Question

The correct statement(s) regarding the periodic properties of elements is(are)

• A. Second ionization enthalpy of carbon atom is less than that of boron atom.
• B. Increasing order of ionic radii: Al$^{3+}$ \textless Mg$^{2+}$ \textless Na$^+$
• C. Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
• D. The H–H bond is weaker than F–F bond.

Hint:

Remember key exceptions and trends:
- IE across period: General increase, but exceptions due to orbital stability (e.g., B \textless Be, O \textless N). For 2nd IE, the electronic configuration of $X^+$ is critical.
- Ionic radii (isoelectronic): Decrease with increasing nuclear charge.
- Density (Group 1): K is less dense than Na.
- Bond energies: Small halogens (F$_2$) have weaker bonds due to lone pair repulsion. H-H is surprisingly strong.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to identify the correct statement(s) among the given options regarding various periodic properties of elements. This question might have multiple correct answers.

Step 2: Key Formula or Approach:
Analyze each statement based on established periodic trends and specific chemical properties.

Step 3: Detailed Explanation:
(A) Second ionization enthalpy of carbon atom is less than that of boron atom.
- Boron (B): $Z=5$. Electronic configuration: $[He] 2s^2 2p^1$.
First ionization: $B \rightarrow B^+ (2s^2) + e^-$.
Second ionization: $B^+ (2s^2) \rightarrow B^{2+} (2s^1) + e^-$. Removing an electron from a fully filled $2s^2$ subshell (which is stable).
- Carbon (C): $Z=6$. Electronic configuration: $[He] 2s^2 2p^2$.
First ionization: $C \rightarrow C^+ (2s^2 2p^1) + e^-$.
Second ionization: $C^+ (2s^2 2p^1) \rightarrow C^{2+} (2s^2) + e^-$. Removing an electron from a $2p^1$ orbital. This is easier than removing from a fully filled $2s^2$ orbital.
- Therefore, the second ionization enthalpy of carbon ($C^+ \rightarrow C^{2+}$) involves removing a $2p$ electron, while for boron ($B^+ \rightarrow B^{2+}$), it involves removing a $2s$ electron from a stable $2s^2$ configuration. Removing a $2p$ electron is energetically less demanding than removing a $2s$ electron in this case.
- So, second IE of C \textless second IE of B. This statement is Correct.
(B) Increasing order of ionic radii: Al$^{3+$ \textless Mg$^{2+}$ \textless Na$^+$}
- These are isoelectronic ions, each having 10 electrons (like Neon).
- For isoelectronic species, ionic radius decreases with increasing nuclear charge (Z).
- $Na^+$: Z=11. $Mg^{2+}$: Z=12. $Al^{3+}$: Z=13.
- The increasing order of nuclear charge is $Na^+ < Mg^{2+} < Al^{3+}$.
- Therefore, the increasing order of ionic radii should be $Al^{3+}$ \textless $Mg^{2+}$ \textless $Na^+$. This statement is Correct.
(C) Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
- Both Na and K are Group 1 alkali metals.
- Density generally increases down a group due to increasing atomic mass dominating the increasing atomic volume.
- However, there is an exception in Group 1: The density of sodium (Na = $0.968 \text{ g/cm}^3$) is greater than the density of potassium (K = $0.86 \text{ g/cm}^3$). This is due to the unusually large increase in atomic volume from Na to K (and less efficient packing in K's BCC lattice), making potassium less dense than sodium.
- So, the statement that density of K is *more* than Na is Incorrect.
(D) The H–H bond is weaker than F–F bond.
- H-H bond energy: $\approx 436 \text{ kJ/mol}$. (Very strong bond).
- F-F bond energy: $\approx 158 \text{ kJ/mol}$. (Relatively weak bond due to lone pair-lone pair repulsion between the small, highly electronegative fluorine atoms).
- Therefore, the H-H bond is stronger than the F-F bond. The statement says H-H bond is *weaker*. This statement is Incorrect.
So, statements (A) and (B) are correct.

Step 4: Final Answer:
Statements (A) and (B) are correct.