WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Chemistry of the P-Block
Stability of oxidation states down Groups 3, 4 and 5: increasing stability of the inert pair cations
- 3 things to know
- A2 Unit 3
Overview
Recall questions on inert pair effect.
What you need to know
Answer each one before you reveal it.
Explain why carbon and silicon show only the +4 state in their common compounds.
Tin(II) chloride solution turns iron(III) solution from yellow to pale green. Write the ionic equation and explain what this shows about tin.
Questions from this subtopic
Each one with the model answer underneath.
- 1
What is meant by the inert pair effect?
Model answer
It is the tendency of the two outermost s electrons of a heavy p-block atom to stay unused in bonding, so that an oxidation state two lower than the group maximum becomes increasingly stable towards the bottom of Groups 3, 4 and 5.
- 2
Lead has the outer configuration 6s²6p². Use this to explain why Pb²⁺ rather than Pb⁴⁺ is its usual ion.
Model answer
Losing only the two 6p electrons gives Pb²⁺ and leaves the 6s² pair intact, which is easy. Reaching +4 requires the tightly held 6s pair to be removed or unpaired as well, and that extra energy is not repaid by the bonds or lattice formed, so Pb²⁺ is the stable ion.
- 3
Why does the stability of the lower oxidation state increase on going down Group 4?
Model answer
The s electrons of the heavier atoms are held more firmly than expected, because the intervening d and f electrons shield the nuclear charge poorly. At the same time the bonds formed by the heavier elements are weaker, so the energy released by forming two extra bonds no longer compensates for the energy needed to involve the s pair.
- 4
State the most stable oxidation state of thallium in Group 3, and contrast it with aluminium.
Model answer
Thallium is most stable as Tl⁺, in which the 6s² pair is retained, whereas aluminium is found almost exclusively as Al³⁺. Thallium(III) compounds are therefore strong oxidising agents, being readily reduced to thallium(I).
- 5
In Group 5, bismuth(V) compounds such as sodium bismuthate(V) are powerful oxidising agents while bismuth(III) compounds are stable. What does this show?
Model answer
It shows the inert pair effect operating in Group 5: at the bottom of the group the 6s² pair resists bonding, so the +3 state is stable and the +5 state is easily reduced to it, making bismuth(V) strongly oxidising.
- 6
Explain why carbon and silicon show only the +4 state in their common compounds.
Model answer
At the top of Group 4 the s and p electrons are of similar energy and are all readily involved in bonding, and the C–X and Si–X bonds formed are strong. The energy released by forming four bonds easily repays the promotion energy, so the +4 state is the stable one and there is no inert pair effect.
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