Answer
See explanation.
Work Step by Step
Equation 41.45 gives the energy of an atomic level, taking screening into account.
It is given in terms of n and $Z_{eff}$ by $E_n=-\frac{Z_{eff}^2}{n^2}(13.6 eV)$.
Find the effective atomic number for an outer electron.
a. Z=7 for nitrogen, so doubly-ionized N has 5 electrons. The electron configuration for $N^{2+}$ is $1s^2 2s^2 2p$.
b. In this ion, there is one electron outside a set of filled subshells. We assume full screening by the inner-subshell electrons, so there are 7 protons and 4 electrons interior to the 2p electron. $Z_{eff}=7-4=3$ for the 2p electron.
$E_{2p}=-\frac{3^2}{2^2}(13.6 eV)= -30.6 eV$
c. Z=15 for phosphorous, so doubly-ionized P has 13 electrons. The electron configuration for $P^{2+}$ is $1s^2 2s^2 2p^6 3s^2 3p$.
d. In this ion, there is one electron outside a set of filled subshells. We assume full screening by the inner-subshell electrons, so there are 15 protons and 12 electrons interior to the 3p electron. $Z_{eff}=15-12=3$ for the 3p electron.
$E_{3p}=-\frac{3^2}{3^2}(13.6 eV)= -13.6 eV$
