Question

Identify the element which has highest second ionization enthalpy

• A. Carbon
• B. Phosphorus
• C. Nitrogen
• D. Oxygen

Hint:

To compare IE$_2$, write down the configuration of the M$^+$ ion.
Elements in higher periods generally have lower ionization enthalpies than those in lower periods (for similar group positions) due to increased size and screening.
Special stability of half-filled ($p^3, d^5, f^7$) or fully-filled ($s^2, p^6, d^{10}, f^{14}$) configurations in the ion being ionized leads to higher IE.
O$^+$ has configuration $...2s^2 2p^3$. Removing an electron from this stable half-filled $p$-subshell requires high energy.

Answer 1

The Correct Option is D

We need to identify the element with the highest second ionization enthalpy (IE$_2$) among Carbon, Phosphorus, Nitrogen, and Oxygen. Second ionization enthalpy is the energy required to remove an electron from the unipositive ion (M$^+$). M$^+$(g) $\rightarrow$ M$^{2+}$(g) + e$^-$ Let's look at the electronic configurations of the M$^+$ ions: \begin{itemize} \item Carbon (C): Neutral C: $1s^2 2s^2 2p^2$. C$^+$: $1s^2 2s^2 2p^1$. Removing an electron from $2p^1$. \item Phosphorus (P): Neutral P: $1s^2 2s^2 2p^6 3s^2 3p^3$ (or [Ne] $3s^2 3p^3$). P$^+$: [Ne] $3s^2 3p^2$. Removing an electron from $3p^2$. \item Nitrogen (N): Neutral N: $1s^2 2s^2 2p^3$. N$^+$: $1s^2 2s^2 2p^2$. Removing an electron from $2p^2$. \item Oxygen (O): Neutral O: $1s^2 2s^2 2p^4$. O$^+$: $1s^2 2s^2 2p^3$. Removing an electron from a half-filled $2p^3$ configuration. \end{itemize} Key factors affecting ionization enthalpy: 1. Nuclear Charge: Higher nuclear charge generally leads to higher IE. 2. Atomic/Ionic Size: Smaller size generally leads to higher IE. 3. Electronic Configuration: Electrons from stable (half-filled or fully-filled) subshells are harder to remove. 4. Screening Effect: Inner electrons screen outer electrons from the nucleus. Comparing the M$^+$ ions: \begin{itemize} \item C$^+ ([He] 2s^2 2p^1$): Second period. \item N$^+ ([He] 2s^2 2p^2$): Second period. Higher nuclear charge than C$^+$. \item O$^+ ([He] 2s^2 2p^3$): Second period. Higher nuclear charge than N$^+$. Crucially, O$^+$ has a half-filled $2p^3$ subshell. This configuration is relatively stable, so removing an electron from it requires a significant amount of energy. \item P$^+$ ([Ne] $3s^2 3p^2$): Third period. Electrons are in the $n=3$ shell, which are further from the nucleus and better screened than $n=2$ electrons. Thus, IE$_2$ for P is generally expected to be lower than for second-period elements like N or O with similar valence configurations for their ions. \end{itemize} Between C, N, O (all in the second period): O$^+$ needs to lose an electron from a stable $2p^3$ configuration. This makes IE$_2$ for Oxygen exceptionally high for its period. N$^+$ ($2p^2$) and C$^+$ ($2p^1$) do not have this $p^3$ stability for the ion. Generally, IE increases across a period. So, for M$^+$ ions $2p^1, 2p^2, 2p^3, 2p^4$, the IE would generally increase. C$^+$ ($2p^1$), N$^+$ ($2p^2$), O$^+$ ($2p^3$), F$^+$ ($2p^4$). The exceptional stability of the $2p^3$ configuration of O$^+$ means that IE$_2$(O) is very high. As seen in the previous question (Q.123), IE$_2$(O)>IE$_2$(F)>IE$_2$(N)>IE$_2$(C). Actual values (kJ/mol): IE$_2$(C) = 2353 IE$_2$(N) = 2856 IE$_2$(O) = 3388 IE$_2$(P): P$^+$ is [Ne]$3s^2 3p^2$. IE$_2$(P) = 1907 kJ/mol. This is lower than for the second-period elements due to larger size and increased screening. So, comparing 2353 (C), 1907 (P), 2856 (N), 3388 (O). Oxygen has the highest second ionization enthalpy among the given options. \[ \boxed{\text{Oxygen}} \]