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Chapter 14: Problem 133

Which of the following oxygen ions are paramagnetic: \(\mathrm{O}^{+}\). \(\mathrm{O}^{-}, \mathrm{O}^{2-}, \mathrm{O}^{2+} ?\)

Short Answer

Expert verified
\( \text{O}^{+} \), \( \text{O}^{-} \), and \( \text{O}^{2+} \) are paramagnetic.

Step by step solution

01

- Identify the number of electrons in each ion

Oxygen (O) has 8 electrons. Calculate the number of electrons for each ion: - \( \text{O}^{+} \): 8 - 1 = 7 electrons - \( \text{O}^{-} \): 8 + 1 = 9 electrons - \( \text{O}^{2-} \): 8 + 2 = 10 electrons - \( \text{O}^{2+} \): 8 - 2 = 6 electrons
02

- Write the electron configuration for each ion

Using the number of electrons, write the electron configuration for each ion: - \( \text{O}^{+} \): 1s^2 2s^2 2p^3 - \( \text{O}^{-} \): 1s^2 2s^2 2p^5 - \( \text{O}^{2-} \): 1s^2 2s^2 2p^6 - \( \text{O}^{2+} \): 1s^2 2s^2 2p^2
03

- Determine if the ion is paramagnetic or diamagnetic

An ion is paramagnetic if it has at least one unpaired electron. - \( \text{O}^{+} \): 2p^3 - 3 unpaired electrons - \( \text{O}^{-} \): 2p^5 - 1 unpaired electron - \( \text{O}^{2-} \): 2p^6 - 0 unpaired electrons - \( \text{O}^{2+} \): 2p^2 - 2 unpaired electrons
04

- Conclusion

From the electron configurations, it is evident that the ions with unpaired electrons are paramagnetic. \( \text{O}^{+} \), \( \text{O}^{-} \), and \( \text{O}^{2+} \) have unpaired electrons and are therefore paramagnetic. \( \text{O}^{2-} \) has no unpaired electrons and is diamagnetic.

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Key Concepts

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

Paramagnetism
Paramagnetism occurs when an ion or an atom has one or more unpaired electrons. These unpaired electrons create a magnetic dipole moment which causes the substance to be attracted to a magnetic field. This property arises from the intrinsic angular momentum of electrons known as 'spin.' Common examples of paramagnetic materials include oxygen ions such as \(\text{O}^{+}\), \(\text{O}^{-}\), and \(\text{O}^{2+}\). The number of unpaired electrons directly influences the strength of the paramagnetic behavior.
Diamagnetism
Diamagnetism is a property where materials have all their electrons paired, resulting in no net magnetic moment. This means they are repelled by magnetic fields rather than attracted. When all electrons in an atom or ion are paired, their spins cancel out each other鈥檚 magnetic moments. A common example is the oxygen ion \(\text{O}^{2-}\), which is diamagnetic because it has no unpaired electrons. Understanding the difference between paramagnetic and diamagnetic materials is crucial for students to grasp magnetic properties of substances.
Electron Configuration
Electron configuration refers to the distribution of electrons of an atom or ion in its atomic orbitals. In the correct configuration, electrons are arranged in increasing energy levels. Here are some examples of oxygen ions:
  • \(\text{O}^{+}\): \(1s^2 2s^2 2p^3\), has 3 unpaired electrons
  • \(\text{O}^{-}\): \(1s^2 2s^2 2p^5\), has 1 unpaired electron
  • \(\text{O}^{2-}\): \(1s^2 2s^2 2p^6\), all electrons are paired
  • \(\text{O}^{2+}\): \(1s^2 2s^2 2p^2\), has 2 unpaired electrons
Understanding electron configurations helps in determining the magnetic properties of ions. This concept is foundational for both chemistry and physics.
Ionic Charge
Ionic charge results from an atom gaining or losing electrons to become more stable, often resembling the nearest noble gas configuration. For oxygen:
  • Neutral oxygen (\(\text{O}\)) has 8 electrons
  • \(\text{O}^{+}\) has lost 1 electron, resulting in 7 electrons
  • \(\text{O}^{-}\) has gained 1 electron, resulting in 9 electrons
  • \(\text{O}^{2-}\) has gained 2 electrons, resulting in 10 electrons
  • \(\text{O}^{2+}\) has lost 2 electrons, resulting in 6 electrons
The charge affects the electron configuration and is crucial for predicting reactivity and bonding behavior.
Unpaired Electrons
Unpaired electrons are electrons that occupy an orbital alone, without a corresponding electron with opposite spin. They play a critical role in magnetic properties. The presence of unpaired electrons can be detected by examining the electron configuration of an ion. Here's how:
  • For \(\text{O}^{+}\): \(2p^3\), 3 unpaired electrons
  • For \(\text{O}^{-}\): \(2p^5\), 1 unpaired electron
  • For \(\text{O}^{2-}\): \(2p^6\), 0 unpaired electrons
  • For \(\text{O}^{2+}\): \(2p^2\), 2 unpaired electrons
By identifying unpaired electrons, it is easier to classify the paramagnetic or diamagnetic nature of the ion, which is important across various fields like material science and quantum chemistry.

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Most popular questions from this chapter

Assuming acid strength relates directly to number of \(\mathrm{O}\) atoms bonded to the central atom, rank \(\mathrm{H}_{2} \mathrm{~N}_{2} \mathrm{O}_{2}\left[\mathrm{or}(\mathrm{HON})_{2}\right], \mathrm{HNO}_{3}\) (or HONO \(_{2}\) ), and \(\mathrm{HNO}_{2}\) (or HONO) in order of decreasing acid strength.

Zeolite \(\mathrm{A}, \mathrm{Na}_{12}\left[\left(\mathrm{AlO}_{2}\right)_{12}\left(\mathrm{SiO}_{2}\right)_{12}\right] \cdot 27 \mathrm{H}_{2} \mathrm{O},\) is used to soften water because it replaces \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) dissolved in the water with \(\mathrm{Na}^{+}\). Hard water from a certain source is \(4.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ca}^{2+}\) and \(9.2 \times 10^{-4} \mathrm{M} \mathrm{Mg}^{2+},\) and a pipe delivers \(25,000 \mathrm{~L}\) of this hard water per day. What mass (in kg) of zeolite A is needed to soften a week's supply of the water? (Assume zeolite A loses its capacity to exchange ions when \(85 \mathrm{~mol} \%\) of its \(\mathrm{Na}^{+}\) has been lost.)

Given the following information, $$\begin{array}{rlrr}\mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(g) & \Delta H=-720 \mathrm{~kJ} \\\ \mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q) & & \Delta H=-1090 \mathrm{~kJ} \\ \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H= & 40.7 \mathrm{~kJ} \end{array}$$ calculate the heat of solution of the hydronium ion: $$\mathrm{H}_{3} \mathrm{O}^{+}(g) \stackrel{\mathrm{H}_{3} \mathrm{O}}{\longrightarrow} \mathrm{H}_{3} \mathrm{O}^{+}(a q)$$

Give the name and symbol or formula of a Group \(4 \mathrm{~A}(14)\) element or compound that fits each description or use: (a) Hardest known natural substance (b) Medicinal antacid (c) Atmospheric gas implicated in the greenhouse effect (d) Gas that binds to Fe(II) in blood (e) Element used in the manufacture of computer chips

Despite the expected decrease in atomic size, there is an unexpected drop in the first ionization energy between Groups \(2 \mathrm{~A}(2)\) and \(3 \mathrm{~A}(13)\) in Periods 2 through \(4 .\) Explain this pattern in terms of electron configurations and orbital energies.
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