Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?

Question

Statement-I : Consider the following pairs of ions $(\text{Sc}^{3+}, \text{Ti}^{3+}), (\text{Ti}^{4+}, \text{Ni}^{2+}), (\text{Cu}^{2+}, \text{Zn}^{2+})$ and $(\text{Cr}^{3+}, \text{Mn}^{3+})$. Out of these pairs three pairs consist of ions that are both coloured :
Statement-II : Among the lanthanide ions $\text{Eu}^{2+}$, $\text{Gd}^{3+}$, $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$, the ion $\text{Tb}^{4+}$ is the strongest reducing agent.
Choose the correct option.

Answer 1

Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?

Solution:
Step 1: Understanding ionization enthalpy.
Ionization enthalpy (also known as ionization energy) is the amount of energy required to remove an electron from an atom or ion in the gas phase. The second ionization enthalpy refers to the energy required to remove a second electron after the first one has been removed.

Step 2: General trend of ionization enthalpy.
In general, ionization enthalpy increases as we move across a period due to the increasing effective nuclear charge and decreases as we move down a group due to increased electron shielding.

Step 3: Analyzing chromium (Cr) and copper (Cu).
- **Chromium (Cr)**: The electronic configuration of chromium is $[Ar] 3d^5 4s^1$. The half-filled $3d$ sublevel ($3d^5$) is exceptionally stable due to symmetry and exchange energy. When the first electron is removed, the resulting ion has the configuration $[Ar] 3d^4$, which is less stable. To remove a second electron, the ion must lose an electron from the stable $3d^5$ configuration, which requires much more energy, leading to a high second ionization enthalpy.
- **Copper (Cu)**: The electronic configuration of copper is $[Ar] 3d^{10} 4s^1$. The fully filled $3d^{10}$ sublevel is particularly stable. When the first electron is removed, the resulting ion has the configuration $[Ar] 3d^{9}$, which is less stable than the original configuration. To remove a second electron, the ion must lose an electron from the stable $3d^{10}$ configuration, resulting in a high second ionization enthalpy.

Step 4: Comparison with neighbouring elements.
The second ionization enthalpy of chromium and copper is significantly higher than that of their neighbouring elements, such as manganese (Mn) and zinc (Zn), because these neighbouring elements do not have the same level of stability in their electron configurations. For example, manganese ($[Ar] 3d^5 4s^2$) has a stable half-filled $3d$ sublevel, but its second ionization enthalpy is not as high as chromium’s because it does not need to break a stable $3d^5$ configuration.

Final Answer: The second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements because, after the first ionization, they are left with stable electron configurations ($3d^5$ for chromium and $3d^{10}$ for copper), and removing a second electron from these stable configurations requires significantly more energy.

Why second ionization enthalpies of chromium and copper are exceptionally higher than those of their neighbouring elements?

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Solution and Explanation

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