1. Which statement best describes a coordination compound?
ⓐ. A compound that must contain two different simple salts crystallised together
ⓑ. A compound in which the metal is attached only by ordinary ionic bonds to all surrounding species
ⓒ. A compound with a central metal atom or ion bonded to ligands by coordinate bonds
ⓓ. A compound that can exist only when all ligands are neutral molecules
Correct Answer: A compound with a central metal atom or ion bonded to ligands by coordinate bonds
Explanation: A coordination compound contains a central metal atom or ion and surrounding species called ligands. These ligands donate lone pairs to the metal and form coordinate bonds. The resulting coordination entity behaves as a distinct unit. Such compounds are different from ordinary salts and also different from simple double salts.
2. In \([Cu(NH_3)_4]SO_4\), which species is the coordination entity?
ⓐ. \([Cu(NH_3)_4]^{2+}\)
ⓑ. \([Cu(NH_3)_4]SO_4\)
ⓒ. \([Cu(NH_3)_2]^{2+}\)
ⓓ. \([Cu(SO_4)_2]^{2-}\)
Correct Answer: \([Cu(NH_3)_4]^{2+}\)
Explanation: The coordination entity is the part in which the central metal and ligands are directly bound together. Here \(Cu^{2+}\) is surrounded by four \(NH_3\) ligands inside the square brackets. The sulfate ion lies outside the bracket and acts as a counter ion. So the complex cation \([Cu(NH_3)_4]^{2+}\) is the distinct coordination entity.
3. Which pair correctly lists a coordination compound first and a double salt second?
ⓐ. \([Cu(NH_3)_4]SO_4\) and \(K_4[Fe(CN)_6]\)
ⓑ. \(NaCl\) and \([Cu(NH_3)_4]SO_4\)
ⓒ. \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\) and \(K_4[Fe(CN)_6]\)
ⓓ. \(K_4[Fe(CN)_6]\) and \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\)
Correct Answer: \(K_4[Fe(CN)_6]\) and \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\)
Explanation: \(K_4[Fe(CN)_6]\) contains the complex ion \([Fe(CN)_6]^{4-}\), so it is a coordination compound. Mohr’s salt, \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\), is a double salt. In water, a double salt loses its identity and yields simple ions, while a coordination compound keeps its complex ion intact. That difference is the key basis for the classification.
4. When \(K_4[Fe(CN)_6]\) dissolves in water, which species are obtained?
ⓐ. \(4K^+ + Fe^{2+} + 6CN^-\)
ⓑ. \(4K^+ + [Fe(CN)_6]^{4-}\)
ⓒ. \(K^+ + [Fe(CN)_6]^-\)
ⓓ. \(4K^+ + Fe^{3+} + 6CN^-\)
Correct Answer: \(4K^+ + [Fe(CN)_6]^{4-}\)
Explanation: \(\textbf{Given:}\)
Compound \(K_4[Fe(CN)_6]\)
\(\textbf{Required:}\)
Species formed on dissolving it in water
\(\textbf{Relevant Principle:}\)
A coordination compound retains its coordination entity in solution, and only the outer counter ions separate.
\(\textbf{Identify the coordination entity:}\)
The bracketed part is \([Fe(CN)_6]^{4-}\).
\(\textbf{Dissociation in water:}\)
\[K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}\]
\(\textbf{Why free }Fe^{2+}\textbf{ and }CN^-\textbf{ are not written:}\)
The six \(CN^-\) ligands remain attached to iron inside the complex ion.
\(\textbf{Final Answer:}\)
The dissolved species are \(4K^+\) and \([Fe(CN)_6]^{4-}\).
5. Which substance behaves as a double salt in aqueous solution?
ⓐ. tetraamminecopper(II) sulfate
ⓑ. potassium hexacyanidoferrate(II)
ⓒ. Mohr’s salt
ⓓ. potassium dicyanidoargentate(I)
Correct Answer: Mohr’s salt
Explanation: A double salt dissociates completely in water and gives all the constituent simple ions. Mohr’s salt, \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\), behaves in this way. The other listed substances contain a coordination entity that remains intact in solution. So their behaviour matches coordination compounds rather than double salts.
6. In \([Cu(NH_3)_4]SO_4\), which combination correctly identifies the central metal ion and the ligand?
ⓐ. Central metal ion \(SO_4^{2-}\), ligand \(NH_3\)
ⓑ. Central metal ion \(Cu^{2+}\), ligand \(SO_4^{2-}\)
ⓒ. Central metal ion \(NH_3\), ligand \(Cu^{2+}\)
ⓓ. Central metal ion \(Cu^{2+}\), ligand \(NH_3\)
Correct Answer: Central metal ion \(Cu^{2+}\), ligand \(NH_3\)
Explanation: The central metal ion is the species that accepts lone pairs from surrounding donor molecules or ions. In this complex, \(Cu^{2+}\) occupies the centre and each \(NH_3\) molecule donates a lone pair to it. The sulfate ion is not directly bonded to copper inside the bracketed unit. That is why \(NH_3\) is the ligand and \(Cu^{2+}\) is the central metal ion.
7. What kind of bond is formed when a ligand donates a lone pair to the central metal ion in a coordination compound?
ⓐ. Coordinate bond
ⓑ. Hydrogen bond
ⓒ. Metallic bond
ⓓ. Non-polar covalent bond
Correct Answer: Coordinate bond
Explanation: In a coordination compound, the ligand supplies both electrons of the shared pair to the metal ion. This type of metal-ligand linkage is called a coordinate bond or coordinate covalent bond. The bond still has covalent character, but its formation starts from lone-pair donation by the ligand. This idea explains why molecules such as \(NH_3\) and \(H_2O\) can act as ligands.
8. A salt dissolves in water and its solution gives the usual tests for \(Fe^{2+}\), \(NH_4^+\), and \(SO_4^{2-}\) separately. Which salt is most likely to show this behaviour?
ⓐ. potassium hexacyanidoferrate(II)
ⓑ. Mohr’s salt
ⓒ. tetraamminecopper(II) sulfate
ⓓ. potassium dicyanidoargentate(I)
Correct Answer: Mohr’s salt
Explanation: \(\textbf{Given:}\)
The solution shows the usual tests for \(Fe^{2+}\), \(NH_4^+\), and \(SO_4^{2-}\) separately.
\(\textbf{Required:}\)
Identify the salt that dissociates into these simple ions in water
\(\textbf{Relevant Principle:}\)
Double salts ionise completely in water, while coordination compounds retain their complex ion.
\(\textbf{Apply the idea:}\)
Mohr’s salt, \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\), breaks into simple ions on dissolution.
\(\textbf{Interpretation:}\)
That is why the individual ions can be detected by ordinary qualitative tests.
\(\textbf{Final Answer:}\)
The salt is \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\).
9. Which observation most strongly indicates that a substance is a coordination compound rather than a double salt?
ⓐ. It contains more than one ionic species in the solid state
ⓑ. It gives an intact complex ion in water
ⓒ. It contains water of crystallisation in its formula
ⓓ. It is formed by mixing two salts in solution
Correct Answer: It gives an intact complex ion in water
Explanation: The key feature of a coordination compound is the persistence of its coordination entity in solution. When such a compound dissolves, the complex ion usually remains intact and only counter ions separate. A double salt does not behave that way. It breaks into its simple constituent ions and loses the special identity seen in the crystal.
10. Which statement is not true for a double salt in aqueous solution?
ⓐ. It dissociates into simple ions
ⓑ. Its constituent ions can usually be identified by ordinary qualitative tests
ⓒ. It loses the special combined identity seen in the crystal
ⓓ. It retains a complex ion as a separate coordinated unit
Correct Answer: It retains a complex ion as a separate coordinated unit
Explanation: A double salt behaves like a combination of ordinary salts when dissolved in water. Its ions separate completely, so the original crystal arrangement is no longer preserved in solution. That is why the usual tests for the constituent ions can be performed. Retention of a coordinated unit is a property of coordination compounds, not of double salts.
11. In aqueous solution of \(K_4[Fe(CN)_6]\), which species remains attached to iron instead of appearing as a free ion?
ⓐ. \(CN^-\)
ⓑ. \(K^+\)
ⓒ. \(Fe^{2+}\)
ⓓ. \(OH^-\)
Correct Answer: \(CN^-\)
Explanation: In \(K_4[Fe(CN)_6]\), the six \(CN^-\) groups are ligands directly bonded to iron in the complex ion \([Fe(CN)_6]^{4-}\). On dissolving the compound, the \(K^+\) ions separate, but the bracketed complex remains intact. So cyanide does not appear as free \(CN^-\) under ordinary simple dissociation. This is one of the clearest signs that the substance is a coordination compound.
12. In \([Cu(NH_3)_4]SO_4\), which species lies outside the coordination sphere?
ⓐ. \(Cu^{2+}\)
ⓑ. \(NH_3\)
ⓒ. \(SO_4^{2-}\)
ⓓ. \([Cu(NH_3)_4]^{2+}\)
Correct Answer: \(SO_4^{2-}\)
Explanation: The coordination sphere consists of the central metal ion and the ligands directly attached to it. In this compound, \(Cu^{2+}\) and four \(NH_3\) molecules form the complex cation \([Cu(NH_3)_4]^{2+}\). The sulfate ion is not inside the bracketed part. It acts as a counter ion and therefore lies outside the coordination sphere.
13. One formula unit of \(K_4[Fe(CN)_6]\) produces how many ions in water?
ⓐ. \(4\)
ⓑ. \(5\)
ⓒ. \(6\)
ⓓ. \(10\)
Correct Answer: \(5\)
Explanation: \(\textbf{Given:}\)
Compound \(K_4[Fe(CN)_6]\)
\(\textbf{Required:}\)
Total number of ions formed from one formula unit in water
\(\textbf{Relevant Principle:}\)
A coordination compound dissociates into outer-sphere ions, while the coordination entity remains intact.
\(\textbf{Identify the dissociation:}\)
\[K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-}\]
\(\textbf{Count the ions formed:}\)
There are \(4\) potassium ions and \(1\) complex ion.
\(\textbf{Intermediate Simplification:}\)
\[4 + 1 = 5\]
\(\textbf{Why }6\textbf{ or }10\textbf{ are wrong:}\)
The six \(CN^-\) ligands do not separate as free ions during simple dissolution.
\(\textbf{Final Answer:}\)
Total ions formed \(= 5\)
14. The square-bracketed part of a coordination compound, containing the metal and the directly attached ligands, is called the:
ⓐ. crystal lattice unit
ⓑ. double salt unit
ⓒ. solvate shell unit
ⓓ. coordination sphere
Correct Answer: coordination sphere
Explanation: The bracketed part marks the region where the metal ion and ligands are directly connected. That whole unit is referred to as the coordination sphere, and it often behaves as one entity in solution. Species outside the bracket usually serve as counter ions. This bracket-based distinction is essential for identifying what remains intact on dissolution.
15. Which substance is expected to retain a distinct complex ion after dissolving in water?
ⓐ. \(Na_2SO_4 \cdot 10H_2O\)
ⓑ. \(KAl(SO_4)_2 \cdot 12H_2O\)
ⓒ. \(K[Ag(CN)_2]\)
ⓓ. \(CaCl_2 \cdot 6H_2O\)
Correct Answer: \(K[Ag(CN)_2]\)
Explanation: \(K[Ag(CN)_2]\) contains the complex ion \([Ag(CN)_2]^-\). In water, the potassium ion separates, but the complex ion remains as a coordinated unit. The other compounds listed are ordinary salts, so they dissociate only into simple ions. This difference in solution behaviour is what places \(K[Ag(CN)_2]\) in the class of coordination compounds.
16. A student prepares separate aqueous solutions of \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\) and \(K_4[Fe(CN)_6]\). In which solution is the ordinary test for free \(Fe^{2+}\) more directly expected to work?
ⓐ. Only in the Mohr’s salt solution
ⓑ. Only in the potassium ferrocyanide solution
ⓒ. In both solutions equally
ⓓ. In neither solution
Correct Answer: Only in the Mohr’s salt solution
Explanation: \(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\) is a double salt, so it dissociates into simple ions in water and provides free \(Fe^{2+}\). In \(K_4[Fe(CN)_6]\), iron remains bound within the complex ion \([Fe(CN)_6]^{4-}\). Because of that, free \(Fe^{2+}\) is not available in the same direct way. The contrast shows how coordination compounds and double salts differ in qualitative behaviour.
17. A compound dissolves in water and still behaves as if the bracketed part remains a single charged unit. Which feature is being demonstrated?
ⓐ. Hydration of ions only
ⓑ. Complete dissociation into simple ions
ⓒ. Retention of the complex ion
ⓓ. Formation of a double salt in water
Correct Answer: Retention of the complex ion
Explanation: A coordination compound keeps its complex ion intact when it dissolves, although outer counter ions may separate. This means the bracketed part continues to behave as one coordinated unit in solution. That is the main difference from a double salt, which breaks into simple ions. The behaviour of the dissolved species therefore reveals the presence of a coordination entity.
18. Which statement correctly compares a double salt and a coordination compound in aqueous solution?
ⓐ. A double salt gives simple ions; a coordination compound preserves its complex ion
ⓑ. A double salt preserves a bracketed unit, whereas a coordination compound gives only simple ions
ⓒ. Both always give the same ions in water, so they are chemically identical
ⓓ. Both must contain water of crystallisation to show any difference in solution
Correct Answer: A double salt gives simple ions; a coordination compound preserves its complex ion
Explanation: When a double salt dissolves, it ionises completely into its constituent simple ions. A coordination compound does not usually release all species separately, because the metal and ligands remain bonded inside the complex ion. This is why the two classes behave differently in qualitative tests. The distinction depends on solution behaviour, not merely on how the solid formula looks.
19. Which statement about ordinary qualitative tests is most appropriate for \(K_4[Fe(CN)_6]\)?
ⓐ. It should directly give all the usual tests for free \(K^+\), \(Fe^{2+}\), and \(CN^-\) separately
ⓑ. It should fail to show any ionic character at all in water
ⓒ. It should behave exactly like Mohr’s salt in aqueous solution
ⓓ. It shows \(K^+\) and intact \([Fe(CN)_6]^{4-}\), not free \(Fe^{2+}\) or \(CN^-\)
Correct Answer: It shows \(K^+\) and intact \([Fe(CN)_6]^{4-}\), not free \(Fe^{2+}\) or \(CN^-\)
Explanation: In aqueous solution, \(K_4[Fe(CN)_6]\) dissociates into \(K^+\) ions and the complex ion \([Fe(CN)_6]^{4-}\). The iron and cyanide units are not present as free simple ions under ordinary dissolution conditions. Because of that, direct tests for free \(Fe^{2+}\) or free \(CN^-\) are not expected in the same way as for a double salt. This preservation of the complex ion is the important clue.
20. In Werner’s coordination theory, the valency satisfied by groups directly attached inside the coordination sphere is called:
ⓐ. principal valency
ⓑ. secondary valency
ⓒ. ionic valency
ⓓ. residual valency
Correct Answer: secondary valency
Explanation: Werner proposed that a metal can show two types of valency in coordination compounds. The secondary valency corresponds to the number of groups directly attached to the metal within the coordination sphere. These groups are usually not ionisable and occupy fixed positions around the metal. This idea helped explain why some ions remain inside brackets while others stay outside.
