krit.club logo

Coordination Compounds - Bonding in Coordination Compounds (VBT and CFT)

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Valence Bond Theory (VBT): Coordination compounds involve the overlap of vacant hybrid orbitals of the central metal ion with filled orbitals of the ligands. For coordination number 6, hybridization is either d2sp3d^2sp^3 (Inner orbital complex) or sp3d2sp^3d^2 (Outer orbital complex).

Magnetic Properties in VBT: If a complex contains unpaired electrons, it is paramagnetic. If all electrons are paired, it is diamagnetic. Strong field ligands like CNCN^- and COCO often cause electron pairing.

Crystal Field Theory (CFT): This theory treats the metal-ligand bond as purely electrostatic. The five degenerate dd-orbitals of the metal ion split into different energy levels when surrounded by a ligand field.

Octahedral Splitting (ΔoΔ_o): In an octahedral field, the dd-orbitals split into two sets: lower energy t2gt_{2g} (dxy,dyz,dzxd_{xy}, d_{yz}, d_{zx}) and higher energy ege_g (dx2y2,dz2d_{x^2-y^2}, d_{z^2}).

Tetrahedral Splitting (ΔtΔ_t): In a tetrahedral field, the splitting is inverted compared to octahedral. The ee orbitals are lower in energy than the t2t_2 orbitals. The splitting energy is smaller: Δt=49Δo\Delta_t = \frac{4}{9} \Delta_o.

Spectrochemical Series: Ligands are arranged in increasing order of their crystal field splitting energy (Δ\Delta): I<Br<SCN<Cl<S2<F<OH<C2O42<H2O<NCS<edta4<NH3<en<CN<COI^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < edta^{4-} < NH_3 < en < CN^- < CO.

Strong Field vs. Weak Field Ligands: For strong field ligands, Δo>P\Delta_o > P (where PP is pairing energy), resulting in low-spin complexes. For weak field ligands, Δo<P\Delta_o < P, resulting in high-spin complexes.

📐Formulae

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

Δt=49Δo\Delta_t = \frac{4}{9} \Delta_o

CFSEoct=[0.4n(t2g)+0.6n(eg)]Δo+nPCFSE_{oct} = [-0.4n(t_{2g}) + 0.6n(e_g)] \Delta_o + nP

CFSEtet=[0.6n(e)+0.4n(t2)]ΔtCFSE_{tet} = [-0.6n(e) + 0.4n(t_2)] \Delta_t

💡Examples

Problem 1:

Using VBT, predict the hybridization, geometry, and magnetic property of [Co(NH3)6]3+[Co(NH_3)_6]^{3+}. (Atomic number of Co=27Co = 27)

Solution:

Co3+Co^{3+} has 3d63d^6 configuration. NH3NH_3 is a strong field ligand, so it causes pairing of 3d3d electrons. Hybridization: d2sp3d^2sp^3. Geometry: Octahedral. Magnetic property: Diamagnetic.

Explanation:

In Co3+Co^{3+}, the 3d3d electrons are 1,1,1,1,1,11, 1, 1, 1, 1, 1 (unpaired). NH3NH_3 forces them to pair up, leaving two 3d3d orbitals vacant. These mix with one 4s4s and three 4p4p orbitals to form d2sp3d^2sp^3 hybrids. Since no unpaired electrons remain, it is diamagnetic.

Problem 2:

Calculate the spin-only magnetic moment of [Fe(H2O)6]2+[Fe(H_2O)_6]^{2+}.

Solution:

Fe2+Fe^{2+} is 3d63d^6. H2OH_2O is a weak field ligand (Δo<PΔ_o < P), so t2g4eg2t_{2g}^4 e_g^2. Number of unpaired electrons (nn) = 44. μ=4(4+2)=244.90 BM\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}.

Explanation:

In a weak field octahedral environment, electrons occupy orbitals singly before pairing. For d6d^6, four electrons remain unpaired (t2g4t_{2g}^4 has 2 unpaired, eg2e_g^2 has 2 unpaired), leading to the calculated magnetic moment.