Question

A neutral atom will have the lowest ionisation potential when electronic configuration is (1) \(1 \mathrm{~s}^{1}\) (2) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}\) (3) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\) (4) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}\)

Short answer

Option 3: \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6} 3 \text{ s}^{1}\).

Step-by-step solution

Understand ionisation potential

The ionisation potential (or ionisation energy) is the energy required to remove an electron from a neutral atom in its gaseous state. Atoms with stable electronic configurations usually have higher ionisation potentials.

Identify the electronic configurations

The given configurations are: 1. \(1 \text{ s}^{1}\) 2. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6}\) 3. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6} 3 \text{ s}^{1}\) 4. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{2}\)

Assess the stability of each configuration

Evaluate how stable each configuration is: 1. \(1 \text{ s}^{1}\) corresponds to Hydrogen. 2. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6}\) corresponds to Neon, which is a noble gas and highly stable. 3. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6} 3 \text{ s}^{1}\) corresponds to Sodium, which has a single valence electron in the 3s orbital. 4. \(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{2}\) corresponds to carbon, which is less stable compared to Neon.

Identify the lowest ionisation potential

Atoms with a single electron in an outermost shell (like Hydrogen in 1s or Sodium in 3s) will have lower ionisation potentials because these electrons are easier to remove. Neon has a full outer shell, making its ionisation potential high. Between Hydrogen and Sodium, Sodium's single 3s electron will be easier to remove than Hydrogen's 1s electron due to increased distance from the nucleus and shielding effects.

Conclusion

\(1 \text{ s}^{2} 2 \text{ s}^{2} 2 \text{ p}^{6} 3 \text{ s}^{1} \) (option 3) has the lowest ionisation potential among the options given.

Key concepts

Electronic Configuration

An electronic configuration describes how electrons are distributed in an atom's orbitals. For example, the configuration \(1s^2 2s^2 2p^1\) means there are two electrons in the 1s orbital, two in the 2s orbital, and one in the 2p orbital. This arrangement is crucial for predicting an element's chemical behavior. The main shells are represented by the numbers (1, 2, 3, etc.), and subshells are labeled as s, p, d, f.
The most stable configuration is one resembling the noble gases, such as Neon (\(1s^2 2s^2 2p^6\)), where all orbitals are fully occupied by electrons. Stability increases with a complete outer shell. In general, electron configurations determine an atom's reactivity, ionization energy, and other properties.

Ionisation Energy

Ionisation energy, also known as ionisation potential, is the energy required to remove an electron from a neutral atom in its gaseous state. This energy reflects how tightly an electron is bound to the nucleus. Higher ionisation energy means the electron is more tightly bound and harder to remove.
This concept is essential in understanding chemical reactivity. Elements with lower ionisation energy tend to lose electrons easily, making them more reactive if they form positive ions. Consider the given electron configurations:

• \(1s^1\): Hydrogen - Low ionisation energy because it's a single electron close to the nucleus.
• \(1s^2 2s^2 2p^6\): Neon - Very high ionisation energy due to a complete and stable electron shell.
• \(1s^2 2s^2 2p^6 3s^1\): Sodium - Lower ionisation energy (compared to Neon), as the outermost 3s electron is easily removed.
• \(1s^2 2s^2 2p^2\): Carbon - Moderate ionisation energy because it lacks a complete outer shell.

Atomic Stability

Atomic stability is directly linked to the arrangement of electrons around the nucleus. Generally, atoms achieve greater stability when their outer electron shells are full or correspond to a noble gas configuration. For example:

• Neon (\(1s^2 2s^2 2p^6\)): Highly stable due to a complete outer shell with 8 electrons.
• Sodium (\(1s^2 2s^2 2p^6 3s^1\)): Less stable because it has only one electron in its outermost 3s shell, making it more likely to lose that electron and achieve a stable noble gas configuration (Neon-like).

Electrons in outer shells, or valence electrons, are most critical for chemical reactions. Atoms often lose, gain, or share electrons to achieve a full valence shell and greater stability. For example, sodium readily loses its single valence electron (3s) to become more stable.

Neutral Atom

A neutral atom has an equal number of protons and electrons, resulting in no overall charge. For example, Rutherfordium (Rf) with an atomic number of 104 has 104 protons and 104 electrons, making it electrically neutral.
When neutral atoms gain or lose electrons, they become ions. Losing an electron turns the atom into a positive ion (cation), while gaining results in a negative ion (anion). Ionisation energy focuses on removing an electron from a neutral atom, highlighting the ease with which an atom can transform into a cation.
Understanding the behavior of neutral atoms is crucial for predicting how they will interact in various chemical reactions. Consider Sodium (\(1s^2 2s^2 2p^6 3s^1\)), which in its neutral state easily loses its 3s electron because it requires minimal energy, transitioning into a stable positive ion (Na+).