Chapter 14: Problem 210
The atomic numbers of vanadium \((\mathrm{V})\), chromium \((\mathrm{Cr})\), manganese \((\mathrm{Mn})\) and iron \((\mathrm{Fe})\) are \(23,24,25\) and 26 respectively. Which one of these may be expected to have the highest second ionisation enthalpy? [A.I.E.E.E. 2003] (a) \(\mathrm{V}\) (b) \(\mathrm{Cr}\) (c) \(\mathrm{Mn}\) (d) \(\mathrm{Fe}\)
Short Answer
Step by step solution
Understand Ionization Enthalpy
Analyze Electron Configurations
Consider the First Ionization
Determine the Second Ionization Enthalpy
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electron Configuration
- Vanadium : [Ar] 3d\(^{3}\) 4s\(^{2}\)
- Chromium : [Ar] 3d\(^{5}\) 4s\(^{1}\)
- Manganese : [Ar] 3d\(^{5}\) 4s\(^{2}\)
- Iron : [Ar] 3d\(^{6}\) 4s\(^{2}\)
Half-Filled Stability
- Manganese maintains the half-filled 3d\(^{5}\) configuration upon first ionization, resulting in a stable state.
- Iron reaches stability after the second ionization to reach [Ar] 3d\(^{5}\).
Transition Metals
- Vanadium exhibits oxidation states from +2 to +5 due to its varying d electron configuration.
- Chromium is well-known for its +3 and +6 states, both reliant on its electron distribution.
Ionization Energy Trends
- Atomic size: Larger atoms have lower ionization energy.
- Nuclear charge: Higher nuclear charge increases ionization energy due to stronger electron attraction.
- Electron configuration: Stable configurations (half-filled or fully filled) require higher ionization energy to disrupt.