Chapter 10: Problem 82
In group \(4 \mathrm{~A}\) or 14 group of the extended form of the periodic table with increase in atomic number, the oxidizing power of tetravalent species increases in the order: (a) \(\mathrm{Ge}>\mathrm{Sn}>\mathrm{Pb}\) (b) \(\mathrm{Pb}>\mathrm{Sn}>\mathrm{Ge}\) (c) \(\mathrm{Pb}>\mathrm{Ge}>\mathrm{Sn}\) (d) \(\mathrm{Ge}>\mathrm{Pb}>\mathrm{Sn}\)
Short Answer
Step by step solution
Understanding Periodic Trends
Analyzing Oxidizing Power
Ordering the Elements by Oxidizing Power
Selecting the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidizing Power
Group 14 Elements
- As we move down the group from Carbon to Lead, there is an increase in atomic size and a decrease in electronegativity.
- This alteration also leads to a gradual weakening of bonds due to the larger atomic radii and lesser effective nuclear charge experienced by the outer electrons.
Tetravalent Species
- As we go from Carbon down to Lead, the stability of the +4 oxidation state diminishes.
- This is due to the "inert pair effect," where the s electrons become more resistant to participate in bonding for heavier elements.
Inert Pair Effect
- It causes Lead to prefer a +2 oxidation state over a +4 oxidation state because the s-electron pair remains "inert," or unreactive.
- The inert pair thus influences the oxidizing power of the element, making the tetravalent states of the heavier elements stronger oxidizers when they gain electrons to achieve the more stable +2 oxidation state.