Question

The electronic configuration for boron is 1s2 2s2 2p1, and the electronic configuration for aluminum is 1s2 2s2 2p6 3s2 3p1. It requires 8.298 eV to remove the outermost electron (2p1) from boron. What can you say about the energy required to remove the outermost electron from aluminum (3p1)?

Group of answer choices

Since each is the outermost electron around the atom, the energy required will be equal to 8.298 eV.

Since each electron is in a p orbital, the energy required will be equal to 8.298 eV.

Since the outermost electron around boron is in the n = 2 state, and the outermost electron around aluminum is in the n = 3 state, the energy required will be less than 8.298 eV.

Since the outermost electron around boron is in the n = 2 state, and the outermost electron around aluminum is in the n = 3 state, the energy required will be greater than 8.298 eV.

None of the above.

Answer 1

The energy required will be less than 8.298 volt. Since there is 3p orbital for the electron in aluminium so there will be less enegry required to remove it as compared to the boron. Because a fully filled orbit blocks the attractive force between the outer orbit electron and the neucleus and act as a blocking shell. So less energy is required to remove the electron since it is loosely attached to the neucleus.