Coordination compounds are central to both CBSE Class 12 Chemistry and competitive exams because they explain bonding, color, magnetism, stereochemistry, and spectral behavior through concepts like ligand field theory, crystal field splitting, Jahn–Teller effects, and stability constants. Mastering this chapter helps you solve numericals on magnetic moments and stereoisomer counting, and also interpret IR/UV–Vis trends and isomerism patterns that frequently appear in board and JEE/NEET-style questions.
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Marking
Q1. Calculate the spin-only magnetic moment (in Bohr magneton, BM) for the hexaaquamanganese(II) ion assuming a high-spin configuration. Use where is the number of unpaired electrons.
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()
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()
Q2. A solution initially contains and . Given formation constants for as and for as , 0.200 M ethylenediamine (en) is added. Assuming free ligand concentrations remain approximately at their initial values, what fraction of copper will be present as at equilibrium?
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Q3. An octahedral complex of (formal ) has crystal-field splitting and pairing energy . Determine the most likely spin state, the crystal field stabilization energy (CFSE) in , and the approximate spin-only magnetic moment (in BM).
(Low-spin; CFSE ; )
(Low-spin; CFSE ; )
(High-spin; CFSE ; )
(High-spin; CFSE ; )
Q4. Assertion (A): In an elongated octahedral Cu(II) () complex the orbital is lowered in energy relative to . Reason (R): This orbital splitting reduces the energy of the lowest transition, causing the absorption maximum to shift to longer wavelength (red shift). Choose the correct option.
(Both A and R are true and R is the correct explanation of A.)
(A is true but R is false.)
(Both A and R are true but R does not correctly explain why A occurs.)
(A is false but R is true.)
Q5. On replacing one CO ligand by PPh in to give , the average in the IR spectrum is observed to decrease. Which statement best explains this observation?
(PPh is a stronger -acceptor than CO; it competes for metal back-donation, reducing M→CO backbonding and hence increases .)
(PPh is a stronger -donor and a poorer -acceptor than CO; it increases electron density on Fe, enhancing metal→CO back-donation and thereby lowering .)
(Steric bulk of PPh forces CO ligands farther from the metal, weakening M–CO backbonding and thereby increasing .)
(PPh forms strong metal←PPh -backbonding that competes with CO backbonding, thus decreasing back-donation to CO and increasing .)
Q6. Calculate the spin-only magnetic moment (in Bohr magneton) for the complex assuming it is a high-spin octahedral complex. Use where is the number of unpaired electrons.
Q7. How many stereoisomers (including optical isomers) are possible for the complex (en = ethylenediamine, a bidentate ligand)?
2
4
5
3
Q8. Which of the following octahedral complexes is expected to exhibit the strongest Jahn–Teller distortion? (Assume high-spin where relevant)
Q9. For an octahedral complex of the type MABC (three different pairs of identical monodentate ligands), how many stereoisomers (geometrical + optical) are possible?
3
6
4
9
Q10. Which of the following complexes satisfy the 18-electron rule?
Only
Both and
, and
Only