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:

Compare $O^{2-}$ and $F^-$ by electron count first, then check which one has the higher nuclear charge pulling on that same electron cloud. For the ionization energy statements, check which principal shell (value of $n$) each successive electron is removed from, since a jump to a new inner shell always causes a large increase.

Answer 1

The Correct Option is D

Statement I:
Step 1: $O^{2-}$ and $F^-$ are isoelectronic species (10 electrons each).
Step 2: In an isoelectronic series, size decreases with increase in nuclear charge.
Step 3: Nuclear charge: \[ Z(O)=8,\quad Z(F)=9 \] Step 4: Hence $O^{2-}$ is larger than $F^-$.
So, Statement I is incorrect.
Statement II:
Step 5: Second ionization energy means removal of electron from the cation.
Step 6: $Na^+$ has noble gas configuration, so removal of electron is very difficult.
Step 7: $Mg^+$ still has one electron in 3s orbital.
Thus, \[ IE_2(Na)>IE_2(Mg) \] Statement II is correct.
Hence, correct option is (4).

Answer 2

Concept:

• For isoelectronic species (same number of electrons), the effective nuclear charge $Z_{eff}$ felt by the electron cloud depends on the actual nuclear charge $Z$, since the shielding from a fixed electron count stays nearly the same.
• A larger $Z_{eff}$ pulls the same electron cloud in more tightly, giving a smaller ionic radius.
• A sudden jump in ionization energy occurs when the electron being removed comes from a shell closer to the nucleus than the electron removed before it.

Step 1: Compare nuclear charge of the isoelectronic pair.
$O^{2-}$ and $F^-$ both have 10 electrons ($1s^2 2s^2 2p^6$).
Nuclear charge: $Z(O) = 8$, $Z(F) = 9$.

Step 2: Apply the effective nuclear charge idea.
Since both ions hold the same 10 electrons, the one with the higher $Z$ pulls the electron cloud closer to the nucleus.
$F^-$ has the higher $Z_{eff}$, so its electron cloud is pulled in more, giving $F^-$ the smaller radius.
So $O^{2-}$ is larger than $F^-$, and Statement I, which claims $O^{2-}$ is smaller, is incorrect.

Step 3: Track which shell is being emptied for each successive ionization.
$Na: 1s^2 2s^2 2p^6 3s^1$. Removing the first electron (from $3s$) leaves $Na^+$ with configuration $1s^2 2s^2 2p^6$.
The second electron to be removed from $Na^+$ must now come from the $n=2$ shell, a completely different and much more tightly bound shell than the one used for the first removal.

Step 4: Compare with magnesium.
$Mg: 1s^2 2s^2 2p^6 3s^2$. Removing the first electron leaves $Mg^+: 1s^2 2s^2 2p^6 3s^1$.
The second electron removed from $Mg^+$ still comes from the same $n=3$ shell as the first, so no shell change occurs.

Step 5: Conclude on ionization energy.
Since the second ionization of $Na$ requires breaking into a lower, inner shell while the second ionization of $Mg$ does not, $IE_2(Na)$ shows a much bigger jump than $IE_2(Mg)$.
Hence $IE_2(Na) > IE_2(Mg)$, and Statement II is correct.

Final Answer: Statement I is incorrect while Statement II is correct.