Arrange the following atoms in order of increasing atomic size: \(\mathrm{Ne}, \mathrm{As}, \mathrm{Se}, \mathrm{K}, \mathrm{S}, \mathrm{Cl}\)

Short Answer

Expert verified
The atoms arranged in order of increasing atomic size are: \(\mathrm{Ne} < \mathrm{S} < \mathrm{Cl} < \mathrm{K} < \mathrm{As} < \mathrm{Se}\).

Step by step solution

01

Identify the elements on the periodic table

Locate each of the atoms in the periodic table: Ne is in Group 18 and Period 2 As is in Group 15 and Period 4 Se is in Group 16 and Period 4 K is in Group 1 and Period 4 S is in Group 16 and Period 3 Cl is in Group 17 and Period 3
02

Evaluate atomic size within the same period

Going from left to right within the same period, the atomic size generally decreases. This is because the effective nuclear charge increases moving left to right, and consequently, the electron cloud is pulled closer to the nucleus, leading to a smaller atomic size. In this exercise, we have Se and S in the same Group (16), but in different periods (4 and 3 respectively). We also have As in the same period (4) as Se and K.
03

Evaluate atomic size within the same group

Going from top to bottom within the same group, the atomic size generally increases. This is because the energy level increases, resulting in a larger distance between the nucleus and the electron cloud. In this exercise, we have Ne, Cl, and S within the same Period (2, 3, and 3 respectively) but in different groups (18, 17, and 16 respectively).
04

Combine the evaluations and arrange the atoms

Now we can arrange the atoms based on the evaluations we made in Steps 2 and 3: 1. Ne has the smallest atomic size since it is in period 2. 2. Within period 3, the atomic size decreases going left to right, so we have S and then Cl. 3. Within period 4, the atomic size decreases going left to right, so we have K, As, and then Se. Putting it all together, we have the following arrangement in order of increasing atomic size: \(\mathrm{Ne} < \mathrm{S} < \mathrm{Cl} < \mathrm{K} < \mathrm{As} < \mathrm{Se}\)

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