Option 1 - Applying valence bond theory
VBT explains bonding by hybrid orbitals of the metal ion overlapping with ligand lone pairs; whether pairing happens depends on the field strength of the ligand. The magnetic moment follows \(\mu = \sqrt{n(n+2)}\) BM.
Step 1: Find the metal ion in [CoF6]3-. Six F- plus overall -3 charge means Co3+. From cobalt (Z = 27) removing three electrons leaves the \(3d^6\) core.
Step 2: Fluoride sits low in the spectrochemical series (weak field), so it cannot pair the d-electrons. The \(3d^6\) set remains spread out with four unpaired spins. Because the 3d orbitals are not vacated, the metal must accept the six ligand pairs in its outer \(4s\,4p\,4d\) orbitals, i.e. \(sp^3d^2\) hybridisation. This is a high-spin, outer-orbital, octahedral complex.
Step 3: Put n = 4 into the formula: \(\mu = \sqrt{4 \times 6} = \sqrt{24}\).
\[\boxed{\mu = 4.90\ \text{BM (paramagnetic)}}\]
Step 4: Now [Ni(CN)4]2-. Four CN- with a -2 charge give Ni2+, which is \(3d^8\) (nickel Z = 28 minus two electrons).
Step 5: Cyanide is a strong-field ligand, so it compresses the eight d-electrons into four filled orbitals and empties one 3d orbital. That inner 3d, together with 4s and two 4p orbitals, forms \(dsp^2\) hybrids pointing to the corners of a square, giving a square-planar inner-orbital complex with no unpaired electrons.
Step 6: With n = 0, \(\mu = \sqrt{0 \times 2} = 0\).
\[\boxed{\mu = 0\ \text{BM (diamagnetic)}}\]
Option 2 - Same-formula, different-shape isomers
Stereoisomerism means the connectivity is identical but the spatial layout differs. In coordination chemistry it splits into two families.
Geometrical isomerism: Here ligands occupy different relative sites. A square-planar example is \([Pt(NH_3)_2Cl_2]\), whose cis form (like groups adjacent) is the anticancer drug cisplatin, while the trans form places like groups opposite. Octahedral \([Co(NH_3)_4Cl_2]^+\) likewise gives cis and trans forms, and an \(MA_3B_3\) type such as \([Rh(py)_3Cl_3]\) gives facial and meridional forms. Complexes with only one type of ligand, or tetrahedral complexes, do not show this.
Optical isomerism: When a complex lacks any plane or centre of symmetry it becomes chiral, so it and its mirror image cannot be overlapped. These enantiomers (labelled d and l) rotate plane-polarised light equally but oppositely. Chelating complexes show this well: \([Co(en)_3]^{3+}\) and cis-\([CoCl_2(en)_2]^+\) are optically active, whereas the corresponding trans-\([CoCl_2(en)_2]^+\) has a symmetry plane and is optically inactive. Thus coordination compounds show both geometrical and optical stereoisomerism.