Class 12 Chemistry · Chapter 2 · Test 4
Stability Constants and Stereoisomerism
36 MCQs
36 questions · 30 min · 36 marks
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Sample questions
The first 5 of 36. Try each one, then tap to see the answer.
Question 1
The stability constant Kstab is:
- AThe equilibrium constant for the formation of a complex ion
- BThe rate constant for the decomposition of a complex
- CThe solubility product
- DThe acid dissociation constant
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Answer: A
Kstab is the equilibrium constant for complex formation.
Question 2
Why is [H2O] omitted from the Kstab expression?
- AWater is a gas
- BWater is the solvent and its concentration is effectively constant
- CWater does not take part in the reaction
- DWater is a product
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Answer: B
Since the solvent concentration stays constant it is left out.
Question 3
Which is the Kstab expression for [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O?
- A[Cu(NH3)4(H2O)2]2+ / [NH3]^4
- B[Cu(H2O)6]2+[NH3]^4 / [Cu(NH3)4(H2O)2]2+
- C[Cu(NH3)4(H2O)2]2+ / ([Cu(H2O)6]2+[NH3]^4)
- D[Cu(NH3)4(H2O)2]2+[H2O]^4 / ([Cu(H2O)6]2+[NH3]^4)
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Answer: C
Kstab = [complex] / ([aqua complex][ligand]^4) with water omitted.
Question 4
A higher log10 Kstab means the complex is:
- ALess stable and easily decomposed
- BInsoluble
- CColourless
- DMore stable with the equilibrium lying far to the right
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Answer: D
A large Kstab means the complex is strongly favoured.
Question 5
A low log10 Kstab shows that:
- AThe complex is less stable and its ligands are easily displaced
- BThe complex is extremely stable
- CThe complex is neutral
- DThe metal is a non-metal
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Answer: A
Low stability lets other ligands displace those present.
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