Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Period 2 has two s-block elements, lithium () and beryllium (), and six p-block elements: boron (), carbon (), nitrogen (), oxygen (), fluorine () and neon (), all in kJ/mol. What does p2.groupby("block")["ie"].mean().round(1) print for each block?
df.groupby("block")["ie"].mean() gives the mean of each block. df["a"].corr(df["b"]) gives the correlation , from to . df.groupby("period")["ie"].diff() gives the change from the element before, inside each period; a negative value is a break in the rise.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. s-block. kJ/mol.
Step 2. p-block. , and kJ/mol.
Answer: s-block kJ/mol, p-block kJ/mol.
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