Amongst the following elements whose electronic configurations are given below, the one having the highest ion

NCERT Exemplar · Q22

Q. Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

(i) [Ne]3s^2 3p^1

(ii) [Ne]3s^2 3p^3

(iii) [Ne]3s^2 3p^2

(iv) [Ar]3d^10 4s^2 4p^3

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Concept understanding — Isoelectronic Species

Isoelectronic Species – From Intuition to Precision

Imagine you are building atoms with LEGO blocks. Each block is a proton (positive charge) or an electron (negative charge). The number of protons decides which element you have — that's the atomic number $Z$ . The number of electrons decides the charge on the particle.

Now, here is the key idea: two different particles can have the same number of electrons. When that happens, their electron clouds are arranged in the same way. They become isoelectronic.

The Intuition

Think of a neutral neon atom. It has 10 protons and 10 electrons. Now take a sodium atom (11 protons, 11 electrons) and remove one electron. You get $\text{Na}^+$ , which has 11 protons but only 10 electrons. The electron count of $\text{Na}^+$ is exactly the same as that of neutral neon.

Even though $\text{Na}^+$ and $\text{Ne}$ are different elements with different nuclear charges, their electron configurations are identical: $1s^2 2s^2 2p^6$ . They are iso (same) electronic (electron arrangement).

Note

Isoelectronic species share the same number of electrons and therefore the same electronic configuration. They differ in nuclear charge ( $Z$ ).

The Precise Statement

Definition: Two or more atoms, ions, or molecules are said to be isoelectronic if they have the same number of electrons.

That is the entire definition. But the real power comes from what follows: because their electron clouds are identical in structure, their properties — like ionic radii, ionization energy, and chemical behaviour — show clear, predictable trends when you compare them.

How to Identify Isoelectronic Species

Count the electrons in each species.
- For a neutral atom: electrons = atomic number $Z$ .
- For a positive ion: electrons = $Z$ - (charge magnitude).
- For a negative ion: electrons = $Z$ + (charge magnitude).
Compare the counts. If they match, the species are isoelectronic.

Example: Which of these are isoelectronic? $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ .

$\text{O}^{2-}$ : $Z=8$ , electrons = $8+2=10$
$\text{F}^-$ : $Z=9$ , electrons = $9+1=10$
$\text{Na}^+$ : $Z=11$ , electrons = $11-1=10$
$\text{Mg}^{2+}$ : $Z=12$ , electrons = $12-2=10$
$\text{Ne}$ : $Z=10$ , electrons = $10$

All five have 10 electrons. They form an isoelectronic series.

\text{Electrons in an ion} = Z - (\text{charge})

The Critical Consequence: Size Trends

Here is where the concept becomes exam-relevant. In an isoelectronic series, as nuclear charge ( $Z$ ) increases, the ionic radius decreases.

Why? The same number of electrons is pulled more strongly by a larger positive nucleus. The electron cloud shrinks.

For the series above ( $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ ):

Species	$Z$	Electrons	Relative Radius
$\text{O}^{2-}$	8	10	Largest
$\text{F}^-$	9	10	↓
$\text{Ne}$	10	10	↓
$\text{Na}^+$	11	10	↓
$\text{Mg}^{2+}$	12	10	Smallest

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The key idea is that ionisation enthalpy increases across a period and is highest for a half-filled or stable configuration. Among the given options, (B) [Ne]3s²3p³ has a half-filled p-subshell, giving it the highest ionisation enthalpy.

The question asks you to compare ionisation enthalpies of four elements based on their electronic configurations. Ionisation enthalpy is the energy needed to remove the most loosely bound electron from a gaseous atom. The higher the ionisation enthalpy, the more difficult it is to remove an electron.

The trend across a period is that ionisation enthalpy generally increases from left to right. But there are exceptions due to extra stability of half-filled and fully-filled subshells. So you need to look at both the position in the periodic table and the stability of the configuration.

Let’s decode each configuration:

Option (A): [Ne]3s²3p¹

This is aluminium (Al). The outermost electron is in the 3p orbital. It’s a single electron in a p-orbital, so it’s relatively easy to remove. Ionisation enthalpy is moderate.
Option (B): [Ne]3s²3p³

This is phosphorus (P). The 3p subshell is exactly half-filled (three electrons, one in each p-orbital). Half-filled subshells have extra stability due to exchange energy and symmetry. Removing an electron disrupts this stable arrangement, so the ionisation enthalpy is higher than expected from the general trend.
Option (C): [Ne]3s²3p²

This is silicon (Si). The 3p subshell has two electrons. It’s less stable than a half-filled configuration, so its ionisation enthalpy is lower than that of phosphorus.
Option (D): [Ar]3d¹⁰4s²4p³

This is arsenic (As). It belongs to the same group as phosphorus (Group 15) but is in the fourth period. Down a group, ionisation enthalpy decreases because the atomic size increases and the outermost electron is farther from the nucleus. So arsenic has a lower ionisation enthalpy than phosphorus. …

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Q.Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

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