Problem 13 - Chemistry in 30 days

Question

\(\mathrm{Rb}\left(\mathrm{ICl}_{2}\right)\) on heating gives (a) \(\mathrm{RbCl}+\mathrm{ICl}\) (b) \(\mathrm{R} \mathrm{\textrm{t }} \mathrm{Cl}+\mathrm{I}_{2}\) (c) \(2 \mathrm{Rb}+\mathrm{I}_{2}+2 \mathrm{Cl}_{2}\) (d) \(2 \mathrm{Rb}+2 \mathrm{ICl}+\mathrm{Cl}\),

Step-by-Step Solution

Answer

(a) \( \mathrm{RbCl} + \mathrm{ICl} \).

Step 1: Understand the Compound

The compound given is \( \mathrm{Rb} \left( \mathrm{ICl}_2 \right) \). This means that rubidium (Rb) is bonded with two chlorine atoms and one iodine atom in the complex \( \mathrm{ICl}_2^- \). The compound is a type of ionic compound where Rb acts as a cation and \( \mathrm{ICl}_2^- \) acts as an anion.

Step 2: Identify Possible Decomposition Products

On heating, compounds can decompose into smaller, simpler substances. Here we need to determine how \( \mathrm{Rb} \left( \mathrm{ICl}_2 \right) \) might decompose. Typically, \( \mathrm{ICl}_2^- \) could break down into \( \mathrm{ICl} \), and \( \mathrm{Rb^+} \) could bond with \( \mathrm{Cl}^- \) resulting in \( \mathrm{RbCl} \).

Step 3: Write the Balanced Equation

Based on the decomposition, \( \mathrm{Rb} \left( \mathrm{ICl}_2 \right) \) can be expected to separate into \( \mathrm{RbCl} \) and \( \mathrm{ICl} \), both of which are stable molecules:\[ \mathrm{Rb} \left( \mathrm{ICl}_2 \right) \rightarrow \mathrm{RbCl} + \mathrm{ICl} \]

Step 4: Verify Correct Option

Now, compare our result from Step 3 with the options provided. The option that matches the compound decomposition \( \mathrm{RbCl} + \mathrm{ICl} \) is (a) \( \mathrm{RbCl} + \mathrm{ICl} \).

Key Concepts

Rubidium Chemistry

Rubidium is a chemical element with the symbol Rb and atomic number 37. It is a soft and silvery-white metal in the alkali metal group. Being an alkali metal, rubidium is highly reactive, similar to its cousins lithium, sodium, and potassium. When it encounters air, rubidium tends to oxidize very quickly, and it reacts violently with water to form rubidium hydroxide and hydrogen gas.

Rubidium's position on the periodic table defines much of its chemistry. It has a single electron in its outermost shell, like all alkali metals. This single electron makes rubidium ready to donate it and form positive ions (cations), specifically rubidium ions ( Rb^+ ). This tendency to form cations defines its chemistry and reactivity.

In terms of ionic compounds, rubidium commonly forms bonds with various nonmetals. It can combine with halogens such as chlorine to form rubidium chloride (RbCl). When rubidium forms ionic compounds like Rb(ICl_2) , it typically acts as a cation, balancing an anionic complex, such as ICl_2^- , present in the compound.

Decomposition Reactions

Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. These reactions often require an external energy source, like heat, light, or electricity, to proceed.

For example, the compound Rb(ICl_2) decomposes upon heating. In this case, the reaction involves breaking down the complex ICl_2^- ion. When this particular compound decomposes, rubidium forms rubidium chloride ( RbCl ), a stable ionic compound, while the ICl_2^- ion breaks into iodine monochloride ( ICl ).

This general process of decomposition is crucial in chemistry. It involves the breaking and reforming of chemical bonds, allowing us to study the compounds' intrinsic properties. It also plays an essential role in various fields, such as materials science and industrial chemistry, where heat can decompose compounds to yield useful products.

Balanced Chemical Equations

A balanced chemical equation represents a chemical reaction with equal numbers of each type of atom on both sides of the equation. Balancing equations is essential because it adheres to the law of conservation of mass—matter is neither created nor destroyed.

In the case of Rb(ICl_2) decomposition, the balanced chemical equation is: \[\mathrm{Rb}(\mathrm{ICl}_2) \rightarrow \mathrm{RbCl} + \mathrm{ICl}\]

Here's how you can verify that the equation is balanced:

• Count the number of Rb atoms: 1 on each side.
• Count the number of Cl atoms: 2 total on the left; 1 in RbCl and 1 in ICl on the right.
• Count the number of I atoms: 1 on each side.

All atoms on the left and right sides are equal, confirming the equation is balanced.

Balancing chemical equations is a fundamental skill in chemistry that ensures we accurately represent chemical reactions and predict the products and reactants' amounts.