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Chapter 7: Problem 61

Would you expect zirconium(II) oxide, \(\mathrm{ZrO},\) to react more readily with \(\mathrm{HCl}(a q)\) or \(\mathrm{NaOH}(a q) ?\)

Short Answer

Expert verified
Zirconium(II) oxide (ZrO) is a basic metal oxide, which generally favorably reacts with acids to form salts and water. Thus, it would react more readily with HCl(aq) in the following acid-base reaction: \(ZrO + 2HCl \rightarrow ZrCl_2 + H_2O\).

Step by step solution

01

Determine the nature of the given reactants

First, we need to classify the reactants according to their acid-base properties. HCl(aq) is an acid, while NaOH(aq) is a base. Zirconium(II) oxide (ZrO) is a metal oxide, which generally exhibit basic properties as they can react with acids to form salts and water.
02

Predict the possible reaction products

Next, we predict the products of the possible reactions between ZrO and the given reactants (HCl and NaOH). For the reaction with HCl, since ZrO is a basic oxide, it reacts with the acid to form a salt and water: \(ZrO + 2HCl \rightarrow ZrCl_2 + H_2O\) For the reaction with NaOH, Zirconium(II) oxide (ZrO) can react with a base in a dissolution process, where a complex ion (zirconate ion) is formed: \(ZrO + NaOH + H_2O \rightarrow Na[Zr(OH)_4]\)
03

Determine which reaction is more favorable

Now we need to determine which reaction is more favorable. The reaction between ZrO and HCl forms {ZrCl_2} and water, indicating a typical acid-base reaction. This indicates that ZrO reacts more readily with HCl(aq) because it behaves as a basic oxide, thus favoring the formation of a salt and water.

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Key Concepts

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

Acid-Base Properties
Chemical reactions between different substances often depend on their acid-base properties. To understand this, it's crucial to know that acids and bases are two fundamental categories of chemicals.
  • Acids, like hydrochloric acid (HCl), donate protons (H鈦� ions) and have a sour taste.
  • Bases, such as sodium hydroxide (NaOH), accept protons and feel slippery.
When these substances react, they typically undergo neutralization. This process results in the formation of water and a salt.

In our exercise, zirconium(II) oxide (ZrO) is considered to be a basic oxide. This means it behaves like a base when it interacts with acids. Understanding this property is key to predicting how it will react with other compounds, specifically acids like HCl or bases like NaOH.
Metal Oxides
Metal oxides are compounds consisting of a metal element and oxygen. These are often basic in nature. This means they will more readily react with acids to form salts and water.

For zirconium(II) oxide (ZrO), it fits into this category of basic metal oxides. Basic oxides can react with acids in a chemical reaction to neutralize them.
  • When ZrO reacts with HCl, it forms zirconium chloride (ZrCl鈧�) and water.
This characteristic is common for metal oxides, which typically derive from metals that can lose electrons to oxygen. Understanding the nature of metal oxides helps predict their reactions with acids and bases effectively.
Reaction Products
When substances react with each other, they form new compounds known as reaction products. The types of products formed depend largely on the reactants.

In the case of zirconium(II) oxide (ZrO), several products can arise depending on the other reactant. For instance:
  • Reacting ZrO with hydrochloric acid (HCl) results in the formation of zirconium chloride (ZrCl鈧�) and water.
  • In a different scenario, if ZrO is exposed to sodium hydroxide (NaOH), it can undergo a more complex reaction that forms a zirconate ion, Na[Zr(OH)鈧刔.

These reaction products provide insight into the nature of the chemical interactions, indicating whether a straightforward acid-base reaction or a complex ion formation has occurred. Understanding these products allows us to predict the most probable and favorable outcome between different chemicals.

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Most popular questions from this chapter

Write the electron configurations for the following ions, and determine which have noble-gas configurations: (a) \(\mathrm{Ti}^{2+},(\mathbf{b})\) (d) \(\mathrm{PO}^{2-}\), (f) \(\mathrm{V}^{3+}\) \(\mathrm{Br}^{-}\) (c) \(\mathrm{Mg}^{2+}\) (e) \(\mathrm{Pt}^{2+}\)

(a) The five most abundant elements in the Earth's crust are \(\mathrm{O}, \mathrm{Si}, \mathrm{Al}, \mathrm{Fe},\) and Ca. Referring to Figure \(7.1,\) are any of these elements among those known before \(1700 ?\) If so which ones? (b) Seven of the nine elements known since ancient times are metals. Referring to Table \(4.5,\) are these metals mostly found at the bottom or top of the activity series?

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}{ }^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S .(\mathbf{d})\) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Consider the following equation: $$ \mathrm{Al}^{3+}(g)+e^{-} \longrightarrow \mathrm{Al}^{2+}(g) $$ Which of the following statements are true? (i) The energy change for this process is the second electron affinity of Al atom since \(\mathrm{Al}^{2+}(g)\) is formed. (ii) The energy change for this process is the negative of the third ionization energy of the Al atom. (iii) The energy change for this process is the electron affinity of the \(\mathrm{Al}^{2+}\) ion.
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