Coordination Chemistry — Practice Quiz

Linkage isomerism

Step 1: Both complexes have the same molecular formula and the same set of ligands (5 NH₃ and 1 NO₂⁻). Step 2: The difference is HOW the nitrite ion (NO₂⁻) is bonded. It is an ambidentate ligand — it can coordinate through either N or O. Step 3: In [Co(NH₃)₅(NO₂)]²⁺, nitrite binds through nitrogen (N-bonded, called nitrito-κN). In [Co(NH₃)₅(ONO)]²⁺, it binds through oxygen (O-bonded, called nitrito-κO). Step 4: This is the definition of linkage isomerism — same ligands, different bonding atoms. Ionisation isomers give different ions in solution. Coordination isomers involve interchange between

dsp², Square planar

Step 1: Determine the electronic configuration of Ni²⁺. Ni is [Ar] 3d⁸ 4s², so Ni²⁺ is [Ar] 3d⁸ (loses 2 electrons from 4s). Step 2: CN⁻ is a strong field ligand. It causes pairing of 3d electrons. The 3d⁸ configuration rearranges: electrons pair up, leaving one 3d orbital empty. Step 3: The one empty 3d orbital, one 4s orbital and two 4p orbitals hybridize to form dsp² hybridisation (inner orbital complex). Step 4: dsp² hybridisation gives a square planar geometry. sp³ (option B) applies to weak-field tetrahedral complexes like [NiCl₄]²⁻. d²sp³ and sp³d² apply to octahedral complexes (coordin

Δₜ = (4/9)Δ₀

Step 1: In an octahedral complex, all 6 ligands approach directly along the x, y, z axes, causing strong repulsion with d orbitals lying along axes (dx²-y² and dz²). Step 2: In a tetrahedral complex, only 4 ligands approach along alternate corners of a cube — none points directly at any d orbital. This means the interaction is weaker. Step 3: Also, there are fewer ligands (4 vs 6) in a tetrahedral complex. Step 4: Both factors together result in Δₜ being much smaller than Δ₀. The mathematical relationship derived is Δₜ = (4/9)Δ₀, approximately 0.44 times the octahedral splitting. This is why m

[Zn(NH₃)₄]²⁺

Step 1: The colour of coordination compounds arises due to d-d transitions — electrons absorb light and jump from lower energy t₂g orbitals to higher energy eg orbitals. Step 2: For d-d transitions to occur, the central metal ion must have partially filled d orbitals (i.e., d¹ to d⁹ configuration). Step 3: Zn²⁺ has the configuration [Ar] 3d¹⁰ — a completely filled d subshell. Since all d orbitals are filled, no d-d transition is possible. Step 4: Therefore [Zn(NH₃)₄]²⁺ is colourless. Ti³⁺ is d¹ (coloured — purple/violet), Fe³⁺ is d⁵ (coloured), Cu²⁺ is d⁹ (coloured — blue). Complexes with d⁰ (