Answer
First IE of K < Second IE of Be < First IE of Li < First IE of BE < Second IE of Na
Work Step by Step
First IE of K < Second IE of Be < First IE of Li < First IE of BE < Second IE of Na
-First IE of K: Potassium is one away from being a Noble gas, so it makes sense that it has the smallest ionization energy as it wants to lose an electron.
-Second IE of Be: After losing an electron, Beryllium is one away from being a Noble gas, so it makes sense that it has the second smallest ionization energy as it wants to lose an electron. Note, this is slightly higher than the first ionization energy of Potassium, for Be has more electrons.
- First IE of Li: Li has fewer electrons than Potassium, so even though it is one away from being a noble gas as well, the stronger attraction of the protons increases ionization energy.
-First IE of BE: This is not one away from being a noble gas, so it is harder to rip the electron off.
-Second IE of Na: $Na^+$ has a noble gas configuration of electrons, so it is hard to rip an electron off. Thus, the second ionization energy is much higher than its first ionization energy.
