Board Previous Year Questions

How to Score Full Marks in the Board Exam

A complete bank of board-style questions with model answers for Coordination Compounds. Use exact IUPAC spellings (ammine, ferrate), state the metal's d-configuration before predicting hybridisation, and draw clear cis/trans or enantiomer structures. For VBT/CFT questions, name the hybridisation and number of unpaired electrons explicitly.

1-Mark Questions (Definitions & Direct)

Q1. Define a ligand. Answer: A ligand is an ion or molecule that donates a lone pair of electrons (a Lewis base) to the central metal atom/ion, forming a coordinate bond.

Q2. What is the coordination number of a metal ion in a complex? Answer: It is the number of donor atoms directly bonded to the central metal ion.

Q3. Write the IUPAC name of [Co(NH3)6]Cl3. Answer: Hexaamminecobalt(III) chloride.

Q4. What is a chelate complex? Answer: A complex in which a bidentate or polydentate ligand binds the metal through two or more donor atoms, forming a ring (e.g. with ethylenediamine).

1-Mark Questions (continued)

Q5. What are ambidentate ligands? Give an example. Answer: Ligands that can coordinate through either of two different donor atoms, one at a time; for example, NO$_2^-$ (through N or O) and SCN$^-$ (through S or N).

Q6. Give the IUPAC name of K4[Fe(CN)6]. Answer: Potassium hexacyanidoferrate(II).

Q7. What is the oxidation state of nickel in [Ni(CO)4]? Answer: Zero (CO is neutral and the complex is neutral).

Q8. State the type of hybridisation and geometry of [Ni(CN)4]2-. Answer: dsp$^2$ hybridisation, square planar geometry.

2-Mark Reasoning Questions

Q9. Explain, using valence bond theory, why [Ni(CN)4]2- is square planar and diamagnetic. Answer: Ni$^{2+}$ is $d^8$. The strong-field CN$^-$ ligand causes the 8 d electrons to pair up, leaving one $(n-1)d$ orbital vacant. This vacant d orbital is used in dsp$^2$ hybridisation, giving a square planar geometry. As all electrons are paired, the complex is diamagnetic.

Q10. Why is [NiCl4]2- paramagnetic while [Ni(CN)4]2- is diamagnetic, though both contain Ni2+? Answer: Cl$^-$ is a weak-field ligand: it does not pair the d electrons, so Ni$^{2+}$ uses sp$^3$ hybridisation (tetrahedral) with 2 unpaired electrons — paramagnetic. CN$^-$ is a strong-field ligand: it pairs the d electrons, giving dsp$^2$ (square planar) with 0 unpaired electrons — diamagnetic.

2-Mark Reasoning Questions (continued)

Q11. Why is [Fe(H2O)6]3+ strongly paramagnetic while [Fe(CN)6]3- is weakly paramagnetic? Answer: Both contain Fe$^{3+}$ ($d^5$). H$_2$O is a weak-field ligand, so the complex is high spin with 5 unpaired electrons (strongly paramagnetic). CN$^-$ is a strong-field ligand, so the complex is low spin with only 1 unpaired electron (weakly paramagnetic).

Q12. What is the chelate effect? Why is [Ni(en)3]2+ more stable than [Ni(NH3)6]2+? Answer: The chelate effect is the extra stability of complexes containing chelating (polydentate) ligands compared with similar complexes of unidentate ligands. Replacing six unidentate NH$_3$ ligands with three chelating en ligands increases the number of free particles in solution, giving a favourable increase in entropy that makes [Ni(en)$_3$]$^{2+}$ more stable.

Q13. Why do coordination compounds show colour? Answer: In a ligand field the d orbitals of the metal split into two sets separated by $\Delta_o$. A d electron absorbs visible light to undergo a d-d transition from the lower to the higher set; the colour observed is complementary to the light absorbed. ($d^0$ and $d^{10}$ ions are usually colourless.)

2-3 Mark Reasoning Questions

Q14. State the postulates of Werner's theory of coordination compounds. Answer: (i) Metals show two kinds of valency — primary (ionisable) and secondary (non-ionisable). (ii) The primary valency is satisfied by negative ions and corresponds to the oxidation state. (iii) The secondary valency is satisfied by ligands, is directional and fixes the geometry, and corresponds to the coordination number. (iv) The secondary valencies are directed toward fixed positions in space, giving the complex a definite geometry.

Q15. What is crystal field splitting? Draw the splitting of d orbitals in an octahedral field. Answer: When ligands approach a metal ion, the five degenerate d orbitals split into a lower-energy set of three orbitals ($t_{2g}$: $d_{xy}$, $d_{yz}$, $d_{zx}$) and a higher-energy set of two orbitals ($e_g$: $d_{x^2-y^2}$, $d_{z^2}$), separated by the crystal field splitting energy $\Delta_o$. The $e_g$ orbitals point along the axes, toward the ligands, and so are raised more in energy.

Q16. Distinguish between a double salt and a complex compound with one example each. Answer: A double salt dissociates completely into all its ions in solution (e.g. Mohr's salt, FeSO$_4\cdot$(NH$_4$)$_2$SO@@GYANGHAR_MATH@@126H$_2$O, gives Fe$^{2+}$, NH$_4^+$ and SO$_4^{2-}$). A complex compound retains its coordination entity in solution (e.g. K$_4$[Fe(CN)$_6$] gives K$^+$ and the intact [Fe(CN)$_6$]$^{4-}$, not free Fe$^{2+}$ or CN$^-$).

3-Mark Numericals & Naming

Q17. Write the IUPAC name of [Co(NH3)5Cl]Cl2. Answer: Ligands in alphabetical order (ammine before chlorido): pentaamminechloridocobalt(III) chloride.

Q18. Determine the oxidation number and coordination number of cobalt in [Co(en)3]3+. Answer: en (ethylenediamine) is a neutral bidentate ligand. Oxidation number of Co = +3. Coordination number = 3 × 2 = 6.

Q19. Calculate the spin-only magnetic moment of [NiCl4]2- (Ni2+, d8, tetrahedral). Answer: Cl$^-$ is weak field, so Ni$^{2+}$ ($d^8$) has 2 unpaired electrons. $\mu = \sqrt{2(2+2)} = \sqrt{8} = \mathbf{2.83\ BM}$.

3-Mark Numericals & Naming (continued)

Q20. Write the formula of potassium hexacyanidoferrate(III) and the oxidation state of iron. Answer: Fe(III) with 6 CN$^-$ gives [Fe(CN)$_6$]$^{3-}$, balanced by 3 K$^+$: K$_3$[Fe(CN)$_6$]; iron is in the +3 state.

Q21. Predict the hybridisation, geometry and magnetic behaviour of [Co(NH3)6]3+ (Co3+, d6). Answer: NH$_3$ is a strong-field ligand, so it pairs the d electrons, freeing two inner d orbitals for d$^2$sp$^3$ hybridisation → octahedral, with 0 unpaired electrons → diamagnetic (an inner-orbital/low-spin complex).

Q22. Calculate the spin-only magnetic moment of [Fe(CN)6]3- (Fe3+, d5, strong field). Answer: CN$^-$ is strong field, so the complex is low spin with 1 unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} = \mathbf{1.73\ BM}$.

3-Mark Isomerism Questions

Q23. Draw and name the geometrical isomers of [Pt(NH3)2Cl2] (square planar). Answer: There are two: the cis isomer (the two Cl ligands adjacent, 90° apart — this is the anticancer drug cisplatin) and the trans isomer (the two Cl ligands opposite, 180° apart).

Q24. What type of isomerism is shown by [Co(NH3)5(NO2)]2+ and [Co(NH3)5(ONO)]2+? Answer: Linkage isomerism — the ambidentate ligand NO$_2^-$ is bonded through nitrogen (nitro) in the first and through oxygen (nitrito) in the second.

Q25. Give evidence that [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are ionisation isomers. Answer: They give different ions in solution. [Co(NH$_3$)$_5$Br]SO$_4$ gives free SO$_4^{2-}$ (white precipitate with BaCl$_2$), whereas [Co(NH$_3$)$_5$SO$_4$]Br gives free Br$^-$ (pale yellow precipitate with AgNO$_3$). The different precipitation tests prove they are ionisation isomers.

3-Mark Reasoning Questions

Q26. Why does a tetrahedral complex of the type [MA2B2] not show geometrical isomerism? Answer: In a tetrahedron all four corner positions are equivalent and mutually adjacent — there is no "opposite" (trans) position. Hence there is no cis/trans distinction, so tetrahedral [MA$_2$B$_2$] complexes do not show geometrical isomerism.

Q27. Why is [Co(en)3]3+ optically active? Answer: It has no plane of symmetry and its mirror image is non-superimposable, so it exists as two enantiomers (d and l forms). It is therefore chiral and optically active.

Q28. On the basis of crystal field theory, explain why CN- is a strong-field ligand and gives a low-spin complex. Answer: CN$^-$ produces a large crystal field splitting $\Delta_o$. When $\Delta_o$ is greater than the pairing energy ($\Delta_o > P$), it is energetically favourable for electrons to pair in the lower $t_{2g}$ set rather than occupy the higher $e_g$ set. This gives a low-spin complex with fewer unpaired electrons.

3-Mark Reasoning & Applications

Q29. Explain the bonding in metal carbonyls such as Ni(CO)4 (synergic bonding). Answer: The metal-carbon bond in a carbonyl is synergic: (i) a sigma bond forms when the carbon lone pair of CO donates into an empty metal orbital (ligand → metal); (ii) a pi back-bond forms when a filled metal d orbital donates electron density into the empty $\pi^*$ antibonding orbital of CO (metal → ligand). The two effects reinforce each other, strengthening the M-C bond.

Q30. Name the central metal ion present in (a) chlorophyll, (b) haemoglobin and (c) vitamin B12. Answer: (a) Magnesium (Mg), (b) Iron (Fe), (c) Cobalt (Co).

Q31. Why is the cis isomer of [Pt(NH3)2Cl2] biologically important? Answer: The cis isomer (cisplatin) is an anticancer drug — it binds to DNA in cancer cells and inhibits their division. The trans isomer is biologically inactive.

5-Mark / Long-Answer Questions

Q32. Using valence bond theory, explain the geometry and magnetic behaviour of (a) [CoF6]3- and (b) [Co(NH3)6]3+. Answer: Both contain Co$^{3+}$ ($d^6$). (a) In [CoF6]3-, F$^-$ is a weak-field ligand, so the d electrons are not paired; Co$^{3+}$ uses outer d orbitals in sp$^3$d$^2** hybridisation → octahedral, with **4 unpaired electrons** → paramagnetic (outer-orbital/high-spin complex). (b) In **[Co(NH3)6]3+**, NH$_3$is stronger and pairs the d electrons, freeing inner d orbitals for **d$^2$sp$^3 hybridisation → octahedral, with 0 unpaired electrons → diamagnetic (inner-orbital/low-spin complex).

Q33. Explain crystal field splitting in octahedral and tetrahedral complexes, and state the high-spin/low-spin condition. Answer: In an octahedral field the d orbitals split into a lower $t_{2g}$ set (−0.4$\Delta_o$) and a higher $e_g$ set (+0.6$\Delta_o$), separated by $\Delta_o$. In a tetrahedral field the splitting is inverted and smaller, with $\Delta_t = \dfrac{4}{9}\Delta_o$. A complex is low spin when $\Delta_o > P$ (strong-field ligand; electrons pair in $t_{2g}$) and high spin when $\Delta_o < P$ (weak-field ligand; electrons occupy $e_g$ singly). Because $\Delta_t$ is small, tetrahedral complexes are almost always high spin.

5-Mark / Long-Answer Questions (continued)

Q34. Write the IUPAC names of (a) [Cr(NH3)6]Cl3, (b) [Co(en)3]Cl3, (c) K2[PtCl4], (d) [Ni(CO)4]. Answer: (a) hexaamminechromium(III) chloride; (b) tris(ethylenediamine)cobalt(III) chloride; (c) potassium tetrachloridoplatinate(II); (d) tetracarbonylnickel(0). (Note the use of tris for en, the -ate ending for the anionic platinum complex, and ammine with two m's for NH$_3$.)

Q35. Describe the types of isomerism shown by coordination compounds, with one example of each. Answer: Structural isomerism: (i) linkage — [Co(NH$_3$)$_5$(NO$_2$)]$^{2+}$ vs (ONO); (ii) coordination — [Co(NH$_3$)$_6$][Cr(CN)$_6$] vs [Cr(NH$_3$)$_6$][Co(CN)$_6$]; (iii) ionisation — [Co(NH$_3$)$_5$Br]SO$_4$ vs [Co(NH$_3$)$_5$SO$_4$]Br; (iv) solvate/hydrate — [Cr(H$_2$O)$_6$]Cl$_3$ vs [Cr(H$_2$O)$_5$Cl]Cl$_2$·H$_2$O. Stereoisomerism: (v) geometrical (cis/trans) — cis- and trans-[Pt(NH$_3$)$_2$Cl$_2$]; (vi) optical — the d and l enantiomers of [Co(en)$_3$]$^{3+}$.

Q36. Give the applications of coordination compounds in (a) biology, (b) analysis, and (c) medicine. Answer: (a) Biology: haemoglobin (Fe complex) carries oxygen, chlorophyll (Mg complex) captures light, and vitamin B$_{12}$ is a cobalt complex. (b) Analysis: EDTA complexometric titrations estimate the hardness of water (Ca$^{2+}$, Mg$^{2+}$); coloured complexes are used in qualitative and quantitative analysis. (c) Medicine: cisplatin [Pt(NH$_3$)$_2$Cl$_2$] is an anticancer drug, and chelation therapy (EDTA, D-penicillamine) removes toxic metals such as lead and mercury from the body.