Question

for each list, place elements in order of least to greatest ionization energy: 1) li lithium o oxygen n nitrogen 2) ba barium si silicon ra radium 3) pb lead sn tin c carbon

Explanation:

Step1: Recall ionization - energy trends

Ionization energy increases across a period and decreases down a group in the periodic table.

Step2: Analyze first set (Li, O, N)

Li is in Group 1, N is in Group 15 and O is in Group 16 of the same period (Period 2). So, Li has the lowest ionization energy. Among N and O, N has a half - filled 2p sub - shell, which gives it extra stability, so O has a lower ionization energy than N. The order is Li < O < N.

Step3: Analyze second set (Ba, Si, Ra)

Si is in Period 3, Ba is in Period 6 and Ra is in Period 7. Si is a non - metal and Ba and Ra are metals. Ba and Ra are in Group 2. Ra is below Ba, so Ra has a lower ionization energy than Ba. The order is Ra < Ba < Si.

Step4: Analyze third set (Pb, Sn, C)

C is in Period 2, Sn is in Period 5 and Pb is in Period 6. They are all in Group 14. As we go down the group, ionization energy decreases. The order is Pb < Sn < C.

Answer:

1. Li, O, N
2. Ra, Ba, Si
3. Pb, Sn, C