Problem 133 - A Complete Resource Book in Chemistry for JEE Main

Question

$$ \text { Match the following } $$$$ \begin{array}{ll} \hline \text { Column-I (Reactions) } & \begin{array}{l} \text { Column-II } \\ \text { (Particles) } \end{array} \\ \hline \text { (a) }{\underline{\phantom{xx}}}_{4} \mathrm{Be}^{9}+{ }_{2} \mathrm{He}^{4} \rightarrow{ }_{6} \mathrm{C}^{12}+\ldots \ldots & \text { (p) }{\underline{\phantom{xx}}}_{2} \mathrm{He}^{4} \\ \text { (b) }{\underline{\phantom{xx}}}_{6} \mathrm{C}^{12}+\ldots \ldots \rightarrow{ }_{5} \mathrm{~B}^{10}+{ }_{2} \mathrm{He}^{4} & \text { (q) }{\underline{\phantom{xx}}}_{0} \mathrm{n}^{1} \\\ \text { (c) }{\underline{\phantom{xx}}}_{7} \mathrm{~N}^{14}+\ldots \ldots \rightarrow{ }_{8} \mathrm{O}^{17}+{ }_{1} \mathrm{H}^{1} & \text { (r) }, \mathrm{D}^{2} \\ \text { (d) }{\underline{\phantom{xx}}}_{20} \mathrm{Ca}^{40}+\ldots . . \rightarrow{ }_{19} \mathrm{~K}^{37}+{ }_{2} \mathrm{He}^{4} & \text { (s) }{\underline{\phantom{xx}}}_{1} \mathrm{H}^{1} \\\ \hline \end{array} $$

Step-by-Step Solution

Answer

(a) - (q), (b) - (r), (c) - (p), (d) - (s)

Step 1: Analyze Reaction (a)

The reaction is \( _4 \text{Be}^9 + _2 \text{He}^4 \rightarrow _6 \text{C}^{12} + \ldots \). For the reaction to be balanced, both the atomic and mass numbers must equal on both sides of the equation. Net mass and atomic numbers on the left are \(9 + 4 = 13\) and \(4 + 2 = 6\), respectively. On the right, we have \(12 + x\) for masses and \(6 + a\) for atomic numbers. Balancing gives \(x = 1\) and \(a = 0\), therefore the missing particle is \(_0\text{n}^1\). Match \((a)\) with \((q)\).

Step 2: Analyze Reaction (b)

The reaction is \( _6 \text{C}^{12} + \ldots \rightarrow _5 \text{B}^{10} + _2 \text{He}^4 \). Initially we have mass 12 on the left. On the right-hand side, the masses are \(10 + 4 = 14 \). Hence, the missing particle must have mass 2. The atomic numbers should also balance to 6 on both sides; currently, \(5 + 2 = 7\) on the right, indicating the missing particle must have an atomic number of 1. Therefore, this is deuterium \(_1\text{H}^2\). Match \((b)\) with \((r)\).

Step 3: Analyze Reaction (c)

The reaction is \( _7 \text{N}^{14} + \ldots \rightarrow _8 \text{O}^{17} + _1 \text{H}^1 \). Looking at mass numbers, the total on the right is \(17 + 1 = 18\). The current mass on the left is 14, so the missing particle must have mass 4. The atomic number right now is 8 + 1 = 9. On the left, to get to 9 from 7, we need \(2\). Hence, the missing particle is \( _2\text{He}^4\). Match \((c)\) with \((p)\).

Step 4: Analyze Reaction (d)

The reaction is \( _{20} \text{Ca}^{40} + \ldots \rightarrow _{19} \text{K}^{37} + _2 \text{He}^4 \). The mass balance on the right is \(37+4 = 41\). Since the initial is 40, we need one more unit of mass, matching \(_1\text{H}^1\) where mass is 1 and atomic number is 1. Match \((d)\) with \((s)\).

Key Concepts

Balancing Atomic Numbers

In nuclear reactions, it's crucial to balance the atomic numbers to ensure that the type of elements remains consistent before and after the reaction. The atomic number, represented as the number at the bottom in nuclear notations such as \(_4\, \text{Be}\, ^9\), indicates the number of protons in the nucleus of an atom. During a nuclear reaction, these numbers must remain the same on both the reactant and product sides, signifying the conservation of charge and identity of elements.
Using step one of the solution as an example: The equation is \(_4 \, \text{Be}^9 + _2 \, \text{He}^4 \rightarrow \, _6 \, \text{C}^{12} + \ldots\). The atomic numbers for the reactants add up to 6 (4 from beryllium and 2 from helium), so the right side also must add up to 6, which allows us to deduce any missing atomic numbers.

Balancing Mass Numbers

Balancing the mass numbers in nuclear equations ensures the mass is conserved across the reaction. The mass number is the sum of protons and neutrons in an atomic nucleus. It’s usually the number located at the top of nuclear notations like \(_4 \, \text{Be}^{9}\).
To achieve a balanced mass number, the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products. For instance, in the reaction \(_4 \, \text{Be}^{9} + _2 \, \text{He}^{4} \rightarrow _6 \, \text{C}^{12} + \ldots\), the total initial mass number is 13 (9 + 4), which means the outputs also need to sum up to 13. This implies that if carbon has a mass of 12, the missing particle must have a mass of 1.

Nuclear Equations

Nuclear equations are used to represent nuclear reactions, which involve changes in an atom’s nucleus and can result in changes in the atomic and mass numbers. Each nuclear reaction is written as an equation, detailing the initial reactants and resulting products. They are similar to chemical equations, yet they focus on the atomic nucleus instead of electron exchange.
For example, \(_6 \, \text{C}^{12} +\ldots \rightarrow _5 \, \text{B}^{10} + _2 \, \text{He}^{4}\) illustrates the transformation involving carbon, boron, and helium, with missing variables determined by balancing the mass and atomic numbers, making it crucial to deduce any missing particle or nucleus.

Particles in Nuclear Reactions

Particles like neutrons, protons, and deuterium play a vital role in nuclear reactions by facilitating changes in atomic structures. Each particle has specific identifying properties:

• Protons have an atomic number of 1 and a mass number of 1.
• Neutrons, symbolized as \(_0 \, \text{n}^1\), have no charge but a mass of 1.
• Deuterium or "D", a form of hydrogen, has both an atomic and mass number of 1.
• Helium Nuclei, also called alpha particles \(_2 \, \text{He}^4\), are common in these reactions.

Each type of particle contributes to the process by either adding mass or changing element identity. This participation is critical for maintaining balance in atomic and mass numbers, a fundamental requirement of nuclear reactions.