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Explain why Xe, and to a limited extent \(\mathrm{Kr},\) form compounds, whereas He, Ne, and Ar do not.

Short Answer

Expert verified
Xe and Kr can form compounds due to their larger atomic size and higher polarizability, unlike smaller and less polarizable He, Ne, and Ar.

Step by step solution

01

- Understand Noble Gases

Noble gases are elements in Group 18 of the periodic table. They are known for their lack of reactivity due to having complete electron shells.
02

- Recognize the Role of Electron Configuration

Xe, Kr, He, Ne, and Ar have complete electron shells, making them stable. However, their ability to form compounds is influenced by their electron configurations and energy levels.
03

- Identify Differences in Atomic Size and Polarizability

Xe has a larger atomic size and higher polarizability compared to He, Ne, and Ar. Polarizability is the ease with which the electron cloud can be distorted to form temporary dipoles.
04

- Explain Formation of Compounds with Xe and Kr

Xe and to some extent Kr can form compounds because their larger atomic sizes and higher polarizability allow their electron clouds to overlap with other atoms more readily, enabling bond formation.
05

- Discuss He, Ne, and Ar Inertness

He, Ne, and Ar are much smaller and less polarizable, which makes their electron clouds less likely to distort and overlap with other atoms. This reduces their ability to form stable compounds.

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Key Concepts

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

Electron Configuration
Noble gases are unique due to their electron configurations. Each of these gases has a complete set of electrons in their outermost shell, achieving a stable configuration. This full valence shell is what makes them so stable and generally non-reactive.

  • Helium (He) has 2 electrons, filling its 1s orbital.
  • Neon (Ne) has 10 electrons, filling the 2s and 2p orbitals.
  • Argon (Ar) has 18 electrons, filling the 3s and 3p orbitals.
  • Krypton (Kr) has 36 electrons, filling the 4s, 4p, and partially 4d orbitals.
  • Xenon (Xe) has 54 electrons, filling up to the 5p orbitals and partly the 5d orbitals.

Since these levels are full, there are very few 'empty spots' for bonds to form. However, the further down the group you go, the more complex the electron configurations become. This increases the energy levels and the potential for interactions with other atoms.
Atomic Size
Atomic size plays a critical role in the reactivity of noble gases. As you move down the group in the periodic table, the atomic size increases.

  • Helium has the smallest atomic radius.
  • Neon, Argon, and Krypton follow in increasing size.
  • Xenon has the largest atomic radius among common noble gases.

The larger the atom, the more 'room' it has for its electron cloud to interact with other electrons. In Xe, the large atomic size means the outer electrons are further from the nucleus and can more easily overlap with the electron clouds of other atoms. This increases the likelihood of forming compounds compared to smaller noble gases like He, Ne, and Ar.
Polarizability
Polarizability refers to how easily the electron cloud around an atom can be distorted. Larger atoms with more electrons tend to have higher polarizability.

  • Helium, with its small size and tight electron cloud, is barely polarizable.
  • Neon and Argon have slightly more polarizability but are still very stable.
  • Krypton and Xenon, being larger, have electron clouds that are more easily polarizable.

This distortion allows for greater interaction with other atoms, leading to temporary dipoles. These dipoles can enable the formation of compounds. Therefore, Xe and Kr, due to their larger size and higher polarizability, can interact more readily with other elements, forming compounds like XeF4 (Xenon Tetrafluoride).

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Most popular questions from this chapter

The oxygen and nitrogen families have some obvious similarities and differences. (a) State two general physical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (b) State two general chemical similarities between Group \(5 \mathrm{~A}(15)\) and \(6 \mathrm{~A}(16)\) elements. (c) State two chemical similarities between \(\mathrm{P}\) and \(\mathrm{S}\). (d) State two physical similarities between \(\mathrm{N}\) and \(\mathrm{O}\). (e) State two chemical differences between \(\mathrm{N}\) and \(\mathrm{O}\).

When an alkaline earth carbonate is heated, it releases \(\mathrm{CO}_{2},\) leaving the metal oxide. The temperature at which each Group \(2 \mathrm{~A}(2)\) carbonate yields a \(\mathrm{CO}_{2}\) partial pressure of 1 atm is listed below: $$ \begin{array}{lc} \text { Carbonate } & \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) \\\ \hline \mathrm{Mg} \mathrm{CO}_{3} & 542 \\\\\mathrm{CaCO}_{3} & 882 \\\\\mathrm{SrCO}_{3} & 1155 \\\\\mathrm{BaCO}_{3} & 1360\end{array}$$ (a) Suggest a reason for this trend. (b) Mixtures of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgO}\) are used to absorb dissolved silicates from boiler water. How would you prepare a mixture of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgO}\) from dolomite, which contains \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3} ?\)

Write a balanced equation for the formation from its elements of sodium peroxide, an industrial bleach.

Disulfur decafluoride is intermediate in reactivity between \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{6}\). It disproportionates at \(150^{\circ} \mathrm{C}\) to these monosulfur fluorides. Write a balanced equation for this reaction, and give the oxidation state of \(\mathrm{S}\) in each compound.

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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