitals are used: sp3 (tetrahedral).
In both complexes nickel is +2: \( x + 4(-1) = -2 \Rightarrow x = +2 \). Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \), with two unpaired electrons.
\( \ce{[Ni(CN)4]^2-} \): \( \ce{CN-} \) pairs the eight 3d electrons into four orbitals; the empty 3d, 4s and two 4p orbitals form dsp2 hybrids. The complex is square planar, with no unpaired electron: low spin (diamagnetic).
\( \ce{[NiCl4]^2-} \): \( \ce{Cl-} \) cannot pair the electrons; the 4s and three 4p orbitals form sp3 hybrids. The complex is tetrahedral, with two unpaired electrons: high spin (paramagnetic, \( \mu = \sqrt{2 \times 4} = 2.83\ \mathrm{BM} \)).
So the answer is square planar, low spin and tetrahedral, high spin, respectively.
NCERT (2026-27 edition):
Valence Bond Theory – “… 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. … Each of the hybridised orbitals receives a pair of electrons from a cyanide ion. The compound is diamagnetic as evident from the absence of unpaired electron.”
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.”