Question

The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s2 2p6 352 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)? 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 n2 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

In aluminium the outermost electron is in n=3 state, which is at a larger distance from the nucleus when compared to boron where n=2 state. As distance increased the attractive force between the nucleus and the outermost electron in aluminium is less. Therefore the energy required will be less than 8.298 eV.