Question

The electronic configuration for boron is 1s22s22p1, and the electronic configuration for aluminum is 1s22s22p6 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 porbital, 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

Energy required to take out the outermost electron of an atom is known as the ionization energy. Now the outermost orbit, n = 2 in boron atom is closer to the nucleus, compared to the outermost orbit, n = 3 in an aluminium atom, ionization energy of boron is higher for boron, because, the electrons in this atom feel a greater nuclear attraction force.

Another reason for boron atom having a larger ionization energy compared to the aluminium atom is its number of electrons. Since aluminium atom has a greater number of electrons compared to the boron atom, the repulsion forces between the electrons is higher for aluminium, so, electrons are a bit easy to remove from this atom.

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