Of the atoms \(\mathrm{Na}, \mathrm{Cl}, \mathrm{K}, \mathrm{Br}\), which has the largest atomic radius? Which has the largest first ionization energy?

First, let us understand the periodic trends for atomic radius and ionization energy. 1. Atomic Radius: Atomic radius generally increases from top to bottom within a group and decreases from left to right across a period in the periodic table. 2. Ionization Energy: It's the energy required to remove an electron from a neutral atom in the gaseous state. Ionization energy generally decreases from top to bottom within a group and increases from left to right across a period in the periodic table.

Now, let's determine the positions of each atom in the periodic table. 1. \(\mathrm{Na}\) (Sodium) belongs to Group 1 (Alkali Metals) and Period 3. 2. \(\mathrm{Cl}\) (Chlorine) belongs to Group 17 (Halogens) and Period 3. 3. \(\mathrm{K}\) (Potassium) belongs to Group 1 (Alkali Metals) and Period 4. 4. \(\mathrm{Br}\) (Bromine) belongs to Group 17 (Halogens) and Period 4.

We will now analyze the atomic radius trend for the given atoms based on their positions in the periodic table. For \(\mathrm{Na}\) and \(\mathrm{K}\) (both alkali metals), \(\mathrm{K}\) is in a lower period (Period 4) than \(\mathrm{Na}\)(Period 3), so \(\mathrm{K}\) has a larger atomic radius. For \(\mathrm{Cl}\) and \(\mathrm{Br}\) (both halogens), \(\mathrm{Br}\) is in a lower period (Period 4) than \(\mathrm{Cl}\) (Period 3), so \(\mathrm{Br}\) has a larger atomic radius. Comparing \(\mathrm{K}\) and \(\mathrm{Br}\), since they both lie in Period 4, and \(\mathrm{K}\) is in an earlier group (Group 1) than \(\mathrm{Br}\) (Group 17), \(\mathrm{K}\) has a larger atomic radius. So, among \(\mathrm{Na}\), \(\mathrm{Cl}\), \(\mathrm{K}\), and \(\mathrm{Br}\), \(\mathrm{K}\) has the largest atomic radius.

Next, we will analyze the ionization energy trend for the given atoms based on their positions in the periodic table. For \(\mathrm{Na}\) and \(\mathrm{K}\) (both alkali metals), \(\mathrm{K}\) is in a lower period (Period 4) than \(\mathrm{Na}\) (Period 3), so \(\mathrm{Na}\) has a higher ionization energy. For \(\mathrm{Cl}\) and \(\mathrm{Br}\) (both halogens), \(\mathrm{Br}\) is in a lower period (Period 4) than \(\mathrm{Cl}\) (Period 3), so \(\mathrm{Cl}\) has a higher ionization energy. Comparing \(\mathrm{Na}\) and \(\mathrm{Cl}\), since they both lie in Period 3, and \(\mathrm{Na}\) is in an earlier group (Group 1) than \(\mathrm{Cl}\) (Group 17), \(\mathrm{Cl}\) has a higher ionization energy. So, among \(\mathrm{Na}\), \(\mathrm{Cl}\), \(\mathrm{K}\), and \(\mathrm{Br}\), \(\mathrm{Cl}\) has the largest first ionization energy.