Chemical Energetics: Question 4

Worked solution

Part (a): First electron affinity of chlorine

The first electron affinity of chlorine is the enthalpy change when one mole of gaseous chlorine atoms each gain one electron to form one mole of gaseous $1-$ ions: $\text{Cl(g)} + e^- \rightarrow \text{Cl}^-\text{(g)}$

Part (b): Why $IE_2(\text{Sr}) > IE_1(\text{Sr})$

The first ionisation energy removes an electron from a neutral strontium atom, $\text{Sr(g)} \rightarrow \text{Sr}^+\text{(g)} + e^-$. The second ionisation energy removes an electron from the resulting $\text{Sr}^+\text{(g)}$ ion, which already carries a $1+$ charge. The same nuclear charge now attracts one fewer electron than before, so the remaining electrons experience a greater effective nuclear charge per electron and are held more tightly. More energy is therefore needed to remove the second electron than the first.

Part (c): Born-Haber cycle and lattice energy

The Born-Haber cycle for $\text{SrCl}_2$ links the direct formation of the solid from its elements with an indirect route through gaseous atoms and ions:

$$\text{Sr(s)} + \text{Cl}_2\text{(g)} \xrightarrow{\ \Delta H_f^{\ominus}\ } \text{SrCl}_2\text{(s)}$$

Breaking the indirect route into steps:

1. Atomise strontium: $\text{Sr(s)} \rightarrow \text{Sr(g)}$, $\Delta H = +166\ \text{kJ mol}^{-1}$
2. First ionisation of strontium: $\text{Sr(g)} \rightarrow \text{Sr}^+\text{(g)} + e^-$, $\Delta H = +548\ \text{kJ mol}^{-1}$
3. Second ionisation of strontium: $\text{Sr}^+\text{(g)} \rightarrow \text{Sr}^{2+}\text{(g)} + e^-$, $\Delta H = +1060\ \text{kJ mol}^{-1}$
4. Atomise two moles of chlorine atoms: $\Delta H = 2 \times (+121) = +242\ \text{kJ mol}^{-1}$
5. Add an electron to two moles of chlorine atoms: $\Delta H = 2 \times (-349) = -698\ \text{kJ mol}^{-1}$
6. Form the lattice from gaseous ions: $\text{Sr}^{2+}\text{(g)} + 2\text{Cl}^-\text{(g)} \rightarrow \text{SrCl}_2\text{(s)}$, $\Delta H = LE$ (unknown)

By Hess’s law, the sum of steps 1–6 equals the direct enthalpy of formation: $\Delta H_f^{\ominus} = \Delta H_{at}(\text{Sr}) + IE_1(\text{Sr}) + IE_2(\text{Sr}) + 2\Delta H_{at}(\text{Cl}) + 2EA_1(\text{Cl}) + LE$

Substituting the known values: $-828 = 166 + 548 + 1060 + 2(121) + 2(-349) + LE$

First total the known steps: $166 + 548 = 714$ $714 + 1060 = 1774$ $2 \times 121 = 242, \quad 1774 + 242 = 2016$ $2 \times (-349) = -698, \quad 2016 - 698 = 1318$

So: $-828 = 1318 + LE$

$LE = -828 - 1318 = -2146\ \text{kJ mol}^{-1}$

Check (independent recomputation): $166+548+1060 = 1774$; $2(121)=242 \Rightarrow 1774+242=2016$; $2(-349)=-698 \Rightarrow 2016-698=1318$; then $LE = -828-1318 = -(828+1318) = -2146\ \text{kJ mol}^{-1}$, consistent.

Part (d): Why the cycle value may differ from a purely ionic model

A Born-Haber lattice energy is an experimental value (built from measurable quantities), whereas a purely ionic model assumes the ions are perfect, undistorted point charges. In reality, the small, doubly-charged $\text{Sr}^{2+}$ ion can polarise the larger, more easily distorted $\text{Cl}^-$ ion, giving the bonding some covalent character. This extra electron-sharing makes the real lattice more stable than the point-charge model predicts, so the Born-Haber value is typically more exothermic (more negative) than the purely ionic calculated value.

Final answers

• (a) $\text{Cl(g)} + e^- \rightarrow \text{Cl}^-\text{(g)}$, the enthalpy change per mole.
• (b) The second electron is removed from $\text{Sr}^+$, where the remaining electrons feel a greater effective nuclear charge per electron than in the neutral atom.
• (c) $LE(\text{SrCl}_2) = \boxed{-2146\ \text{kJ mol}^{-1}}$
• (d) Covalent character (polarisation of $\text{Cl}^-$ by $\text{Sr}^{2+}$) not accounted for in the purely ionic model.