In terms of period and group where would you locate the element with Z = 114?

Exercises · 3.5

Z = 114

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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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Element 114 (flerovium) sits in Period 7, Group 14 — directly below lead in the carbon family, found by filling electrons through the 7th shell and counting four valence electrons in the p-block.

Why electron configuration determines position

The periodic table is organized by how electrons fill atomic orbitals. The period number tells you the highest principal quantum level ( $n$ ) occupied by electrons, while the group number reflects the valence electron count and the block (s, p, d, f) tells you which subshell is being filled last.

For superheavy elements like $Z = 114$ , we follow the Aufbau principle through the entire sequence of orbital filling, then read off the period from the outermost shell and the group from the valence configuration.

Step-by-step location

Write the full electron configuration for $Z = 114$

Following the standard filling order (1s → 2s → 2p → 3s → 3p → 4s → 3d → … → 7p), we build up 114 electrons:

[\text{Rn}] \, 5f^{14} \, 6d^{10} \, 7s^2 \, 7p^2

Here radon ( $Z = 86$ ) accounts for the first 86 electrons, then we fill the 5f subshell (14 electrons), the 6d subshell (10 electrons), the 7s subshell (2 electrons), and finally place 2 electrons in 7p.

Check: $86 + 14 + 10 + 2 + 2 = 114$ ✓

Identify the period

The outermost electrons occupy the $n = 7$ shell (7s and 7p orbitals). The period number equals the highest principal quantum number with electrons.

Period = 7
Identify the block

The last electron enters a 7p orbital, so element 114 belongs to the p-block.
Determine the group number

For p-block elements, the group number is $10 + \text{(number of s and p valence electrons)}$ .

Valence electrons: $7s^2 \, 7p^2$ gives us $2 + 2 = 4$ valence electrons. …

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Q.In terms of period and group where would you locate the element with Z=114Z = 114Z=114?

Why electron configuration determines position

Step-by-step location

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Q.In terms of period and group where would you locate the element with $Z = 114$ ?

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