An excited state hydrogen atom whose electron is 2p will be easier to ionize than an excited state hydrogen atom whose electron is 2s. The answer is FALSE. Why is this false? To my understanding, a 2p electron is higher in energy and is therefor easier to remove...right? So why is the answer false?

The statement is false because the energy required to ionize an electron from an excited state in a hydrogen atom does not depend on the specific orbital (2p or 2s) that the electron occupies. The ionization energy is determined by the energy difference between the initial state and the final state, regardless of the orbital.

In a hydrogen atom, the energy levels are determined by the principal quantum number (n). The energy of an electron in the nth energy level is given by the formula E = -13.6 eV/n^2 . The ionization energy is the energy required to remove an electron from the atom, which is equal to the energy difference between the initial state and the final state.

Therefore, the ionization energy of an electron in the 2p orbital and an electron in the 2s orbital would be the same if they are in the same excited state. The specific orbital does not affect the ionization energy.

It is important to note that the energy levels of the hydrogen atom are determined by the quantum mechanical behavior of the electron and are not solely based on the distance from the nucleus. The energy levels are a result of the wave-like nature of the electron and the solutions to the Schrödinger equation.

In summary, the statement that an excited state hydrogen atom with a 2p electron is easier to ionize than an excited state hydrogen atom with a 2s electron is false. The ionization energy is determined by the energy difference between the initial and final states, regardless of the specific orbital occupied by the electron.