Problem 127
Question
The atomic number of \(\mathrm{V}, \mathrm{Cr}, \mathrm{Mn}\) and \(\mathrm{Fe}\) are respectively \(23,24,25\) and 26 . which one of these may be expected to have the highest second ionization enthalpy? (a) Mn (b) \(\mathrm{Fe}\) (c) \(\mathrm{Cr}\) (d) \(\mathrm{V}\)
Step-by-Step Solution
Verified Answer
Therefore, chromium (Cr) is expected to have the highest second ionization enthalpy.
1Step 1: Understanding Second Ionization Enthalpy
Second ionization enthalpy refers to the energy required to remove a second electron from a singly charged cation of an element. This typically requires more energy than the first because it involves an electron from a positively charged ion.
2Step 2: Looking at Electron Configurations
Examine the electron configurations for these elements: \( \mathrm{V}: [\mathrm{Ar}] 3d^3 4s^2 \), \( \mathrm{Cr}: [\mathrm{Ar}] 3d^5 4s^1 \), \( \mathrm{Mn}: [\mathrm{Ar}] 3d^5 4s^2 \), \( \mathrm{Fe}: [\mathrm{Ar}] 3d^6 4s^2 \). After the first ionization, vanadium becomes \( [\mathrm{Ar}] 3d^3 4s^1 \), chromium becomes \( [\mathrm{Ar}] 3d^5 \), manganese becomes \( [\mathrm{Ar}] 3d^5 4s^1 \), and iron becomes \( [\mathrm{Ar}] 3d^6 4s^1 \).
3Step 3: Evaluate Stability After First Ionization
Stability of the electronic configuration following first ionization is crucial for determining the second ionization enthalpy. You should consider that half-filled and fully-filled electron configurations are more stable. Post-first ionization, chromium has a stable half-filled \(3d^5\) configuration, which means the second ionization will require significantly more energy as it disrupts this stability.
4Step 4: Conclusion on Second Ionization Enthalpy
Given chromium's stable \(3d^5\) configuration after the first ionization, removing another electron would disrupt this stable state, leading to higher second ionization enthalpy compared to vanadium, manganese, and iron. Thus, chromium is expected to have the highest second ionization enthalpy.
Key Concepts
Electron ConfigurationIonization EnergyTransition Metals
Electron Configuration
Electron configuration is a way to represent the arrangement of electrons around the nucleus of an atom. For transition metals like vanadium (V), chromium (Cr), manganese (Mn), and iron (Fe), we utilize the Aufbau principle, which helps us determine their ground state electron configurations.
- Vanadium (V) has an electron configuration of: \( [\text{Ar}] 3d^3 4s^2 \).
- Chromium (Cr) uniquely prefers a half-filled arrangement: \( [\text{Ar}] 3d^5 4s^1 \).
- Manganese (Mn) aligns with a full 4s orbital and half-filled 3d: \( [\text{Ar}] 3d^5 4s^2 \).
- Iron (Fe) adds up to: \( [\text{Ar}] 3d^6 4s^2 \).
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. The first ionization energy removes one electron; the second continues with the removal of another from an already charged ion.
This process demands even more energy because:
This process demands even more energy because:
- The cation after the first ionization has a positive charge, increasing nuclear attraction.
- Removing an electron often leaves a configuration closer to stability, making further removal more difficult.
Transition Metals
Transition metals, situated in the d-block of the periodic table, are characterized by their partially filled d orbitals and are known for their unique chemical and physical properties. These metals exhibit several distinct traits:
- They often form colored compounds due to d-d electronic transitions within their d orbitals.
- They can have multiple oxidation states, enabling a variety of chemical reactions.
- Many possess magnetic properties owing to unpaired electrons.
Other exercises in this chapter
Problem 125
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