Answer:
K / Potassium
Step-by-step explanation:
Out of the list of elements provided, the element that requires the least amount of energy to remove the most loosely held electron from a gaseous atom in the ground state is K (potassium).
The amount of energy needed to remove an electron from an atom is known as the ionization energy. The ionization energy of an element depends on the atomic structure of the element, specifically the arrangement of electrons in the atom's outermost energy level (valence shell). Elements with a low ionization energy have valence electrons that are held less tightly to the nucleus, making them easier to remove.
In general, the ionization energy of an element increases as you move from left to right across a period (row) of the periodic table. This trend is due to the increasing nuclear charge (number of protons in the nucleus) as you move across the period, which leads to a stronger attraction between the nucleus and the valence electrons. However, there are some exceptions to this trend, such as K (potassium), which has a lower ionization energy than its neighbors on the periodic table (Ca and Sr).
Therefore, out of the elements listed (K, Br, Ti, and Cu), K has the lowest ionization energy and would require the least amount of energy to remove the most loosely held electron from a gaseous atom in the ground state.