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(a) Why is cesium more reactive toward water than is lithium? (b) One of the alkali metals reacts with oxygen to form a solid white substance. When this substance is dissolved in water, the solution gives a positive test for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\). When the solution is tested in a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal? (c) Write a balanced chemical equation for reaction of the white substance with water.

Short Answer

Expert verified
Cesium is more reactive toward water than lithium because it has a larger atomic size and lower ionization energy due to its position in the periodic table. The likely identity of the alkali metal is potassium (K), as it produces a lilac-purple flame and forms potassium peroxide (K2O2) when reacting with oxygen. The balanced chemical equation for the reaction of potassium peroxide with water is: \(K_{2}O_{2} + 2H_{2}O \rightarrow 2KOH + H_{2}O_{2}\).

Step by step solution

01

Cesium vs Lithium Reactivity

The reactivity of alkali metals with water increases down the group in the periodic table. This is due to the increasing size of the atoms and the decrease in ionization energy. As the size of an atom increases, the outer electron is further away from the nucleus and held less strongly by the positive charge of the protons in the nucleus. Therefore, it requires less energy to remove the outer electron and form a positive ion. As the ionization energy decreases down the group, the reactivity of alkali metals with water increases. Cesium is located below lithium in the periodic table, which means it has a larger atomic size and lower ionization energy. That's why cesium is more reactive towards water than lithium. #Step 2: Identify the alkali metal based on its properties#
02

Alkali Metal Identification

The given alkali metal reacts with oxygen to form a solid white substance. When this substance is dissolved in water, it gives a positive test for hydrogen peroxide (H2O2). When the metal is tested in a burner flame, it produces a lilac-purple flame. Based on the flame color, it is likely that the metal is potassium (K), as potassium is known for giving a lilac-purple color in a flame test. #Step 3: Write the balanced chemical equation#
03

Balanced Chemical Equation

To write the balanced chemical equation for the reaction of the white substance with water, we need to consider the information given. The white substance, when dissolved in water, gives a positive test for hydrogen peroxide (H2O2). This suggests that the white substance is potassium peroxide (K2O2). The balanced chemical equation for the reaction of potassium peroxide with water can be written as follows: \(K_{2}O_{2} + 2H_{2}O \rightarrow 2KOH + H_{2}O_{2}\)

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Key Concepts

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

Periodic Table Trends
In chemistry, periodic table trends are the predictable patterns seen in certain measures when moving across or down the periodic table. These trends occur because atomic structure dictates the properties of the elements.
For alkali metals, their reactivity increases as you move down the group. This is primarily because, as atoms get larger, the outermost electron is further from the nucleus. Consequently, it is less tightly held by the nucleus's protons.
  • Atomic size: Each subsequent element in a group has atoms with additional electron shells, increasing the atomic radius.
  • Reactivity increase: Greater atomic size leads to lower ionization energy, making it easier for alkali metals to lose their outer electron and react.
This results in elements like cesium, which are lower down in the group, being more reactive than those like lithium that are further up.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. This is a critical factor in determining how easily an element can participate in chemical reactions.
As you progress down a group in the periodic table, ionization energy tends to decrease.
  • Electrons are further from the nucleus: The outer electrons are less affected by the attractive force of the protons.
  • Shielding effect: Inner electrons reduce the full nuclear charge felt by outer electrons, making them easier to remove.
For alkali metals, this declining ionization energy means that elements like cesium can lose their outermost electron more easily than lithium, enhancing their reactivity.
Flame Test
Flame tests are a simple method to identify alkali metals based on the color they produce when burned. This test leverages the fact that heating an element emits distinctive wavelengths of light that correspond to specific colors.
  • Lithium burns with a red flame.
  • Sodium produces a bright yellow flame.
  • Potassium gives a lilac or light purple flame.
The color comes from electrons being excited to higher energy levels by heat and then dropping back down, releasing energy as light. For the exercise, a lilac-purple flame suggests the presence of potassium, supporting the identification of the metal as potassium in the reaction.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of bonds, resulting in new materials with different properties. Understanding these processes is vital for predicting and balancing reactions.
In the exercise, reacting a white solid with water forms new substances. From the given clues and balanced reaction, potassium peroxide reacts with water to produce potassium hydroxide and hydrogen peroxide:
\[K_{2}O_{2} + 2H_{2}O \rightarrow 2KOH + H_{2}O_{2}\]
  • Reactants: Potassium peroxide and water.
  • Products: Potassium hydroxide and hydrogen peroxide.
This example demonstrates the conversion of reactants to products through a balanced chemical equation, showcasing the conservation of mass and charge.

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Most popular questions from this chapter

(a) Which will have the lower energy, a 4 s or a \(4 p\) electron in an As atom? (b) How can we use the concept of effective nuclear charge to explain your answer to part (a)?

The experimental \(\mathrm{Bi}-\mathrm{I}\) bond length in bismuth triiodide, \(\mathrm{BiI}_{3}\), is \(2.81 \AA\). Based on this value and data in Figure \(7.7\), predict the atomic radius of \(B\) i.

Chlorine reacts with oxygen to form \(\mathrm{Cl}_{2} \mathrm{O}_{7} .\) (a) What is the name of this product (see Table 2.6)? (b) Write a balanced equation for the formation of \(\mathrm{Cl}_{2} \mathrm{O}_{7}(l)\) from the elements. (c) Under usual conditions, \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) is a colorless liquid with a boiling point of \(81^{\circ} \mathrm{C}\). Is this boiling point expected or surprising? (d) Would you expect \(\mathrm{Cl}_{2} \mathrm{O}_{7}\) to be more reactive toward \(\mathrm{H}^{+}(a q)\) or \(\mathrm{OH}^{-}(a q) ?\) Explain.

Potassium superoxide, \(\mathrm{KO}_{2}\), is often used in oxygen masks (such as those used by firefighters) because \(\mathrm{KO}_{2}\) reacts with \(\mathrm{CO}_{2}\) to release molecular oxygen. Experiments indicate that \(2 \mathrm{~mol}\) of \(\mathrm{KO}_{2}(\mathrm{~s})\) react with each mole of \(\mathrm{CO}_{2}(g) .\) (a) The products of the reaction are \(\mathrm{K}_{2} \mathrm{CO}_{3}(s)\) and \(\mathrm{O}_{2}(g) .\) Write a balanced equation for the reaction between \(\mathrm{KO}_{2}(\mathrm{~s})\) and \(\mathrm{CO}_{2}(\mathrm{~g}) .\) (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced? (c) What mass of \(\mathrm{KO}_{2}(s)\) is needed to consume \(18.0 \mathrm{~g}\) \(\mathrm{CO}_{2}(g) ?\) What mass of \(\mathrm{O}_{2}(g)\) is produced during this reaction?

Detailed calculations show that the value of \(Z_{\text {eff }}\) for \(\mathrm{Na}\) and \(K\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Na and \(K\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group?

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