(a) The average distance from the nucleus of a 3 s electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a 3 s electron from the chlorine atom, as compared with a $2 p$ electron?

We are given that the average distance from the nucleus of a 3s electron in a chlorine atom is smaller than that for a 3p electron. According to the concept of shielding effect, the closer an electron is to the nucleus, the lower its energy level due to increased electrostatic attraction towards the nucleus. Therefore, the 3s orbital is lower in energy than the 3p orbital.

To determine which electron requires more energy to be removed, we will consider their principal quantum numbers and shielding effect. The 3s electron is in the third energy level while the 2p electron is in the second energy level. Generally, electrons in higher energy levels require less energy to be removed. However, we also need to consider the shielding effect. In the case of chlorine, the 3s electrons shield the 2p electrons from the nucleus. Consequently, the effective nuclear charge experienced by the 2p electrons is less than the 3s electrons. Considering both the principal quantum number and the shielding effect, we come to the conclusion that it requires less energy to remove a 3s electron than a 2p electron from the chlorine atom.