Question

Question 22 5 pts The electronic configuration for boron is 1s2 2s22p1, and the electronic configuration for aluminum is 152 252 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 (3p 1)? 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. C None of the above,

Answer 1

Answer :

• Boron (5)& Al (13) are of same group .
• going top to bottom in a group , atomic number increases so the atomic radius increases hence the force of attraction between nucleus and electron decreases i.e. force of repulsion between electron increases which help in ionization as a result the less ionization energy is required .
• that is why the ionization energy of Al is less than B .
• option (c) is correct.