To solve this problem, we need to analyze the given nuclear reaction:$\\ ^{22}_{11}$Na$\rightarrow X + e^{+} + v \\$This reaction is a beta-plus decay, where a proton in the nucleus is converted into a neutron,$\\$emitting a positron $(e^{+})$ and a neutrino $(v). \\$In beta-plus decay, the atomic number decreases by 1, while the mass number remains unchanged.$\\$Given:$\\$• Initial element: $^{22}_{11}$Na$\\$• Final element: $X \\$Since the mass number remains the same, the mass number of $X$ is 22.$\\$The atomic number of $X$ is $11 - 1 = 10. \\$Therefore, $X$ is $^{22}_{10}$Ne$. \\$This corresponds to Option 3.