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

Question

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

Step-by-Step Solution

Answer

(a) matches with (q), (b) with (r), (c) with (r), and (d) with (s).

Step 1 - Match Reaction (a)

In reaction (a), we have \( {}_4 \mathrm{Be}^{9} + {}_2 \mathrm{He}^{4} \rightarrow {}_6 \mathrm{C}^{12} + X \). To find what \( X \) is, balance the mass number and atomic number. The total initial mass number is 9 + 4 = 13. The total atomic number is 4 + 2 = 6. On the right side, the mass and atomic number of \( {}_6 \mathrm{C}^{12} \) are 12 and 6, respectively. So, for particle \( X \), we need mass number 13 - 12 = 1 and atomic number 6 - 6 = 0, which corresponds to the neutron \( {}_0 \mathrm{n}^{1} \). Match it with (q) from Column-II.

Step 2 - Match Reaction (b)

For reaction (b), \( {}_6 \mathrm{C}^{12} + X \rightarrow {}_5 \mathrm{B}^{10} + {}_2 \mathrm{He}^{4} \). Initial mass number is 12. Final mass numbers: 10 for \( {}_5 \mathrm{B}^{10} \) and 4 for \( {}_2 \mathrm{He}^{4} \), total 14. Therefore, we need a particle \( X \) with mass number 2 and atomic number 1 + 2 = 3, making it deuterium \( {}_1 \mathrm{D}^2 \). Match it with (r) from Column-II.

Step 3 - Match Reaction (c)

For reaction (c), \( {}_7 \mathrm{N}^{14} + X \rightarrow {}_8 \mathrm{O}^{17} + {}_1 \mathrm{H}^{1} \). Initial mass number is 14. Final mass numbers: 17 - 1 = 16, so we need \( X \) with mass number 16 - 14 = 2. Initially, atomic number is 7. Final atomic numbers: 8 - 1 = 7, so \( X \) has atomic number 8 - 7 = 1, indicating deuterium \( {}_1 \mathrm{D}^2 \). Match it with (r) from Column-II.

Step 4 - Match Reaction (d)

For reaction (d), \( {}_{20} \mathrm{Ca}^{40} + X \rightarrow {}_{19} \mathrm{~K}^{37} + {}_2 \mathrm{He}^{4} \). Start mass number is 40. Finish mass number: 37 for \( {}_{19} \mathrm{~K}^{37} \) and 4 for \( {}_2 \mathrm{He}^{4} \), totaling 41. \( X \) needs mass number 41 - 40 = 1. Initially atomic number is 20, final is 19 due to \( {}_{18} \mathrm{K} \), so \( X \) has an atomic number of 20 - 19 = 1, identifying it as a proton \( {}_1 \mathrm{H}^1 \). Match it with (s) from Column-II.

Key Concepts

Balancing Mass Number

Understanding how to balance mass numbers in nuclear reactions is crucial. The mass number is the total number of protons and neutrons in an atomic nucleus. Therefore, when solving nuclear reactions, pay close attention to the sum of the mass numbers on both sides of the reaction equation.
For example, in reaction (a), you have \( {}_4 \mathrm{Be}^{9} + {}_2 \mathrm{He}^{4} \rightarrow {}_6 \mathrm{C}^{12} + X \). The left side starts with \(9 + 4 = 13\). On the right side, the mass number of \( {}_6 \mathrm{C}^{12} \) is 12. This means particle \( X \) must have a mass number of \(1\) to balance the equation to 13, confirming that \( X \) is a neutron \( {}_0 \mathrm{n}^{1} \). Balancing mass numbers is similar for the other reactions, ensuring the total mass before and after the reaction remains the same. Remember, disrupted balance can indicate the omission of particles or miscalculated numbers.

Balancing Atomic Number

In addition to mass number, atomic numbers must also be balanced in nuclear reactions. The atomic number represents the number of protons in an atom, which determines the element type. Maintaining the same total atomic number on both sides of the equation preserves elemental identity.
Taking reaction (b) as an example: \( {}_6 \mathrm{C}^{12} + X \rightarrow {}_5 \mathrm{B}^{10} + {}_2 \mathrm{He}^{4} \). The initial atomic number is \(6\), and after reaction, \(5 + 2 = 7\). To stay balanced, \( X \) must have an atomic number of \(1\), which corresponds to deuterium \( {}_1 \mathrm{D}^2 \). Ensure every reaction balances not only in mass number but also in atomic number to reflect a credible transformation of particles.

Particle Identification

Identifying unknown particles \( X \) in nuclear reactions involves comparing mass and atomic numbers, often revealing common subatomic particles like neutrons, protons, or deuterium. Once balance checks are complete, matching the properties of \( X \) aids in correctly identifying it.
For instance, in reaction (c): \( {}_7 \mathrm{N}^{14} + X \rightarrow {}_8 \mathrm{O}^{17} + {}_1 \mathrm{H}^{1} \), the calculated mass number for \( X \) is \(2\) and the atomic number is \(1\). These quantities perfectly match deuterium \( {}_1 \mathrm{D}^{2} \). Similarly, discerning \( {}_1 \mathrm{H}^{1} \) in reaction (d) stems from achieving an atomic \(1\), aligning with a proton. Knowing general properties of these fundamental particles simplifies the identification task.