(i) Write the electronic configurations of the following ions: (a) H^- (b) Na^+ (c) O²⁻ (d) F^- (ii) What are

Exercises · 2.23

Q. (i) Write the electronic configurations of the following ions:

(a)

H^-

(b)

Na^+

(c)

O^{2-}

(d)

F^-

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a)

3s^1

(b)

2p^3

and

(c)

3p^5

(iii) Which atoms are indicated by the following configurations?

(a)

[He]\ 2s^1

(b)

[Ne]\ 3s^2\ 3p^3

(c)

[Ar]\ 4s^2\ 3d^1

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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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This problem covers three core ideas: writing electronic configurations for ions, deducing atomic numbers from outer-electron configurations, and identifying elements from noble-gas-core configurations. The key is to remember that ions gain or lose electrons from the neutral atom’s configuration, and that the outermost electrons determine the element’s group and period.

(i) Electronic configurations of the ions

The trick with ions is simple: cations lose electrons, anions gain electrons. Always start from the neutral atom’s configuration, then adjust.

1. (a) $H^-$

Neutral hydrogen has one electron: $1s^1$ . The $H^-$ ion has gained one extra electron, so it has two electrons total. Both go into the 1s orbital.

Configuration: $1s^2$

This is the same as the noble gas helium.

2. (b) $Na^+$

Neutral sodium (atomic number 11) has the configuration $1s^2\ 2s^2\ 2p^6\ 3s^1$ . To form $Na^+$ , it loses the single 3s electron.

Configuration: $1s^2\ 2s^2\ 2p^6$

This is the neon configuration.

3. (c) $O^{2-}$

Neutral oxygen (atomic number 8) is $1s^2\ 2s^2\ 2p^4$ . The $O^{2-}$ ion gains two electrons, which fill the 2p subshell completely.

Configuration: $1s^2\ 2s^2\ 2p^6$

Again, the neon configuration.

4. (d) $F^-$

Neutral fluorine (atomic number 9) is $1s^2\ 2s^2\ 2p^5$ . Gaining one electron completes the 2p subshell.

Configuration: $1s^2\ 2s^2\ 2p^6$

Also neon.

Note

All three ions $Na^+$ , $O^{2-}$ , and $F^-$ have the same electronic configuration — they are isoelectronic with neon. The $H^-$ ion is isoelectronic with helium.

(ii) Atomic numbers from outermost electron configurations

The outermost electrons tell you which shell and subshell the element’s last electron(s) occupy. The atomic number is the total number of electrons in the neutral atom.

1. (a) $3s^1$

This means the outermost electron is in the 3s subshell. The element is in period 3, group 1 (alkali metals). The only element with a single 3s electron is sodium.

Atomic number: 11

(Configuration: $1s^2\ 2s^2\ 2p^6\ 3s^1$ )

2. (b) $2p^3$

The outermost electrons are in the 2p subshell, with three electrons. This is period 2, group 15 (the nitrogen family). The element is nitrogen.

Atomic number: 7

(Configuration: $1s^2\ 2s^2\ 2p^3$ )

3. (c) $3p^5$

Outermost electrons in the 3p subshell, with five electrons. Period 3, group 17 (halogens). The element is chlorine.

Atomic number: 17

(Configuration: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^5$ ) …

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Q.(i) Write the electronic configurations of the following ions:

(i) Electronic configurations of the ions

(ii) Atomic numbers from outermost electron configurations

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