Question

Question 22 5 pts The electronic configuration for boron is 152 2s22p1, and the electronic configuration for aluminum is 1s2 252 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 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

in boron outer most orbit is 2p1

in aluminum outer most orbit is 3p1

so, the attractive force in aluminum nuecleus to 3p1 is less as compared
to boron. so less amount of energy is required
so correct answer is

since the outer electron around boron is n=2 state and in aluminum is

n=3 state, the energy required will be less than 8.298 ev..................... c