(b) The last element of the p-block in 6th period is represented by the outermost electronic configuration. (i

NCERT Exemplar · Q13

Q. (b) The last element of the p-block in 6th period is represented by the outermost electronic configuration.

(i) 7s^2 7p^6

(ii) 5f^14 6d^10 7s^2 7p^0

(iii) 4f^14 5d^10 6s^2 6p^6

(iv) 4f^14 5d^10 6s^2 6p^4

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Concept understanding — Aufbau Principle

The Aufbau Principle: Building Atoms from the Ground Up

Imagine you're building a house. You don't start with the roof — you lay the foundation first, then the walls, then the roof. Electrons in an atom behave the same way: they fill the lowest-energy "rooms" (orbitals) before moving to higher-energy ones. That's the core idea behind the Aufbau Principle.

The word Aufbau comes from German, meaning "building up" or "construction." It describes how electrons are added one by one to an atom as you move across the periodic table.

The Intuition: Why Lowest Energy First

An atom is most stable when its electrons are in the lowest possible energy states. Think of it like a ball rolling down a hill — it naturally settles at the bottom, not halfway up. If you try to put an electron into a high-energy orbital while a lower-energy orbital is empty, the atom would be unstable and the electron would "fall" down to the empty spot, releasing energy.

So nature follows the path of least energy: fill from the bottom up.

The Precise Statement

Aufbau Principle: In the ground state of an atom, electrons fill atomic orbitals in order of increasing orbital energy. Lower-energy orbitals are completely filled before electrons occupy higher-energy orbitals.

This is not a guess — it's a rule that emerges from quantum mechanics and is confirmed by experimental data (like atomic spectra and ionization energies).

The Order of Filling: The (n + l) Rule

But here's where it gets interesting: orbital energies don't simply increase with principal quantum number $n$ . For example, the 4s orbital ( $n=4, l=0$ ) actually has lower energy than the 3d orbital ( $n=3, l=2$ ). So electrons fill 4s before 3d.

To predict the correct order, use the (n + l) rule:

Tip

For an orbital with quantum numbers $n$ and $l$ , its energy increases with $(n + l)$ . If two orbitals have the same $(n + l)$ value, the one with lower $n$ fills first.

Let's check:

4s: $n + l = 4 + 0 = 4$
3d: $n + l = 3 + 2 = 5$

Since $4 < 5$ , 4s fills first. That's why potassium (atomic number 19) has the electron configuration $[\text{Ar}]\,4s^1$ , not $[\text{Ar}]\,3d^1$ .

The Complete Filling Order (Memory Aid)

Here's the sequence you'll use for most elements up to atomic number 118:

1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p \rightarrow 6s \rightarrow 4f \rightarrow 5d \rightarrow 6p \rightarrow 7s \rightarrow 5f \rightarrow 6d \rightarrow 7p

A common trick is to write the orbitals in rows by $n$ and draw diagonal arrows:

1s
2s  2p
3s  3p  3d
4s  4p  4d  4f
5s  5p  5d  5f
6s  6p  6d
7s  7p

Follow the diagonals from top-right to bottom-left — that gives you the filling order.

Important Exceptions (Don't Panic)

The Aufbau principle works beautifully for most elements, but there are a few well-known exceptions where a half-filled or fully-filled subshell provides extra stability:

Chromium ( $Z=24$ ): Expected $[\text{Ar}]\,4s^2 3d^4$ , but actual is $[\text{Ar}]\,4s^1 3d^5$ (half-filled d-subshell is more stable) …

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The last element of the p-block in the 6th period has its $6p$ subshell completely filled. Following the Aufbau principle, the orbitals filled before $6p$ in this period are $6s$ , $4f$ , and $5d$ . The outermost electronic configuration is $\boxed{4f^{14} 5d^{10} 6s^2 6p^6}$ .

The electronic configuration of an element describes how its electrons are distributed among atomic orbitals. To determine the outermost electronic configuration for an element, we use the Aufbau principle, which states that electrons fill atomic orbitals of the lowest available energy levels before occupying higher energy levels.

For elements in the periodic table:

The period number corresponds to the principal quantum number ( $n$ ) of the outermost $s$ and $p$ orbitals being filled.
The block (s, p, d, f) indicates the type of orbital that receives the last electron (the differentiating electron).
The "last element of the p-block" implies that the p-subshell for that period is completely filled with 6 electrons.

Let's apply this to the given problem:

Identify the Period and Block:

The element is in the 6th period and is the last element of the p-block.
- Being in the 6th period means the principal quantum number for the outermost $s$ and $p$ orbitals is $n=6$ . So, we expect $6s$ and $6p$ orbitals to be involved.
- Being the "last element of the p-block" means the $6p$ subshell is completely filled, i.e., it will have $6p^6$ electrons.
Determine the Order of Orbital Filling for the 6th Period:

According to the Aufbau principle and the $(n+l)$ rule, the general order of filling orbitals for elements in the 6th period is:

$1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p \rightarrow \mathbf{6s} \rightarrow \mathbf{4f} \rightarrow \mathbf{5d} \rightarrow \mathbf{6p} \rightarrow \dots$

We are interested in the orbitals filled starting from $6s$ up to $6p$ .
$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, \dots$
Fill the Orbitals for the Last p-block Element of the 6th Period:
- The 6th period begins with the filling of the $6s$ orbital. It holds 2 electrons: $6s^2$ .
- After $6s$ , the $4f$ orbitals are filled. The $f$ -subshell holds a maximum of 14 electrons: $4f^{14}$ .
- Next, the $5d$ orbitals are filled. The $d$ -subshell holds a maximum of 10 electrons: $5d^{10}$ .
- Finally, the $6p$ orbitals are filled. Since it's the last element of the p-block, the $6p$ subshell is completely filled with 6 electrons: $6p^6$ . …

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Q.(b) The last element of the p-block in 6th period is represented by the outermost electronic configuration.

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