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Write electron configurations to show the first two ionization steps for potassium. Explain why the second ionization energy is much larger than the first.

Short Answer

Expert verified
The second ionization energy is much larger because it disturbs a stable electron configuration.

Step by step solution

01

Understand the Ionization Process

Ionization energy refers to the energy required to remove an electron from an atom or ion. The first ionization energy involves removing the first, outermost electron, while the second ionization energy involves removing a second electron from the newly formed ion.
02

Write the Electron Configuration of Neutral Potassium

Potassium (K) is element 19, so its neutral atom configuration starts with the electrons filling the lowest available energy levels in order of increasing energy: \[ \text{Neutral K: 1s}^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \] This configuration shows that potassium has one electron in the 4s subshell, which is the outermost electron.
03

First Ionization of Potassium

Removing one electron from a neutral potassium atom results in the formation of a K+ ion:\[ \text{K}^+: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \]This removal happens from the 4s subshell.
04

Second Ionization of Potassium

The second ionization involves the removal of an electron from the K+ ion, resulting in a K2+ ion. However, after the first ionization, the K+ ion has a stable noble gas configuration (same as Argon):\[ \text{K}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^5 \]This removal happens from the 3p subshell, breaking the stable configuration.
05

Explain the Energy Difference

The first ionization energy is low because it involves removing one loosely held 4s electron. The second ionization energy is much larger because it requires removing an electron from a full and stable 3p shell, thereby breaking its stable electron configuration similar to a noble gas.

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Key Concepts

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

Electron Configuration
In the world of chemistry, understanding the electron configuration of an element is crucial. Electron configuration provides a method to describe the distribution of electrons in an atom's orbitals. Electrons fill orbitals in a specific order, following the "Aufbau principle", Pauli exclusion principle, and Hund's rule.

The Aufbau principle states that electrons occupy the lowest energy orbitals first. For instance, the 1s orbital is filled before the 2s orbital. Pauli exclusion principle says no two electrons can have identical sets of quantum numbers in an atom. Lastly, Hund's rule notes that electrons will fill degenerate orbitals (orbitals of the same energy) singly before pairing occurs.

In the electron configuration, this is often represented as series of numbers and letters, such as 1s, 2s, 2p, and so on. Here, the numbers depict the energy level or shell, and the letters correspond to the type of orbital (s, p, d, f). The superscript number identifies the number of electrons in those orbitals.
Potassium
Potassium, symbolized as K, is the 19th element on the periodic table. It's a member of the alkali metal group, denoting that it's highly reactive and exists primarily in ionic form in nature.

The neutral electron configuration of Potassium is written as \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^1 \].

This means potassium has an electron filling the outermost 4s orbital. The loosely held nature of this 4s electron makes Potassium highly reactive. Upon first ionization, when this 4s electron is removed, Potassium forms a K+ ion.
  • The first ionization energy of Potassium is relatively low due to the ease with which the outer 4s electron is removed.
  • Once the 4s electron is not present, the resulting K+ ion achieves a stable, noble gas configuration similar to Argon.
Noble Gas Configuration
Noble gas configuration is a critical concept in understanding why certain elements or ions are especially stable. Noble gases, located in Group 18 of the periodic table, are noted for their complete electron shells, which render them chemically inert.

For an element like potassium, achieving a noble gas configuration is pivotal after ionization. When Potassium loses its single 4s electron, the resulting K+ ion experiences a complete octet, matching the electron configuration of Argon: \[ 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \].

This configuration makes the K+ ion energetically favorable and chemically stable.
  • Any further ionization (removal of an electron) from this stable state requires a substantial amount of energy, hence the second ionization energy of Potassium is significantly higher.
  • The removal of a second electron disturbs the stability, breaking into the filled p orbital structure.
Understanding noble gas configurations is essential in predicting the likelihood of ion formation and the energy needed to achieve such transformations.

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Most popular questions from this chapter

Spectroscopists have observed \(\mathrm{He}^{+}\) in outer space. This ion is a one-electron species like a neutral hydrogen atom. Calculate the energy of the photon emitted for the transition from the \(n=5\) to the \(n=3\) state in this ion using the equation: \(E_{n}=-Z^{2} / n^{2}\left(2.179 \times 10^{-18} \mathrm{~J}\right) . \mathrm{Z}\) is the positive charge of the nucleus and \(n\) is the principal quantum number. In what part of the electromagnetic spectrum does this radiation lie?

The ionization energy of \(\mathrm{H}(\mathrm{g})\) is \(1312 \mathrm{~kJ} / \mathrm{mol}\). The second ionization energy of \(\mathrm{He}(\mathrm{g})\) is almost exactly four times this value, and the third ionization energy of \(\mathrm{Li}(\mathrm{g})\) is almost exactly nine times that of hydrogen. What relationship, if any, do these data suggest between ionization energy and a fundamental characteristic of elements?

Write electron configurations for these elements. (a) Zirconium (Zr). This metal is exceptionally resistant to corrosion and so has important industrial applications. Moon rocks show a surprisingly high rirconium content compared with rocks on Earth. (b) Rhodium (Rh), which is used in jewelry and in industrial catalysts.

The Bohr equation for hydrogen can be modified to apply to one-electron species other than uncharged hydrogen atoms, for example \(\mathrm{Li}^{2+},\) to calculate the energy of electron transitions in the ion. The modified equation is \(E_{n}=-Z^{2} / n^{2}\left(2.179 \times 10^{-18} \mathrm{~J}\right) . Z\) is the pos- itive charge of the nucleus and \(n\) is the principal quantum number. Calculate the energy of the photon emitted for the transition from the \(n=4\) to the \(n=1\) state in this ion. In what region of the electromagnetic spectrum does it lie?

Place these atoms and ions in order of decreasing size: \(\mathrm{Ar}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}, \mathrm{Ca}^{2+} .\) Briefly explain your reasoning.

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