Chlorine has two stable nuclides, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl} .\) In contrast, \(^{36} \mathrm{Cl}\) is a radioactive nuclide that decays by beta emission. (a) What is the product of decay of \(^{36} \mathrm{Cl} ?\) (b) Based on the empirical rules about nuclear stability, explain why the nucleus of \(^{36} \mathrm{C}\) is less stable than either \(^{35}\mathrm{Cl}\) or \(^{37} \mathrm{Cl}\).

The decay we are considering is beta emission (β- decay), which occurs when a neutron in an unstable nucleus converts into a proton, an electron (β- particle), and an electron antineutrino. The general equation for beta decay is: \(n \rightarrow p + e^- + \bar{\nu}_e \) Now, applying this rule to \(^{36}\mathrm{Cl}\): \( _{17}^{36}\mathrm{Cl} \rightarrow _{Z}^{A}\mathrm{X} + e^- + \bar{\nu}_e \) Since one neutron transforms into a proton, the mass number (A) remains the same but the atomic number (Z) increases by one. So the product of decay (\(\mathrm{X}\)) has: atomic number (Z) = 17+1 = 18 mass number (A) = 36 By looking at the periodic table, the element with atomic number 18 is Argon (\(\mathrm{Ar}\)). Thus, the product of decay for \(^{36} \mathrm{Cl}\) is: \( _{17}^{36}\mathrm{Cl} \rightarrow _{18}^{36}\mathrm{Ar} + e^- + \bar{\nu}_e \)

To analyze the stability of the given nuclides, we can use the two empirical rules of nuclear stability: 1. Even-odd rule: Isotopes with even number of protons and neutrons are more stable than isotopes with odd number of protons and neutrons. 2. Magic number rule: Isotopes with a "magic number" of protons or neutrons (2, 8, 20, 28, 50, 82, or 126) tend to be more stable. For \(^{35} \mathrm{Cl}\): - Number of protons (Z) = 17 (odd) - Number of neutrons (N) = 35 - 17 = 18 (even) For \(^{37} \mathrm{Cl}\): - Number of protons (Z) = 17 (odd) - Number of neutrons (N) = 37 - 17 = 20 (even and magic number) For \(^{36} \mathrm{Cl}\): - Number of protons (Z) = 17 (odd) - Number of neutrons (N) = 36 - 17 = 19 (odd) According to the even-odd rule, both stable isotopes \(^{35}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}\) have an even number of neutrons, while the unstable isotope \(^{36}\mathrm{Cl}\) has an odd number of neutrons. Furthermore, \(^{37}\mathrm{Cl}\) has a magic number of neutrons (20), making it more stable. In conclusion, the product of decay of \(^{36} \mathrm{Cl}\) undergoing beta emission is \(^{36}\mathrm{Ar}\). The radioactive nuclide \(^{36} \mathrm{Cl}\) is less stable than the stable nuclides \(^{35}\mathrm{Cl}\) and \(^{37}\mathrm{Cl}\) due to having an odd number of neutrons and not obeying the empirical rules for nuclear stability.