Step 1: Understanding the Concept:
We need to correlate specific electronic configurations in the second period ($n=2$) to their relative 1st Ionization Energy (IE) values. The general trend of IE across a period increases, but there are distinct anomalies due to subshell stability (full $s$-orbital vs $p^1$, and half-filled $p^3$ vs $p^4$). Step 2: Key Formula or Approach:
Identify the elements based on the configuration:
$ns^2 \to \text{Beryllium (Be)}$
$ns^2np^1 \to \text{Boron (B)}$
$ns^2np^3 \to \text{Nitrogen (N)}$
$ns^2np^6 \to \text{Neon (Ne)}$
Compare their IE based on trends: Noble gases>half-filled $p$>full $s$>partial $p$. Step 3: Detailed Explanation:
Let's list the expected order of Ionization Energies for these elements:
1. Neon (Ne, $2s^2 2p^6$) is a noble gas with a completely filled valence shell. Removing an electron is extremely difficult, making it have the highest IE among the group.
2. Nitrogen (N, $2s^2 2p^3$) has a stable exactly half-filled $2p$ subshell, requiring significant energy to break this symmetry.
3. Beryllium (Be, $2s^2$) has a completely filled $2s$ subshell. Its electrons are highly penetrating and tightly held compared to an unpaired $2p$ electron.
4. Boron (B, $2s^2 2p^1$) has a single electron in a higher-energy $2p$ orbital. It is easier to remove this lone $p$ electron to reach the stable $2s^2$ state.
Thus, the expected order from lowest to highest IE is: Boron<Beryllium<Nitrogen<Neon.
Now map the given numerical values: 800, 899, 1402, 2080.
- Neon (D) = highest = 2080 (I)
- Nitrogen (C) = second highest = 1402 (IV)
- Beryllium (A) = third highest = 899 (II)
- Boron (B) = lowest = 800 (III)
Summary of matches: A $\to$ II, B $\to$ III, C $\to$ IV, D $\to$ I. Step 4: Final Answer:
The correct matching is A-II, B-III, C-IV, D-I.
