Question

$K_2Cr_2O_7 + KI \xrightarrow{H^+} I_2 + Cr^{3+}$
Statement-I : Size of $O^{2-}$ is smaller than $F^-$.
Statement-II: Second ionization energy of $\text{Na}$ is greater than second ionization energy of $\text{Mg}$.

• A. Both statements are correct.
• B. Both statements are incorrect.
• C. Statement I is correct while Statement II is incorrect.
• D. Statement I is incorrect while Statement II is correct.

Hint:

Always analyze isoelectronic species based on nuclear charge ($Z$): higher $Z$ means stronger pull on electrons, resulting in smaller size. Ionization energy jumps drastically when an electron is removed from a completely filled inner shell (noble gas configuration).

Answer 1

The Correct Option is D

Let's analyze the given statements one by one to determine their correctness. 

1. Statement I: The size of \(O^{2-}\) is smaller than \(F^−\).
   • The ionic radius depends on the effective nuclear charge and the number of electrons. \(O^{2-}\) has 10 electrons and \(F^−\) has 9 electrons, making \(O^{2-}\) larger due to increased electron-electron repulsions and a lesser effective nuclear charge compared to \(F^−\).
   • Hence, \(O^{2-}\) is larger than \(F^−\), which makes the statement incorrect.
2. Statement II: The second ionization energy of \(\text{Na}\) is greater than the second ionization energy of \(\text{Mg}\).
   • The second ionization energy involves the removal of a second electron from a cation. For \(\text{Na}^+\), removing the second electron means taking an electron from the noble gas configuration \(\text{Ne}\), which requires a lot of energy.
   • On the other hand, \(\text{Mg}^+\) loses its second electron from a less stable configuration, requiring less energy than \(\text{Na}^+\).
   • This implies that the second ionization energy of \(\text{Na}\) is indeed greater than that of \(\text{Mg}\), making the statement correct.

Based on the above analysis, the correct answer is: Statement I is incorrect while Statement II is correct.