Question

1. for each of the following elements, circle the element with the largest atomic radius and put a square around the element with the smallest atomic radius.

a) o c be ne b) na rb cs he
c) pb c sn si d) au w s sr

Explanation:

Step1: Recall atomic radius trends

Atomic radius decreases across a period (from left to right) and increases down a group. Noble gases (except He) have slightly larger atomic radii due to van - der - Waals radius measurement.

Step2: Analyze part (a)

• Period 2 elements: Be (Group 2), C (Group 14), O (Group 16), Ne (Group 18).
• From left to right: Be > C > O. Ne (noble gas) has a larger radius than O (but smaller than Be). So, largest is Be, smallest is O.

Step3: Analyze part (b)

• Na (Period 3, Group 1), Rb (Period 5, Group 1), Cs (Period 6, Group 1), He (Period 1, Group 18).
• Down Group 1: Cs > Rb > Na. He (Period 1) has the smallest radius. So, largest is Cs, smallest is He.

Step4: Analyze part (c)

• Group 14 elements: C (Period 2), Si (Period 3), Sn (Period 5), Pb (Period 6).
• Down the group: Pb > Sn > Si > C. So, largest is Pb, smallest is C.

Step5: Analyze part (d)

• Au (Period 6, Group 11), W (Period 6, Group 6), Sr (Period 5, Group 2), S (Period 3, Group 16).
• Period 6 elements (Au and W) are larger than Period 5 (Sr) and Period 3 (S). Among Au and W: In the same period, from left (Sr is in lower period but compare based on general trends, in Period 6, Group 6 (W) and Group 11 (Au), atomic radius: Au < W (as we move left to right radius decreases). Sr (Period 5) is larger than S (Period 3). So, largest is W, smallest is S.

Answer:

a) \(\textcircled{Be}\), \(\boxed{O}\)
b) \(\textcircled{Cs}\), \(\boxed{He}\)
c) \(\textcircled{Pb}\), \(\boxed{C}\)
d) \(\textcircled{W}\), \(\boxed{S}\)