Atomic Structure - Electrons in atoms (HL only)

The convergence limit of the Lyman series for hydrogen is observed at a wavelength of $91.16 \text{ nm}$. Calculate the first ionization energy ($IE_1$) of hydrogen in $kJ \cdot mol^{-1}$. (Use $h = 6.63 \times 10^{-34} \text{ J s}$, $c = 3.00 \times 10^8 \text{ m s}^{-1}$, $L = 6.02 \times 10^{23} \text{ mol}^{-1}$)

$E_{photon} = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{91.16 \times 10^{-9} \text{ m}} = 2.182 \times 10^{-18} \text{ J}$ $IE_1 = E_{photon} \times L = 2.182 \times 10^{-18} \text{ J} \times 6.02 \times 10^{23} \text{ mol}^{-1} = 1.313 \times 10^6 \text{ J mol}^{-1} = 1313 \text{ kJ mol}^{-1}$

First, the energy of a single photon at the convergence limit is calculated using the wavelength. This energy corresponds to the energy required to remove one electron from one atom. Multiplying by Avogadro's constant ($L$) gives the ionization energy per mole.

Explain why the first ionization energy of Sulfur ($S$) is lower than that of Phosphorus ($P$).

$P: [Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1$ $S: [Ne] 3s^2 3p_x^2 3p_y^1 3p_z^1$

Phosphorus has a half-filled $3p$ subshell with three singly occupied orbitals. Sulfur has four electrons in the $3p$ subshell, meaning one $3p$ orbital contains a pair of electrons. The repulsion between these two electrons in the same orbital makes it easier to remove one of them, resulting in a lower $IE_1$ for $S$ despite its higher nuclear charge.

Write the full electron configuration for the $Fe^{2+}$ ion.

$Fe: [Ar] 3d^6 4s^2 \implies Fe^{2+}: [Ar] 3d^6$

When transition metals form ions, electrons are always removed from the $4s$ subshell before the $3d$ subshell. Therefore, $Fe$ ($Z=26$) loses its two $4s$ electrons first.