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Chapter 5: Problem 70

Give the electron configurations of these ions, and indicate which ones are isoelectronic. (a) \(\mathrm{Ca}^{2+}\) (b) \(\mathrm{K}^{+}\) (c) \(\mathrm{O}^{2-}\)

Short Answer

Expert verified
\(\mathrm{Ca}^{2+}\) and \(\mathrm{K}^{+}\) are isoelectronic with \([Ar]\) configuration.

Step by step solution

01

Determine Electron Configuration for Ca

The electron configuration for a neutral calcium atom (Ca) is \([Ar] 4s^2\). When it becomes \(\mathrm{Ca}^{2+}\), it loses two electrons, resulting in an electron configuration of \([Ar]\), meaning it is now balanced at the argon electron configuration.
02

Determine Electron Configuration for K

The electron configuration for a neutral potassium atom (K) is \([Ar] 4s^1\). When potassium loses an electron to become \(\mathrm{K}^{+}\), it results in an electron configuration of \([Ar]\).
03

Determine Electron Configuration for O

A neutral oxygen atom has the electron configuration \(1s^2 2s^2 2p^4\). When it gains two electrons to become \(\mathrm{O}^{2-}\), the configuration becomes \(1s^2 2s^2 2p^6\), which is equivalent to \([Ne]\).
04

Identify Isoelectronic Ions

The ions \(\mathrm{Ca}^{2+}\) and \(\mathrm{K}^{+}\) have the same electron configuration \([Ar]\) and are hence isoelectronic with each other. \(\mathrm{O}^{2-}\) has a different configuration \([Ne]\) and is not isoelectronic with the others.

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Key Concepts

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

Isoelectronic Ions
Isoelectronic ions are ions that have the same number of electrons and the same electron configuration, even though they might differ in size and charge due to differences in the number of protons. This concept helps us understand why certain ions have similar chemical properties.
  • Isoelectronic ions may belong to different elements.
  • The chemical behavior of these ions can be similar because they have the same electron structure.
  • For example, in our exercise, both \( \mathrm{Ca}^{2+} \) and \( \mathrm{K}^{+} \) have the electron configuration \([Ar]\).
Understanding isoelectronicity is crucial in predicting the reactivity and formation of compounds.
Calcium Ion
The calcium ion \( (\mathrm{Ca}^{2+}) \) is formed when a calcium atom loses two electrons. In its neutral state, calcium has the electron configuration \([Ar] 4s^2\). After losing two electrons, it becomes \([Ar]\), achieving a stable and noble gas configuration.
  • Calcium is located in Group 2 of the periodic table.
  • The process of losing two electrons makes it a cation with a 2+ charge.
  • This change gives it the same configuration as argon, making it isoelectronic with \( \mathrm{K}^{+} \).
Achieving a noble gas configuration often leads to greater stability.
Potassium Ion
The potassium ion \( (\mathrm{K}^{+}) \) forms when a potassium atom loses one electron. Initially, a neutral potassium atom has the configuration \([Ar] 4s^1\).
  • By losing that one 4s electron, it transforms into \([Ar]\) like a noble gas.
  • Potassium is a Group 1 element in the periodic table.
  • It becomes a cation with a +1 charge.
  • It is isoelectronic with \( \mathrm{Ca}^{2+} \).
This makes the potassium ion stable and less reactive.
Oxygen Ion
The oxygen ion \( (\mathrm{O}^{2-}) \) is produced when an oxygen atom gains two electrons. The initial configuration of oxygen is \(1s^2 2s^2 2p^4\). Gaining two additional electrons changes this to \(1s^2 2s^2 2p^6\), which corresponds to the noble gas configuration of neon \([Ne]\).
  • Oxygen naturally tends to gain electrons due to its high electronegativity.
  • It becomes an anion with a 2- charge.
  • Unlike \( \mathrm{Ca}^{2+} \) and \( \mathrm{K}^{+} \), \( \mathrm{O}^{2-} \) is not isoelectronic with them as it has a different electron structure \([Ne]\).
The stability gained mimics that of the noble gases, which are inert.

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Most popular questions from this chapter

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?

Suppose two electrons in the same system each have \(n=3, \ell=0\) (a) How many different electron arrangements would be possible if the Pauli exclusion principle did not apply in this case? (b) How many would apply if it is operative?

Nearly all first-row transition elements form \(2+\) ions. (a) For which of these elements are the \(2+\) ions paramagnetic? (b) For which element do compounds containing \(2+\) ions and chloride ions have the greatest paramagnetism? (Chloride ions have no unpaired electrons.)

Write the formula for a compound formed by chlorine and element \(X\), if element \(X\) has the electronic configuration \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\).

The lanthanides, or rare earths, are only "medium rare," because all can be purchased for a reasonable pricc. Give clectron configurations for atoms of thesc lanthanides. (a) Europium (Eu), the most expensive of the rare earth elements; \(1 \mathrm{~g}\) ean be purchased for ahout $$\$ 1.40$$. (b) Ytterbium (Yb). Less expensive than Eu, Yb costs only about $$\$ 0.35$$ per gram. It was named for the village of Yuerby in Sweden, where a mineral source of the element was found.
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