1. Which description captures the essential structure of a coordination compound?
ⓐ. A substance containing only simple cations and anions joined by electrostatic attraction
ⓑ. A substance containing a central metal atom or ion bonded to surrounding ligands
ⓒ. A molecular compound in which every atom contributes one electron to each bond
ⓓ. A salt formed only when two independently stable salts crystallise together
Correct Answer: A substance containing a central metal atom or ion bonded to surrounding ligands
Explanation: A coordination compound contains a central metal atom or ion surrounded by ions or molecules called ligands. Each ligand attaches to the metal through at least one donor atom. The ligand normally supplies an electron pair for formation of the metal–ligand coordinate bond. Counter ions may also be present, but they are not part of the immediate metal–ligand arrangement. Merely containing a metal ion does not make an ordinary ionic salt a coordination compound.
2. In the coordination entity \(\mathrm{[Co(NH_3)_6]^{3+}}\), the central species and the surrounding ligands are respectively:
ⓐ. \(\mathrm{NH_3}\) and \(\mathrm{Co^{3+}}\)
ⓑ. \(\mathrm{Co^{3+}}\) and six nitrogen atoms
ⓒ. \(\mathrm{Co^{3+}}\) and six \(\mathrm{NH_3}\) molecules
ⓓ. \(\mathrm{Co}\) atoms and three \(\mathrm{NH_3}\) molecules
Correct Answer: \(\mathrm{Co^{3+}}\) and six \(\mathrm{NH_3}\) molecules
Explanation: Cobalt is written first inside the square brackets and acts as the central metal ion. The six \(\mathrm{NH_3}\) molecules surround cobalt and function as ligands. Each \(\mathrm{NH_3}\) ligand coordinates through the lone pair on its nitrogen atom. The subscript \(6\) gives the number of ammonia ligands, not the oxidation state or charge of cobalt. The overall charge \(3+\) belongs to the complete coordination entity.
3. The electron pair used to form a metal–ligand coordinate bond is normally:
ⓐ. supplied by the ligand and accepted by the metal centre
ⓑ. supplied by the metal centre and accepted by the ligand
ⓒ. formed by equal donation of one electron from each participant
ⓓ. transferred completely from the metal to form an ionic ligand
Correct Answer: supplied by the ligand and accepted by the metal centre
Explanation: A ligand possesses at least one donor atom with an available lone pair. It donates this electron pair into a suitable vacant or partly available orbital of the metal centre. Since both bonding electrons initially come from the ligand, the interaction is described as a coordinate bond. After bond formation, the bond is treated as a metal–ligand covalent interaction rather than as a permanently labelled one-way transfer. Complete electron transfer would describe ionic formation rather than coordinate-bond formation.
4. In \(\mathrm{K_4[Fe(CN)_6]}\), the species enclosed by the coordination sphere is:
ⓐ. \(\mathrm{[K_2Fe(CN)_4]^{2-}}\)
ⓑ. \(\mathrm{[KFe(CN)_6]^{3-}}\)
ⓒ. \(\mathrm{[K_4(CN)_6]^{2-}}\)
ⓓ. \(\mathrm{[Fe(CN)_6]^{4-}}\)
Correct Answer: \(\mathrm{[Fe(CN)_6]^{4-}}\)
Explanation: The square brackets enclose the central iron ion together with all directly attached cyanido ligands. This entire bracketed unit is the coordination entity. The four \(\mathrm{K^+}\) ions remain outside the brackets and balance its \(4-\) charge. Iron alone is only the central metal, while \(\mathrm{CN^-}\) represents an individual ligand. Bracket placement separates the persistent complex ion from its counter ions.
5. For \(\mathrm{[Co(NH_3)_6]Cl_3}\), the counter ions are:
ⓐ. six \(\mathrm{NH_3}\) molecules
ⓑ. three \(\mathrm{Cl^-}\) ions
ⓒ. one \(\mathrm{Co^{3+}}\) ion
ⓓ. the complete \(\mathrm{[Co(NH_3)_6]^{3+}}\) ion
Correct Answer: three \(\mathrm{Cl^-}\) ions
Explanation: The bracketed species \(\mathrm{[Co(NH_3)_6]^{3+}}\) is the coordination entity. The three chloride ions written outside the square brackets are counter ions. Their total charge of \(3-\) balances the \(3+\) charge of the complex cation. The ammonia molecules are directly bonded ligands and belong inside the coordination sphere. Moving chloride inside the brackets would change its chemical role from counter ion to coordinated ligand.
6. The distinction between a coordination entity and the complete coordination compound is shown by:
ⓐ. \(\mathrm{Co^{3+}}\) being the entity and \(\mathrm{NH_3}\) being the compound
ⓑ. \(\mathrm{NH_3}\) being the entity and \(\mathrm{Cl^-}\) being the compound
ⓒ. Entity: \(\mathrm{[Co(NH_3)_6]^{3+}}\); compound: \(\mathrm{[Co(NH_3)_6]Cl_3}\)
ⓓ. \(\mathrm{Cl^-}\) being the entity and \(\mathrm{Co^{3+}}\) being the compound
Correct Answer: Entity: \(\mathrm{[Co(NH_3)_6]^{3+}}\); compound: \(\mathrm{[Co(NH_3)_6]Cl_3}\)
Explanation: A coordination entity consists of the central metal and the ligands directly attached to it. In this case, that unit is \(\mathrm{[Co(NH_3)_6]^{3+}}\). The complete electrically neutral compound also includes the three chloride counter ions and is written as \(\mathrm{[Co(NH_3)_6]Cl_3}\). The charge of the coordination entity need not be the same as the charge of the entire compound. Confusing the bracketed ion with the complete formula later leads to errors in charge balance and nomenclature.
7. Assertion: A ligand behaves as a Lewis base during formation of a coordination bond.
Reason: The ligand donates an available electron pair to the central metal atom or ion.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A Lewis base is an electron-pair donor, and a ligand performs this role when it binds to a metal centre. Its donor atom supplies a lone pair to a suitable metal orbital. The metal acts as the corresponding Lewis acid because it accepts the pair. The Reason directly states the electron-pair donation that defines the ligand’s Lewis-base behaviour. A ligand without an available donor pair cannot form this type of coordinate bond through that atom.
8. The claim that every coordination entity must carry an electrical charge is disproved by:
ⓐ. \(\mathrm{[Ag(NH_3)_2]^+}\)
ⓑ. \(\mathrm{[Fe(CN)_6]^{4-}}\)
ⓒ. \(\mathrm{[Ni(CO)_4]}\)
ⓓ. \(\mathrm{[Co(NH_3)_6]^{3+}}\)
Correct Answer: \(\mathrm{[Ni(CO)_4]}\)
Explanation: A coordination entity may be cationic, anionic, or electrically neutral. In \(\mathrm{[Ni(CO)_4]}\), the carbonyl ligands are neutral and nickel has oxidation state \(0\), so the complete entity has no charge. The silver and cobalt examples are complex cations, while the iron example is a complex anion. Neutrality does not remove the defining metal–ligand coordinate bonds. Charge classification and identification as a coordination entity are separate decisions.
9. Match each item in Column I with its role in Column II.
| Column I | Column II |
|---|
| P. Central metal ion | 1. Electron-pair donor |
| Q. Ligand | 2. Receives the donated pair in a suitable orbital |
| R. Available lone pair | 3. Lewis acid centre |
| S. Vacant or partly available metal orbital | 4. Pair of electrons used for donation |
ⓐ. P-1, Q-3, R-2, S-4
ⓑ. P-3, Q-1, R-4, S-2
ⓒ. P-2, Q-4, R-1, S-3
ⓓ. P-4, Q-2, R-3, S-1
Correct Answer: P-3, Q-1, R-4, S-2
Explanation: The central metal ion accepts electron density and therefore acts as a Lewis acid, so P matches \(3\). A ligand donates an electron pair and acts as a Lewis base, so Q matches \(1\). The available lone pair is the pair used for donation, giving R-\(4\). A suitable vacant or partly available metal orbital receives that pair, giving S-\(2\). The donor–acceptor roles refer to electron pairs rather than to complete transfer of a charged particle.
10. Ammonia can coordinate through nitrogen, whereas \(\mathrm{NH_4^+}\) normally cannot coordinate through nitrogen. The best explanation is that:
ⓐ. ammonia is negatively charged but \(\mathrm{NH_4^+}\) is neutral
ⓑ. ammonia has more nitrogen atoms than \(\mathrm{NH_4^+}\)
ⓒ. \(\mathrm{NH_4^+}\) has an empty \(d\) orbital but ammonia does not
ⓓ. \(\mathrm{NH_3}\) has a free lone pair; \(\mathrm{NH_4^+}\) has none
Correct Answer: \(\mathrm{NH_3}\) has a free lone pair; \(\mathrm{NH_4^+}\) has none
Explanation: The nitrogen atom of \(\mathrm{NH_3}\) possesses a lone pair that can be donated to a metal centre. During formation of \(\mathrm{NH_4^+}\), this lone pair is used to bond with \(\mathrm{H^+}\). Nitrogen in \(\mathrm{NH_4^+}\) therefore has no comparable free lone pair available for ordinary coordination through nitrogen. Ligand behaviour depends on donor-pair availability rather than simply on whether a species is neutral or charged. A molecule containing a possible donor atom is not necessarily a ligand if its electron pair is unavailable.
11. In the charge-balance relation
\[
x+\sum(\text{ligand charges})=\text{charge on the coordination entity},
\]
the symbol \(x\) represents ______.
ⓐ. the oxidation state of the central metal
ⓑ. the number of ligands outside the brackets
ⓒ. the coordination number in every complex
ⓓ. the charge on a single counter ion
Correct Answer: the oxidation state of the central metal
Explanation: The unknown \(x\) is assigned to the oxidation state of the central metal atom or ion. The charges of all coordinated ligands are then added algebraically to it. Their sum must equal the net charge written on the coordination entity. Coordination number is obtained by counting directly bonded donor atoms and is not represented by \(x\) in this relation. Keeping oxidation state, coordination number, and complex charge separate prevents three different quantities from being treated as interchangeable.
12. A learner treats oxidation state and valency as identical in every situation. The most suitable correction is:
ⓐ. Both are always signed charges written on the complete compound
ⓑ. Oxidation state is signed electron bookkeeping; valency is unsigned combining capacity
ⓒ. Valency is the charge on the coordination entity, whereas oxidation state is the number of ligands
ⓓ. Oxidation state and valency both equal the coordination number of the metal
Correct Answer: Oxidation state is signed electron bookkeeping; valency is unsigned combining capacity
Explanation: Oxidation state is a formal electron-bookkeeping value and may be positive, negative, or zero. Valency expresses combining capacity and is traditionally written without an algebraic sign. Their numerical magnitudes may coincide in some simple substances, but that does not make the concepts universally identical. Neither quantity is defined as the number of ligands or donor atoms attached to the metal. Coordination chemistry requires oxidation state, complex charge, and coordination number to be determined independently.
13. An iron atom has the ground-state configuration \(\mathrm{[Ar]\,3d^6\,4s^2}\). The \(d\)-electron configuration of \(\mathrm{Fe^{3+}}\) is:
ⓐ. \(\mathrm{3d^3}\)
ⓑ. \(\mathrm{3d^4}\)
ⓒ. \(\mathrm{3d^6}\)
ⓓ. \(\mathrm{3d^5}\)
Correct Answer: \(\mathrm{3d^5}\)
Explanation: \( \textbf{Neutral-atom configuration:} \) Iron is given as \(\mathrm{[Ar]\,3d^6\,4s^2}\).
\( \textbf{Ion required:} \) Formation of \(\mathrm{Fe^{3+}}\) requires removal of three electrons.
\( \textbf{Removal order:} \) For transition-metal cations, the \(\mathrm{4s}\) electrons are removed before the \(\mathrm{3d}\) electrons.
\( \textbf{First two electrons removed:} \)
\[
\mathrm{[Ar]\,3d^6\,4s^2\rightarrow[Ar]\,3d^6}
\]
\( \textbf{Third electron removed:} \)
\[
\mathrm{[Ar]\,3d^6\rightarrow[Ar]\,3d^5}
\]
\( \textbf{Resulting ion configuration:} \) \(\mathrm{Fe^{3+}}\) is a \(d^5\) ion.
\( \textbf{Final answer:} \) The required configuration is \(\mathrm{3d^5}\). Subtracting all three electrons directly from the written \(3d\) exponent would ignore the earlier removal of the two \(4s\) electrons.
14. In the complex ion \(\mathrm{[Fe(CN)_6]^{4-}}\), the oxidation state of iron is:
ⓐ. \(+4\)
ⓑ. \(-2\)
ⓒ. \(+2\)
ⓓ. \(+6\)
Correct Answer: \(+2\)
Explanation: \( \textbf{Complex charge:} \) The complete coordination entity has charge \(4-\).
\( \textbf{Ligand charge:} \) Each cyanido ligand, \(\mathrm{CN^-}\), contributes a charge of \(-1\).
\( \textbf{Number of ligands:} \) There are six cyanido ligands.
\( \textbf{Let the iron oxidation state be:} \) \(x\).
\( \textbf{Charge-balance equation:} \)
\[
x+6(-1)=-4
\]
\( \textbf{Simplification:} \)
\[
x-6=-4
\]
\( \textbf{Solve for the metal:} \)
\[
x=+2
\]
\( \textbf{Final answer:} \) Iron has oxidation state \(+2\). The \(4-\) charge belongs to the entire complex ion and is not the oxidation state of iron.
15. The dissolution of \(\mathrm{K_4[Fe(CN)_6]}\) in water is best represented by:
ⓐ. \(\mathrm{K_4[Fe(CN)_6]\rightarrow4K^++[Fe(CN)_6]^{4-}}\)
ⓑ. \(\mathrm{K_4[Fe(CN)_6]\rightarrow2K^++[Fe(CN)_6]^{2-}}\)
ⓒ. \(\mathrm{K_4[Fe(CN)_6]\rightarrow4K^{2+}+[Fe(CN)_6]^{4-}}\)
ⓓ. \(\mathrm{K_4[Fe(CN)_6]\rightarrow4K^++Fe^{2+}+6CN^-}\)
Correct Answer: \(\mathrm{K_4[Fe(CN)_6]\rightarrow4K^++[Fe(CN)_6]^{4-}}\)
Explanation: Potassium ions lie outside the coordination sphere and separate as four \(\mathrm{K^+}\) ions in water. The bracketed ferrocyanide ion remains intact as \(\mathrm{[Fe(CN)_6]^{4-}}\). Four positive charges and one \(4-\) complex charge preserve electrical neutrality. Complete separation into \(\mathrm{Fe^{2+}}\) and \(\mathrm{CN^-}\) would incorrectly treat the coordination compound as a double salt. The alternatives with two potassium ions or \(\mathrm{K^{2+}}\) fail the stoichiometric or ionic-charge check.
16. Separate aqueous solutions of Mohr’s salt and \(\mathrm{K_4[Fe(CN)_6]}\) are treated with a reagent that tests for free \(\mathrm{Fe^{2+}}\). The expected observation is that:
ⓐ. both solutions respond identically because both formulas contain iron
ⓑ. only \(\mathrm{K_4[Fe(CN)_6]}\) responds because iron lies inside square brackets
ⓒ. Mohr’s salt releases free \(\mathrm{Fe^{2+}}\); ferrocyanide does not
ⓓ. neither solution responds because iron-containing salts never form simple ions
Correct Answer: Mohr’s salt releases free \(\mathrm{Fe^{2+}}\); ferrocyanide does not
Explanation: Mohr’s salt is a double salt and dissociates in water to give simple ions, including \(\mathrm{Fe^{2+}}\). Potassium ferrocyanide gives the stable complex ion \(\mathrm{[Fe(CN)_6]^{4-}}\), so its iron is not present as free \(\mathrm{Fe^{2+}}\). Merely finding iron in the formula does not show that the same iron species exists in solution. Qualitative tests depend on the actual dissolved ions rather than on elemental composition alone. This difference in ion availability is evidence for retention of the coordination entity.
17. Ignoring water of crystallisation, the number of moles of simple ions produced by complete dissociation of \(1\,\mathrm{mol}\) of Mohr’s salt, \(\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}\), is:
ⓐ. \(3\,\mathrm{mol}\)
ⓑ. \(4\,\mathrm{mol}\)
ⓒ. \(6\,\mathrm{mol}\)
ⓓ. \(5\,\mathrm{mol}\)
Correct Answer: \(5\,\mathrm{mol}\)
Explanation: \( \textbf{Formula considered:} \) \(\mathrm{(NH_4)_2Fe(SO_4)_2\cdot6H_2O}\).
\( \textbf{Role of water:} \) The \(6\mathrm{H_2O}\) molecules are water of crystallisation and are not counted as ions.
\( \textbf{Dissociation of the ionic part:} \)
\[
\mathrm{(NH_4)_2Fe(SO_4)_2\rightarrow2NH_4^++Fe^{2+}+2SO_4^{2-}}
\]
\( \textbf{Ammonium-ion count:} \) \(1\,\mathrm{mol}\) of the salt gives \(2\,\mathrm{mol}\) of \(\mathrm{NH_4^+}\).
\( \textbf{Iron-ion count:} \) It gives \(1\,\mathrm{mol}\) of \(\mathrm{Fe^{2+}}\).
\( \textbf{Sulfate-ion count:} \) It gives \(2\,\mathrm{mol}\) of \(\mathrm{SO_4^{2-}}\).
\( \textbf{Total simple ions:} \)
\[
2+1+2=5\,\mathrm{mol}
\]
\( \textbf{Charge verification:} \) Total positive charge is \(+4\), and total negative charge is \(-4\).
\( \textbf{Final answer:} \) The salt produces \(5\,\mathrm{mol}\) of simple ions per mole, excluding water molecules.
18. Study the comparison below.
| Substance | Main dissolved species proposed | Identity in solution |
|---|
| P. Mohr’s salt | \(\mathrm{NH_4^+}\), \(\mathrm{Fe^{2+}}\), \(\mathrm{SO_4^{2-}}\) | Original double-salt identity lost |
| Q. \(\mathrm{K_4[Fe(CN)_6]}\) | \(\mathrm{K^+}\), \(\mathrm{[Fe(CN)_6]^{4-}}\) | Complex-ion identity retained |
| R. \(\mathrm{K_4[Fe(CN)_6]}\) | \(\mathrm{K^+}\), \(\mathrm{Fe^{2+}}\), \(\mathrm{CN^-}\) | Complex-ion identity retained |
| S. Mohr’s salt | Only one undissociated neutral particle | Simple ions absent |
The two chemically consistent rows are:
ⓐ. P and R
ⓑ. P and Q
ⓒ. Q and S
ⓓ. R and S
Correct Answer: P and Q
Explanation: Row P represents the normal behaviour of a double salt because Mohr’s salt gives its simple constituent ions in water. Row Q correctly shows the dissociation of potassium ferrocyanide into potassium ions and an intact complex anion. Row R is inconsistent because it shows free \(\mathrm{Fe^{2+}}\) and \(\mathrm{CN^-}\) while simultaneously claiming that the complex identity is retained. Row S is inconsistent because a soluble double salt does not remain entirely as one neutral undissociated particle. The decisive comparison is between simple-ion formation and preservation of the bracketed coordination entity.
19. One mole each of Mohr’s salt and \(\mathrm{K_4[Fe(CN)_6]}\) can both produce five moles of ions on ideal dissociation. This observation shows that:
ⓐ. total ion count alone cannot distinguish a double salt from a coordination compound
ⓑ. both substances must be double salts because their ion counts are equal
ⓒ. both substances must release free \(\mathrm{Fe^{2+}}\) because each contains iron
ⓓ. equal ion counts prove that the dissolved ions have identical chemical identities
Correct Answer: total ion count alone cannot distinguish a double salt from a coordination compound
Explanation: Mohr’s salt gives \(2\mathrm{NH_4^+}\), \(\mathrm{Fe^{2+}}\), and \(2\mathrm{SO_4^{2-}}\), amounting to five ions per formula unit. Potassium ferrocyanide gives \(4\mathrm{K^+}\) and one \(\mathrm{[Fe(CN)_6]^{4-}}\) ion, which also totals five ions. The numerical count is therefore the same, but the nature of the particles is very different. Mohr’s salt produces simple constituent ions, whereas potassium ferrocyanide preserves a complex ion. Identification must use ionic identity or characteristic reactions rather than ion count alone.
20. Aqueous Mohr’s salt gives a precipitate when tested for sulfate ions because:
ⓐ. sulfate remains permanently enclosed in a coordination sphere
ⓑ. iron changes its oxidation state from \(+2\) to \(+6\)
ⓒ. ammonium ions are converted into sulfate ions
ⓓ. free \(\mathrm{SO_4^{2-}}\) ions are released by the double salt
Correct Answer: free \(\mathrm{SO_4^{2-}}\) ions are released by the double salt
Explanation: Mohr’s salt dissociates to give freely hydrated sulfate ions. These ions can participate in ordinary qualitative reactions, such as formation of an insoluble sulfate with a suitable reagent. Their availability reflects the loss of the double salt’s separate identity after dissolution. The sulfate ions are not hidden inside a persistent metal–ligand coordination sphere. Qualitative analysis therefore detects the simple ions represented in the constituent salts.
