Question

Question 22 5 pts The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s2 2s22p6 352 3p1. It requires 8.298 eV to remove the outermost electron (21) 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 ap 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 then - 2 state, and the outermost electron around aluminum is in the n - 3 state the energy required will be greater than 8.298 eV O None of the above

Answer 1

The third option is correct, for explanation please find the attachment.

We know from Bhor's moo atomic model, the binding energy of an electron in 13.6 H-atom t En n2 so so Where in= principle quantum number. m. inereases the binding emeregy decreases. of energy eleefrom from a heigher quantem number et orbiting electron. We need Less amount to remove Since n=2 for boron, and n=5 for Al require less energi for Al, compaired it so born. " Third optim is correet: