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Among the following elements the configuration having highest ionisation energy is (1) \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{\prime}\) (2) \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\) (3) \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}\) (4) \(|\mathrm{Ar}| 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\)

Short Answer

Expert verified
Configuration 2 (\(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\)) has the highest ionisation energy.

Step by step solution

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01

Review Definitions

Understand that ionisation energy is the energy required to remove one electron from a neutral atom in the gaseous phase.
02

Examine Electron Configurations

Look at the given electron configurations:1. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{\prime}\)2. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\)3. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}\)4. \(|\mathrm{Ar}| 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\)
03

Identify Stable Configurations

Elements with half-filled or fully-filled subshells have higher ionisation energies due to their extra stability. Compare the given configurations:2. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\) is half-filled and thus stable.4. \(|\mathrm{Ar}| 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\) is half-filled and thus stable.
04

Compare Other Configurations

Among the remaining configurations:1. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{\prime}\)3. \(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{2}\), both have less stability compared to 2 and 4.
05

Determine Atomic Number

Elements with a higher atomic number typically have lower ionisation energy due to increased distance from the nucleus. Compare the elements:- 2 (\(|\mathrm{Ne}| 3 \mathrm{~s}^{2} 3 \mathrm{p}^{3}\)) corresponds to Phosphorus.- 4 (\(|\mathrm{Ar}| 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{3}\)) corresponds to Arsenic.Since Phosphorus has a lower atomic number than Arsenic, it will have a higher ionisation energy.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
The term 'electron configuration' refers to the distribution of electrons in an atom's orbitals. It's like a map showing where each electron 'lives.' Each electron configuration is built around the atomic number, which equals the number of protons and also the number of electrons in a neutral atom. Electrons are arranged in specific orbitals (s, p, d, f) and fill up these orbitals in a sequence determined by the Aufbau principle. For example, the configuration \(\text{|Ne| 3s^{2} 3p^{3}}\) means that the element has a total electron count matching Neon (10 electrons) plus 5 more electrons in the 3rd shell. This structure indicates how electrons populate different shells and subshells around the nucleus. Understanding electron configuration is essential because the arrangement of electrons determines an element's chemical behavior. Note that more stable configurations, like those that are half-filled (e.g., p^3) or fully filled, often result in higher ionisation energy.
Atomic Stability
Atomic stability depends greatly on how electrons are arranged in an atom's orbitals. Atoms strive to achieve a stable electronic configuration, usually symbolized by a filled or half-filled shell. This stability is reflected in their chemical properties and their resistance to change. For example:
  • Fully-filled subshells (e.g., p^6, d^10) provide maximum stability.
  • Half-filled subshells (like p^3 or d^5) also confer a significant degree of stability due to symmetrical electron distribution and minimized electron repulsion.
So when comparing different electron configurations, ones that are either fully or half-filled will result in greater atomic stability. This stability then correlates with the ionisation energy because the more stable the atom is, the harder it is to remove an electron from it. In essence, a stable electron configuration means a higher ionisation energy.
Periodic Trends
Periodic trends in the periodic table, such as ionisation energy, are not random; they follow predictable patterns. Ionisation energy generally increases as you move from left to right across a period. This is because the effective nuclear charge (the pull electrons feel from the nucleus) increases, making it harder to remove an electron. On the other hand, ionisation energy decreases as you move down a group due to an increase in the atomic radius. Electrons are farther from the nucleus and shielded by inner-shell electrons, making them easier to remove. It's important to understand this for questions involving elements like Phosphorus \(\text{|Ne| 3s^2 3p^3}\) and Arsenic \(\text{|Ar| 3d^{10} 4s^{2} 4p^{3}}\). Although both have stable, half-filled p^3 subshells, Phosphorus is higher on the periodic table, meaning it has a smaller atomic radius and higher ionisation energy than Arsenic.

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