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Chapter 10: Problem 27

A sudden jump between the values of second and third ionization energies of element would be associated with which of the following electronic configurations. (a) \(1 s^{2} 2 s^{2} 2 v^{6} 3 s^{1}\) (b) \(1 s^{2}=^{2} 2 p^{6} 3 s^{2} 3 p^{\prime}\) (c) \(1 \mathrm{~s}^{2} 2 s^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{1} 3 \mathrm{p}^{2}\) (d) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}\)

Short Answer

Expert verified
The large ionization energy jump is associated with configuration (d).

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. A large jump in ionization energy occurs when an electron is removed from a stable, fully filled or half-filled energy level, as these configurations are more stable.
02

Evaluating Each Electronic Configuration

We need to identify which electronic configuration would display a large jump between the second and third ionization energies. This often occurs when the electron being removed after the second ionization is from a more stable, lower energy level.
03

Analyze Option (a)

Configuration: \(1s^2 2s^2 2v^6 3s^1\). Here, the first and second ionization energies will involve removing two 3s electrons. No jump is likely as both are similar in energy.
04

Analyze Option (b)

Configuration: \(1s^2 2p^6 3s^2 3p'\). This is malformed and not directly understandable. Seems incorrect syntax for electronic configuration.
05

Analyze Option (c)

Configuration: \(1s^2 2s^2 2p^6 3s^1 3p^2\). The third ionization involves removing a 2p electron after two outer 3s or 3p electrons are removed, leading to no significant jump between second and third ionizations.
06

Analyze Option (d)

Configuration: \(1s^2 2s^2 2p^6 3s^2\). Here, after removing two 3s electrons (the first two ionizations), the third ionization will remove an electron from the core level (2p), which is a stable and lower energy level, causing a significant jump in ionization energy.
07

Final Decision

Based on the analysis, option (d) \( 1s^2 2s^2 2p^6 3s^2 \) would show a large jump between the second and third ionization energies due to the removal of a core electron after the outer 3s electrons are removed.

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Key Concepts

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

Electronic Configuration
Electronic configuration is the arrangement of electrons in an atom's orbitals. Electrons are filled into orbitals based on the Aufbau principle, which dictates the order of filling from lower to higher energy levels.
The order generally follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on. This arrangement affects an element's chemical properties and its behavior during the ionization process.
  • Core Electrons: Found in inner shells like 1s and 2p, often not involved in bonding.
  • Valence Electrons: Located in the outermost shell, play a key role in chemical reactions and bonding.
Understanding the electronic configuration helps predict how an atom might gain, lose, or share electrons.
Energy Levels
Energy levels are regions around an atom's nucleus where electrons are likely to be found. These levels are akin to steps on a ladder, with electrons filling lower energy levels first.
Each level can only hold a certain number of electrons:
  • 1st level holds up to 2 electrons (1s)
  • 2nd level holds up to 8 electrons (2s, 2p)
  • 3rd level holds up to 18 electrons (3s, 3p, 3d)
Adding electrons progresses from one level to the next, influencing the atom鈥檚 stability and reactivity. Electrons closer to the nucleus have lower energy, making them more stable and harder to remove.
Stable Configurations
Atoms are more stable when they have filled or half-filled energy levels. This is due to the uniform distribution of electrons, which minimizes repulsion and maximizes stability.
  • Filled Configurations: These occur, for example, with a completely filled 2p subshell. It minimizes energy and increases stability.
  • Half-Filled Configurations: These are another point of stability, such as a half-filled "3p" orbital.
The stability of an atom's electronic configuration profoundly influences its ionization energy, with electrons in filled or half-filled levels being harder to remove.
Ionization Process
The ionization process involves removing an electron from an atom, which requires energy. The amount of energy needed to ionize depends on the electron's location and the atom鈥檚 electronic configuration.
  • First Ionization Energy: Energy required to remove the outermost electron.
  • Subsequent Ionization Energies: Increase as more electrons are removed, especially when moving from a stable configuration.
  • Significant Energy Jump: Occurs when an electron is removed from a stable configuration, such as moving from 3s to a core level like 2p. This shift requires overcoming the stability of the lower energy level.
In essence, these jumps indicate a change in the electronic configuration's stability, with larger energy required to remove electrons closer to the core, due to their more stable position.

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Most popular questions from this chapter

The successive ionization energies in kJ/mol of an element \(\mathrm{P}\) are \(740,1500,7000,10500\), \(13600,18000,21700 .\) Which ion is the most likely to be formed when \(X\) reacts with chloride ? (a) \(\underline{p^{2}}\) (b) \(\mathrm{P}^{+}\) (c) \(\mathrm{P}^{+2}\) (d) \(\mathrm{P}^{+3}\)

Among \(\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SiO}_{2}, \mathrm{P}_{2} \mathrm{O}_{3}\) and \(\mathrm{SO}_{2}\) the correct order of acid strength is (a) \(\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}\) b) \(\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}<\mathrm{SO}_{2}\) (c) \(\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{P}_{2} \mathrm{O}_{3}\) d) \(\mathrm{SiO}_{2}<\mathrm{SO}_{2}<\mathrm{Al}_{2} \mathrm{O}_{3}<\mathrm{P}_{2} \mathrm{O}_{3}\)

The atomic sizes are not significantly different for which series of elements? (a) \(\mathrm{Bi}, \mathrm{Na}, \mathrm{K}, \mathrm{Rb}\) (b) \(\mathrm{Na}, \mathrm{Mg}, \mathrm{Al}, \mathrm{Si}\) (c) \(\mathrm{O}, \mathrm{S}, \mathrm{Sc}, \mathrm{Te}\) (d) \(\mathrm{Cr}, \mathrm{Mn}, \mathrm{Fe}, \mathrm{Co}\)

An element is called Ununbium. What is its atomic number? (a) 102 (b) 112 (c) 98 (d) 108

Four successive members of the first row transition elements are listed below with then atomic numbers. Which one of them is expected to have the highest third ionization enerig \(?\) (a) Vanadium \((Z=23)\) (b) Chromium \((Z=24)\) (c) Iron \((Z=26)\) (d) Manganese \((Z=35)\)
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