# Which of the following elements would have the largest second ionization energy:?

## Which of the following elements would have the largest second ionization energy: K Rb Sr Ca

May 1, 2016

Potassium.

#### Explanation:

The second ionization energy is defined as the energy needed to remove an electron from one mole of $1 +$ ions in the gaseous state to form one mole of $2 +$ ions in the gaseous state.

${\text{X"_ ((g))^(+) + color(white)(a)"energy"color(white)(a) -> "X"_ ((g))^(2+) + "e}}^{-}$

Much like the first ionization energy, the second ionization energy deals with removing an outermost electron, i.e. an electron located furthest from the nucleus.

The magnitude of the second ionization energy will thus depend on how far away from the nucleus the electron that must be removed is.

Electrons located closer to the nucleus will require more energy to be removed. Likewise, electrons located further away from the nucleus will require less energy to remove.

In your case, you're dealing with two elements located in group 2, calcium, $\text{Ca}$, and strontium, $\text{Sr}$, and two elements located in group 1, potassium, $\text{K}$, and rubidium, $\text{Rb}$.

Right from the start, you can eliminate the two elements located in group 2 because the second ionization energy will be used to remove an electron located on the same energy level as the first electron that was removed to form the $1 +$ ion.

When electrons are being removed from the same energy level, the successive ionization energy increase in a somewhat linear manner. However, when an electron is removed from a lower energy level, a significant jump in ionization energy will take place.

In the case of potassium and rubidium, the first ionization energy will remove their outermost electron. The second ionization energy will have to remove an electron located on a lower energy level.

For rubidium, which is located in period 5, the second ionization energy will remove an electron from the fourth energy level.

For potassium, which is located in period 4, the second ionization energy will remove an electron from the third energy level.

Since the their energy level is closer to the nucleus than the fourth energy level, it follows that the second electron to be removed from potassium will require more energy than the second electron to be removed from rubidium.

Therefore, potassium will have the highest ionization energy from the four elements given to you.

The actual values are

${\text{For Ca: " 2^"nd"" IE" = "1145.4 kJ mol}}^{- 1}$

${\text{For Sr: " color(white)(.)2^"nd" " IE" = "1064.2 kJ mol}}^{- 1}$

${\text{For K: " color(white)(.)2^"nd"" IE" = "3052 kJ mol}}^{- 1}$

${\text{For Rb: " 2^"nd" " IE" = "2633 kJ mol}}^{- 1}$

https://www.webelements.com/periodicity/ionisation_energy_2/