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The alkali metals play virtually the same general chemical role in all their reactions. (a) What is this role? (b) How is it based on atomic properties? (c) Using sodium, write two balanced cquations that illustrate this role.

Short Answer

Expert verified
Alkali metals form cations with a +1 charge. They do this because of their single valence electron and low ionization energy. For sodium:

Step by step solution

01

Identify the General Chemical Role of Alkali Metals

The general chemical role of alkali metals is that they readily lose one electron to form a cation with a +1 charge. This results in the formation of stable ionic compounds.
02

Explain the Atomic Properties Leading to This Role

The atomic properties that contribute to this behavior include their single valence electron, large atomic radius, and low ionization energy. These factors make it easy for the alkali metals to lose their one valence electron.
03

Write a Balanced Equation 1 for Sodium

When sodium reacts with water, it forms sodium hydroxide and hydrogen gas. The balanced equation is:
04

Write a Balanced Equation 2 for Sodium

When sodium reacts with chlorine, it forms sodium chloride. The balanced equation is:

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Key Concepts

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

Formation of Cations
Alkali metals have a unique role in chemical reactions due to their tendency to form cations. This means that they easily lose one electron from their outer shell. By losing this electron, they achieve a stable electronic configuration similar to that of noble gases. This process of losing an electron results in the formation of a positively charged ion, known as a cation.
  • Example: Sodium (Na) loses one electron to become Na+.
  • All alkali metals, like lithium (Li) and potassium (K), exhibit similar behavior.
Understanding the formation of cations is essential for grasping how alkali metals react with other substances, such as forming ionic compounds by bonding with negatively charged ions (anions).
Valence Electron
The key to understanding the reactivity of alkali metals lies in their valence electron. A valence electron is the outermost electron of an atom and is crucial in determining how the atom interacts in chemical reactions.
The alkali metals each have one valence electron. This single electron is relatively loosely bound to the nucleus due to its position in the outermost shell. This makes it very easy for these metals to donate or lose that electron in chemical reactions.
  • This single valence electron results in similar reactivity across the alkali metal group.
  • The loss of this electron leads to the formation of a cation with a +1 charge.
The simplicity of having just one valence electron is a defining feature of alkali metals, making them highly reactive and prone to forming ionic bonds.
Low Ionization Energy
Ionization energy is the energy required to remove an electron from an atom. Alkali metals have a low ionization energy, which means it takes relatively little energy to remove their single valence electron. This property is a major reason why they so readily form cations.
  • Low ionization energy facilitates the loss of the valence electron.
  • As you move down the group in the periodic table, the ionization energy decreases even further.
The combination of low ionization energy and a single valence electron makes alkali metals extremely reactive. They tend to lose their valence electron rapidly in the presence of other elements, leading to the formation of stable ionic compounds.
Atomic Radius
The atomic radius refers to the size of an atom. Alkali metals have a relatively large atomic radius compared to other elements in their periods. This large size is a result of having more electron shells.
The greater the atomic radius, the further the valence electron is from the nucleus. This distance weakens the attractive force between the positive nucleus and the negative electron, making it easier for the electron to be lost.
  • Larger atomic radius means valence electron is less tightly held.
  • This property, along with low ionization energy, contributes to the high reactivity of alkali metals.
Understanding the atomic radius helps in explaining why alkali metals readily lose their valence electron and form cations, engaging in various chemical reactions.

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Most popular questions from this chapter

Give the oxidation state of sulfur in each substance: (a) \(\mathrm{S}_{8}\) (b) \(\mathrm{SF}_{4}\) (c) \(\mathrm{SF}_{6}\) (d) \(\mathrm{H}_{2} \mathrm{~S}\) (e) \(\mathrm{FeS}_{2}\) (f) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (g) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

(a) Give the physical state and color of each halogen at STP. (b) Explain the change in physical state down Group \(7 \mathrm{~A}(17)\) in terms of molecular properties.

An industrial chemist treats solid \(\mathrm{NaCl}\) with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and obtains gaseous \(\mathrm{HCl}\) and \(\mathrm{NaHSO}_{4}\). When she substitutes solid NaI for \(\mathrm{NaCl}\), she obtains gaseous \(\mathrm{H}_{2} \mathrm{~S},\) solid \(\mathrm{I}_{2},\) and \(\mathrm{S}_{8},\) but no \(\mathrm{HI}\). (a) What type of reaction did the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) undergo with NaI? (b) Why does NaI, but not \(\mathrm{NaCl}\), cause this type of reaction? (c) To produce HI \((g)\) by reacting NaI with an acid, how does the acid have to differ from sulfuric acid?

Boron's chemistry is not typical of its group. (a) Cite three ways in which boron and its compounds differ significantly from the other \(3 \mathrm{~A}(13)\) members and their compounds. (b) What is the reason for these differences?

An element forms an oxide, \(\mathrm{E}_{2} \mathrm{O}_{3},\) and a fluoride, \(\mathrm{EF}_{3}\) (a) Of which two groups might \(\mathrm{E}\) be a member? (b) How does the group to which E belongs affect the properties of the oxide and the fluoride?

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