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Chapter 8: Problem 52

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide \(\quad\) c. calcium nitride b. strontium fluoride \(\quad\) d. aluminum bromide

Short Answer

Expert verified
The noble gases that have the same electron configuration as the ions in the given compounds are as follows: a. Cesium sulfide: Cs鈦� - Xenon (Xe) and S虏鈦� - Argon (Ar) b. Strontium fluoride: Sr虏鈦� - Krypton (Kr) and F鈦� - Neon (Ne) c. Calcium nitride: Ca虏鈦� - Argon (Ar) and N鲁鈦� - Neon (Ne) d. Aluminum bromide: Al鲁鈦� - Neon (Ne) and Br鈦� - Argon (Ar)

Step by step solution

01

Find the ions present in the compound

For each compound, identify the elements present, determine the corresponding ions and their charges. Remember that elements tend to gain or lose electrons to achieve a stable electron configuration like a noble gas. a. Cesium sulfide Cesium (Cs) has a +1 charge as it loses one electron, and Sulfide (S) has a -2 charge since it gains two electrons. The ions are Cs鈦� and S虏鈦�. b. Strontium fluoride Strontium (Sr) has a +2 charge when it loses two electrons, and Fluoride (F) has a -1 charge after gaining one electron. The ions are Sr虏鈦� and F鈦�. c. Calcium nitride Calcium (Ca) has a +2 charge after losing two electrons, and Nitride (N) has a -3 charge when it gains three electrons. The ions are Ca虏鈦� and N鲁鈦�. d. Aluminum bromide Aluminum (Al) has a +3 charge after it loses three electrons, and Bromide (Br) has a -1 charge when it gains one electron. The ions are Al鲁鈦� and Br鈦�.
02

Determine the electron configuration of ions

Using the ions' charges, find the electron configuration of each ion, as they gain or lose electrons to achieve a noble gas configuration. For example, Cs鈦� loses one electron from its outer shell. So, we need to find the electron configuration of Cs without its outermost electron to find the configuration of Cs鈦�. Perform this process for all ions present in each compound.
03

Identify the noble gas with the same electron configuration

Compare the electron configurations of the ions obtained in Step 2 to the electron configurations of noble gases, and identify the noble gas with the same electron configuration. a. Cesium sulfide: Cs鈦� has a similar electron configuration to Xenon (Xe). S虏鈦� also has a similar electron configuration to Argon (Ar). b. Strontium fluoride: Sr虏鈦� has a similar electron configuration to Krypton (Kr). F鈦� has a similar electron configuration to Neon (Ne). c. Calcium nitride: Ca虏鈦� has a similar electron configuration to Argon (Ar). N鲁鈦� has a similar electron configuration to Neon (Ne). d. Aluminum bromide: Al鲁鈦� has a similar electron configuration to Neon (Ne). Br鈦� has a similar electron configuration to Argon (Ar). The electron configurations of ions are similar to noble gases as they have full outer electron shells, making them stable.

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Key Concepts

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

Ionic Compounds
Ionic compounds are formed when atoms transfer electrons from one to another, leading to the formation of ions. These ions then attract each other due to opposite charges, forming a bond. For example, in cesium sulfide, cesium atoms lose electrons to become Cs鈦�, and sulfur atoms gain electrons to become S虏鈦�. When these ions combine, they create the ionic compound cesium sulfide.

This electron transfer ensures that each atom achieves a stable electron configuration similar to that of noble gases, achieving electric neutrality in the compound. Ionic compounds generally have high melting and boiling points, and they conduct electricity when melted or dissolved because of the movement of ions.

Key characteristics of ionic compounds:
  • Formation through electron transfer
  • High melting and boiling points
  • Conduct electricity in a liquid state or solution
Electron Configuration
Electron configuration refers to the arrangement of electrons in an atom's electron shells or orbitals. Atoms strive to be stable, often achieving the electron configuration of the nearest noble gas. This stability results when an atom has full outer electron shells, a state usually seen in noble gases.

For ions, achieving noble gas electron configurations means either gaining or losing electrons. For instance, Cs鈦�, in cesium sulfide, has lost an electron from its outer shell, and thus its electron configuration mirrors that of xenon. The sulfur ion S虏鈦�, on the other hand, gains two electrons to reach the electron configuration of argon.

How to determine electron configuration:
  • Identify the atomic number to know the number of electrons.
  • Write the configuration starting from the lowest energy level.
  • Apply rules such as Pauli's exclusion and Hund鈥檚 rule for accurate distribution.
Chemical Stability
Chemical stability means how likely a substance is to undergo a chemical change. Atoms and ions with full electron shells are generally more stable. This desire for stability drives atoms in chemical reactions to form compounds which allow them to reach a noble gas-like electron configuration.

Noble gases are inherently stable, so elements will often gain or lose electrons in reactions to mirror the electron configuration of the closest noble gas. For example, Sr虏鈦� in strontium fluoride resembles krypton, a stable noble gas. This enables the structure's stability, reducing likelihood of further chemical change.

Characteristics of chemical stability:
  • Full outer electron shells
  • Resilience to further reactions
  • Minimal energy configurations
Ion Formation
Ion formation is a process where atoms gain or lose electrons to form charged particles. Metals typically lose electrons to become positively charged cations, while nonmetals gain electrons to become negatively charged anions. This is evident in compounds like aluminum bromide, where Al loses three electrons forming Al鲁鈦�, and Br gains one electron becoming Br鈦�.

The drive behind ion formation is the quest for a stable electron configuration, typically resembling that of the noble gases. This alignment offers ions a balanced and low-energy state, minimizing further electron exchanges.

Steps in ion formation:
  • Identify electrons in the outer shell.
  • Determine if gaining or losing electrons leads to stability.
  • Adjust electrons and note the resulting charge.
Ion formation is key for creating stable compounds, as ions combine to neutralize overall charge.

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Most popular questions from this chapter

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$: \mathrm{N}=\mathrm{N}=\dot{\mathrm{O}}\longleftrightarrow: \mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{Q}} : \longleftrightarrow : \dot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}$$ Given the following bond lengths, \(\mathrm{N}-\mathrm{N} \qquad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\) \(\mathrm{N}=\mathrm{N} \qquad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\) \(\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm}\) rationalize the observations that the N-N bond length in \(\mathrm{N}_{2} \mathrm{O}\) is 112 \(\mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is 119 \(\mathrm{pm}\) . Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\) . Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

What do each of the following sets of compounds/ions have in common with each other? a. \(\mathrm{SO}_{3}, \mathrm{NO}_{3}^{-}, \mathrm{CO}_{3}^{2-}\) b. \(\mathrm{O}_{3}, \mathrm{SO}_{2}, \mathrm{NO}_{2}-\)

Calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) shells are used by mollusks, corals, and snails to form protective coverings. Draw the Lewis structure for \(\mathrm{CaCO}_{3} .\) Be sure to include any resonance structures.

Draw a Lewis structure for the \(N, N\) -dimethylformamide molecule. The skeletal structure is Various types of evidence lead to the conclusion that there is some double bond character to the C-N bond. Draw one or more resonance structures that support this observation.

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for magnesium fluoride. $$\mathrm{Mg}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{MgF}_{2}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-22913 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of } \mathrm{Mg}} & \quad{735 \mathrm{kJ} / \mathrm{mol}} \\ {\text {Second ionization energy of } \mathrm{Mg}} & \quad {1445 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Electron affinity of } \mathrm{F}} & {-328 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Bond energy of } \mathrm{F}_{2}} & \quad {154 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation for } \mathrm{Mg}} & \quad {150 . \mathrm{kJ} / \mathrm{mol}} \end{array}\)
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