Question: Which of the following statements is/are correct?...

Question

Which of the following statements is/are correct?

Answer 1

Solution

The problem asks us to identify the correct statements regarding ionization enthalpies. We will analyze each statement based on electronic configuration, effective nuclear charge, stability of half-filled/fully-filled subshells, and periodic trends.

(A) The second ionization enthalpy of oxygen atom is greater than that of fluorine atom.

Oxygen (O): Electronic configuration: $1s^2 2s^2 2p^4$ $1 s^{2} 2 s^{2} 2 p^{4}$ .
- To determine $IE_2(O)$ , we consider the removal of an electron from $O^+$ .
- $O^+ (g) \rightarrow O^{2+}(g) + e^-$ . The electronic configuration of $O^+$ is $1s^2 2s^2 2p^3$ . This is a half-filled p-subshell, which is a relatively stable configuration.
Fluorine (F): Electronic configuration: $1s^2 2s^2 2p^5$ $1 s^{2} 2 s^{2} 2 p^{5}$ .
- To determine $IE_2(F)$ , we consider the removal of an electron from $F^+$ .
- $F^+ (g) \rightarrow F^{2+}(g) + e^-$ . The electronic configuration of $F^+$ is $1s^2 2s^2 2p^4$ .
Comparison: We are comparing the energy required to remove an electron from $O^+(2p^3)$ and $F^+(2p^4)$ . Due to the extra stability of the half-filled $2p^3$ subshell in $O^+$ , removing an electron from $O^+$ requires more energy than removing an electron from $F^+$ , despite F having a higher nuclear charge.
Conclusion: $IE_2(O) > IE_2(F)$ .
Statement (A) is correct.

(B) The third ionization enthalpy of phosphorus is greater than that of aluminium.

Phosphorus (P): Atomic number 15. Electronic configuration: $[Ne] 3s^2 3p^3$ $[N e] 3 s^{2} 3 p^{3}$ .
- To determine $IE_3(P)$ , we consider the removal of an electron from $P^{2+}$ .
- $P^{2+}(g) \rightarrow P^{3+}(g) + e^-$ . The electronic configuration of $P^{2+}$ is $[Ne] 3s^2 3p^1$ . The electron being removed is a $3p$ electron.
Aluminium (Al): Atomic number 13. Electronic configuration: $[Ne] 3s^2 3p^1$ $[N e] 3 s^{2} 3 p^{1}$ .
- To determine $IE_3(Al)$ , we consider the removal of an electron from $Al^{2+}$ .
- $Al^{2+}(g) \rightarrow Al^{3+}(g) + e^-$ . The electronic configuration of $Al^{2+}$ is $[Ne] 3s^1$ . The electron being removed is a $3s$ electron.
Comparison: Both Al and P are in Period 3. P has a higher nuclear charge (Z=15) than Al (Z=13). While removing a $3s$ electron (from $Al^{2+}$ ) is generally harder than removing a $3p$ electron (from $P^{2+}$ ) due to better penetration, the significantly higher effective nuclear charge of $P^{2+}$ (due to higher Z) makes its electrons more tightly bound. This effect dominates, leading to a higher $IE_3$ for P.
Conclusion: $IE_3(P) > IE_3(Al)$ .
Statement (B) is correct.

(C) The first ionization enthalpy of aluminium is slightly greater than that of gallium.

Aluminium (Al): Atomic number 13. Electronic configuration: $[Ne] 3s^2 3p^1$ . (Period 3, Group 13)
Gallium (Ga): Atomic number 31. Electronic configuration: $[Ar] 3d^{10} 4s^2 4p^1$ . (Period 4, Group 13)
Comparison: Normally, ionization enthalpy decreases down a group due to increasing atomic size and shielding. However, for Group 13 elements, the presence of 10 d-electrons in Ga (which are poor shielders) causes an increase in the effective nuclear charge experienced by the outermost electron. This leads to a smaller than expected atomic radius for Ga and a higher $IE_1$ compared to Al.
Conclusion: $IE_1(Ga) > IE_1(Al)$ . Therefore, the statement that $IE_1(Al)$ is greater than $IE_1(Ga)$ is incorrect.
Statement (C) is incorrect.

(D) The second ionization enthalpy of copper is greater than that of zinc.

Copper (Cu): Atomic number 29. Electronic configuration: $[Ar] 3d^{10} 4s^1$ $[A r] 3 d^{10} 4 s^{1}$ .
- To determine $IE_2(Cu)$ , we consider the removal of an electron from $Cu^+$ .
- $Cu^+ (g) \rightarrow Cu^{2+}(g) + e^-$ . The electronic configuration of $Cu^+$ is $[Ar] 3d^{10}$ . This is a very stable, fully filled d-subshell configuration. Removing an electron from this highly stable configuration requires a very large amount of energy.
Zinc (Zn): Atomic number 30. Electronic configuration: $[Ar] 3d^{10} 4s^2$ $[A r] 3 d^{10} 4 s^{2}$ .
- To determine $IE_2(Zn)$ , we consider the removal of an electron from $Zn^+$ .
- $Zn^+ (g) \rightarrow Zn^{2+}(g) + e^-$ . The electronic configuration of $Zn^+$ is $[Ar] 3d^{10} 4s^1$ . The electron being removed is a $4s$ electron, which leads to a stable $3d^{10}$ configuration.
Comparison: Removing an electron from the highly stable $3d^{10}$ configuration of $Cu^+$ is much harder than removing a $4s^1$ electron from $Zn^+$ , even though Zn has a higher nuclear charge. The stability of the fully filled $3d^{10}$ subshell in $Cu^+$ is the dominant factor.
Conclusion: $IE_2(Cu) > IE_2(Zn)$ .
Statement (D) is correct.

Based on the analysis, statements (A), (B), and (D) are correct.