Understanding Ionization Energy and Electronic Configuration

Ionization energy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. This process results in the formation of a positive ion (cation).

The first ionization energy ($IE_1$) refers to the energy needed to remove the first electron: 

X(g) + IE$_1$ → X$^+$(g) + e$^- $

Several factors influence ionization energy:

• Nuclear Charge: Higher nuclear charge increases the attraction between the nucleus and electrons, making it harder to remove an electron, thus increasing ionization energy.
• Atomic Size: Larger atomic size means the outermost electrons are further from the nucleus and experience less attraction, making them easier to remove, thus decreasing ionization energy.
• Shielding Effect: Inner electrons shield the outer electrons from the full nuclear charge. Increased shielding reduces the effective nuclear charge experienced by the outer electrons, decreasing ionization energy.
• Electronic Configuration Stability: Atoms with stable electronic configurations (like completely filled or half-filled subshells) have higher ionization energies because removing an electron disrupts this stability.

Analyzing Electronic Configurations and Ionization Energy Trend

The given electronic configurations are based on the noble gas Neon ([Ne]), indicating elements in the third period.

• Option 1: [Ne] $3s^2 3p^1$ - This is the electronic configuration of Aluminum (Al), in Group 13.
• Option 2: [Ne] $3s^2 3p^3$ - This is the electronic configuration of Phosphorus (P), in Group 15.
• Option 3: [Ne] $3s^2 3p^2$ - This is the electronic configuration of Silicon (Si), in Group 14.
• Option 4: [Ne] $3s^2 3p^4$ - This is the electronic configuration of Sulfur (S), in Group 16.

These elements are in the same period (Period 3) and appear in the periodic table in the order Al, Si, P, S.

Generally, ionization energy increases across a period from left to right because the nuclear charge increases while the shielding effect from inner electrons remains relatively constant. This leads to a stronger attraction between the nucleus and the valence electrons, making them harder to remove.

Following the general trend, we might expect the ionization energy order to be Al < Si < P < S.

Identifying the Highest Ionization Energy

While the general trend is increasing across a period, there are exceptions, particularly when considering the stability of electronic configurations.

Let's look at the filling of the 3p subshell for each element:

• Al: $3p^1$
• Si: $3p^2$
• P: $3p^3$ (Half-filled p subshell)
• S: $3p^4$

A half-filled subshell ($p^3$, $d^5$, $f^7$) is a particularly stable arrangement. Removing an electron from a stable configuration requires significantly more energy than would be predicted by the general trend.

Comparing Phosphorus ($3p^3$) and Sulfur ($3p^4$):

• Phosphorus has a stable half-filled 3p subshell.
• Sulfur has one electron pair in the 3p subshell ($3p^4$). Electron-electron repulsion between paired electrons in the same orbital makes it slightly easier to remove one electron compared to removing an electron from a stable half-filled or fully-filled subshell.

Due to the extra stability of the half-filled 3p subshell, Phosphorus has a higher first ionization energy than Sulfur, even though Sulfur is to the right of Phosphorus in the periodic table.

Comparing all options based on the trend and the half-filled stability:

• Al ($3p^1$) is furthest left, lowest ionization energy.
• Si ($3p^2$) is to the right of Al, higher ionization energy.
• S ($3p^4$) is to the right of Si, but the paired electron makes its ionization energy lower than P.
• P ($3p^3$) has the stable half-filled subshell, giving it the highest ionization energy among the given options.

The observed trend in ionization energy for these elements is typically Al < Si < S < P.

Therefore, the element with the electronic configuration [Ne] $3s^2 3p^3$ (Phosphorus) will have the highest ionization energy among the given options.

Revision Table: Key Factors Affecting Ionization Energy

Additional Information: Successive Ionization Energies

Successive ionization energies are the energies required to remove subsequent electrons ($IE_1$, $IE_2$, $IE_3$, etc.). Each successive ionization energy is always greater than the previous one ($IE_1 < IE_2 < IE_3 ...$). This is because with each electron removed, the remaining electrons are held more tightly by the now increased positive charge of the ion.

There are particularly large jumps in successive ionization energies when an electron is removed from a stable, filled shell or subshell, as this involves breaking a very stable electronic configuration.

For example, for Sodium (Na), removing the first electron ([Ne] $3s^1$) is relatively easy ($IE_1$ is low). However, removing the second electron from the stable [Ne] core requires a much larger amount of energy ($IE_2$ is very high).