Coordination compounds form a core part of Class 12 Chemistry because they connect electronic structure with observable properties like colour, magnetism, stereochemistry, and stability constants. Board and competitive exams repeatedly test oxidation states, ligand field splitting, spin states, CFSE, and stereoisomerism—so mastering these ideas from this chapter directly improves scoring accuracy in both conceptual and calculation-based questions.
20
Minutes
15
Questions
1 / -0
Marking
Q1. For the complex ion , determine the oxidation state of Fe and the number of unpaired electrons in the complex assuming CN is a strong-field ligand (pairing occurs).
Fe oxidation state +3; 1 unpaired electron
Fe oxidation state +2; 4 unpaired electrons
Fe oxidation state +2; 0 unpaired electrons
Fe oxidation state +1; 1 unpaired electron
Q2. How many stereoisomers exist for the octahedral complex (en = ethylenediamine, bidentate), and which of them are optically active?
Three stereoisomers: a trans form (achiral) and a cis pair of enantiomers (the cis pair are optically active)
Two stereoisomer types: cis (optically active) and trans (achiral)
Four stereoisomers: two cis (optically inactive) and two trans (optically active)
One stereoisomer only, which is optically active
Q3. A hexacoordinate octahedral complex of a first-row transition metal shows a spin-only magnetic moment of . Assuming spin-only behaviour ( BM), identify the metal ion M (oxidation state) and its -electron count.
,
,
,
,
Q4. Assertion (A): The complex is paramagnetic and adopts a tetrahedral geometry.
Reason (R): Chloride is a weak-field ligand, so for Ni () the pairing energy exceeds the tetrahedral splitting , pairing is not favoured and the complex remains high-spin with two unpaired electrons.
A is true, R is true but R does not correctly explain A
Both A and R are true and R correctly explains A
A is true but R is false
A is false but R is true
Q5. A metal ion with configuration forms an octahedral complex for which . The pairing energy is . Using CFSE values (t electron: , e electron: ) determine whether the complex will be high-spin or low-spin and estimate the net energy difference between the two spin states.
Low-spin () preferred; low-spin more stable by
Low-spin () preferred; low-spin more stable by
High-spin () preferred; high-spin more stable by
Both spin states are almost equally stable since CFSE gain nearly equals pairing penalty
Q6. Determine the oxidation state of cobalt and the number of ‑electrons in the complex .
Oxidation state ,
Oxidation state ,
Oxidation state ,
Oxidation state ,
Q7. Consider the octahedral complexes and . Which complex has the larger spin-only magnetic moment and what is its approximate value? (Use , assume is weak-field and is strong-field.)
has the larger moment;
has the larger moment;
Both complexes have similar moments;
has the larger moment;
Q8. An octahedral complex exhibits a spin-only magnetic moment of about . Using and assuming chloride is a weak-field ligand (high-spin), identify the metal ion and its ‑electron count.
; oxidation state ;
; oxidation state ;
; oxidation state ;
; oxidation state ;
Q9. Assertion (A): is diamagnetic and adopts a square‑planar geometry. Reason (R): is a strong‑field ligand that pairs electrons in (), allowing hybridisation and a square‑planar structure.
A is true but R is false.
A is false but R is true.
Both A and R are true and R is the correct explanation of A.
Both A and R are true but R is not the correct explanation of A.
Q10. How many stereoisomers (including enantiomers) are possible for the octahedral complex , where = ethylenediamine (bidentate)?
2
3 (one trans isomer + a pair of cis enantiomers)
4
1
Q11. Calculate the spin‑only magnetic moment (in Bohr magneton, BM) of the complex . Use the spin‑only formula , where is the number of unpaired electrons.
Q12. A metal ion M forms ML and ML with stepwise formation constants and . If initial concentrations are and (ligand in large excess so ), what percentage of total metal is present as at equilibrium? (Use and .)
Q13. For an octahedral complex of type (three different monodentate ligands A, B, C; two of each), how many stereoisomers are possible when enantiomeric pairs are counted separately? (Count distinct spatial isomers under rotation; non‑superposable mirror images are counted separately.)
Q14. A transition metal ion with electronic configuration forms octahedral complexes. Given pairing energy , compare complexes with ligands L1 and L2 that produce octahedral splitting energies and , respectively. Using CFSE contributions per electron and per electron and assuming low‑spin has one extra pairing relative to high‑spin, determine the spin state for each ligand and the magnitude (in kJ mol) by which the favored spin state is more stable than the other.
L1: high‑spin (favoured by ); L2: low‑spin (favoured by )
L1: low‑spin (favoured by ); L2: high‑spin (favoured by )
L1: low‑spin (LS lower in energy by relative to HS); L2: high‑spin (HS lower in energy by )
L1: high‑spin (favoured by ); L2: high‑spin (favoured by )
Q15. Assertion (A): The square‑planar complex is diamagnetic.
Reason (R): Nickel(II) has electronic configuration .
Choose the correct statement.
Both A and R are true and R is the correct explanation for A.
Both A and R are true but R is not the correct explanation for A.
A is true but R is false.
A is false but R is true.