A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three α-particles in succession. The group of the periodic table to which the resulting daughter element would belong is :

When an atom undergoes α-decay, it loses 2 protons and 2 neutrons, effectively moving two groups to the left in the periodic table. If an alkaline earth metal (Group 2) undergoes three successive α-decays, it will lose a total of 6 protons, moving 6 groups to the left in the periodic table. Starting from Group 2 and moving 6 groups to the left would place the resulting daughter element in Group 14 (since 2 - 6 = -4, and considering the periodic table's structure, we move to the right from the beginning, thus 14).