Coordination compounds form a major and scoring chapter because they link core concepts like oxidation state, geometry, crystal field splitting, magnetism, stereochemistry, and stability constants. CBSE and competitive exams commonly test these ideas through calculation-based MCQs and conceptual HSAB/trans-effect/Jahn–Teller questions, so strong command here directly boosts both accuracy and speed.
20
Minutes
15
Questions
1 / -0
Marking
Q1. Consider the complex ion . Determine the oxidation state of Co and the number of ‑electrons on Co.
Oxidation state , configuration
Oxidation state , configuration
Oxidation state , configuration
Oxidation state , configuration
Q2. For the equilibrium the overall formation constant is . If initial concentrations are and (ligand in large excess so ), estimate the fraction of metal present as at equilibrium.
Approximately
Approximately
Approximately
Approximately
Q3. Which complex has the larger spin‑only magnetic moment: or ? Use where is the number of unpaired electrons.
has larger ; ,
has larger ; ,
Both have similar
,
Q4. Consider octahedral Fe(II) (d) complexes A and B with and . Take pairing energy . Neglect other contributions. For a d ion the energy difference (low‑spin minus high‑spin) can be written as . Which statement is correct?
Both A and B are low‑spin; ,
Both A and B are high‑spin; ,
A is low‑spin, B is high‑spin; ,
A is high‑spin, B is low‑spin; ,
Q5. Assertion (A): The complex exhibits optical isomerism. Reason (R): The coordination number of Co in is six.
A is true, R is true and R is a correct explanation of A
A is true, R is true but R is NOT a correct explanation of A
A is true, R is false
A is false, R is true
Q6. (Calculate the spin-only magnetic moment (in Bohr magneton, BM) of the metal ion in the complex . Use where is the number of unpaired electrons.)
Q7. (A solution of is mixed with . The formation constant for is . Calculate the equilibrium concentration of free (in M). Assume ideal behaviour.)
Q8. (For an octahedral metal ion (Fe) the pairing energy is and the octahedral splitting . Which spin-state is predicted and why?)
High-spin; because
Low-spin; because
High-spin; because additional pairing energy (240 kJ,mol) exceeds CFSE gain (280 kJ,mol)
Low-spin; because CFSE gain upon low-spin formation () is greater than extra pairing energy ()
Q9. (Compare (Fe) and (Fe). Which statement about Fe–C bond length and CN stretching frequency is correct?)
Fe–C bond shorter in ; higher in
Fe–C bond shorter in ; lower in
Fe–C bond lengths equal in both complexes; identical
Fe–C bond shorter in ; higher in
Q10. (Thiocyanate ion SCN is ambidentate. Statement A: In complexes of Al with SCN, the ligand binds preferentially through nitrogen (isothiocyanato). Statement R: According to HSAB principle, hard acids prefer hard bases; Al being a hard acid therefore prefers the harder donor site N over the softer S of SCN.)
A and R are true, and R correctly explains A
A and R are true, but R is not the correct explanation for A
A is false but R is true
A is true but R is false
Q11. Calculate the spin-only magnetic moment (in Bohr magneton, BM) for the high-spin octahedral complex . Use where is the number of unpaired electrons.
3.87 BM
4.90 BM
0.00 BM
2.83 BM
Q12. How many stereoisomers (counting enantiomers separately) exist for the complex ?
3
4
2
5
Q13. is . Experimentally is square-planar and diamagnetic, while is tetrahedral and paramagnetic (two unpaired electrons). Which option best explains this contrast?
Because has a larger ionic radius that forces square-planar coordination.
Because is a strong -donor which increases pairing energy and forces electron pairing.
Because is a stronger-field ligand than , giving a larger splitting and pairing.
Because is a strong-field (often -acceptor) ligand producing large crystal-field splitting that favours electron pairing and square-planar arrangement for , whereas is weak-field giving small splitting and a tetrahedral arrangement with two unpaired electrons.
Q14. Assertion (A): In square-planar complexes the ligand trans to a strong trans-effect ligand is displaced more rapidly than when trans to a weak trans-effect ligand.
Reason (R): Strong trans-effect ligands weaken the metal–ligand bond trans to them by donating electron density or stabilizing the transition state, thereby lowering the activation energy for substitution.
Both A and R are true but R does not explain A.
A is true but R is false.
Both A and R are true and R correctly explains A.
A is false but R is true.
Q15. Assertion (A): Octahedral complexes of () typically exhibit a pronounced Jahn–Teller distortion (usually elongation of two opposite bonds).
Reason (R): Uneven occupancy of the set () in a configuration creates electronic degeneracy that is relieved by a tetragonal distortion, lowering the overall electronic energy.
Both A and R are true but R does not 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.