Problem 2

(a) In the NaCl crystal,

\ce{Na+}

and

\ce{Cl-}

each form a face-centred cubic lattice; the ionic radii are

r(\ce{Na+}) = 0.102

nm and

r(\ce{Cl-}) = 0.181

nm. Give the numbers of

\ce{Na+}

and

\ce{Cl-}

per unit cell, their coordination numbers, and calculate the crystal density. (b) From the Born\–Haber data \—

\Delta_fH(\ce{NaCl(s)}) = -411

\Delta_{sub}H(\ce{Na}) = +109

IE(\ce{Na}) = +496

D(\ce{Cl2}) = +242

EA(\ce{Cl}) = -349

kJ mol

^{-1}

\— write the equations for the formation step and the direct dissociation

\ce{NaCl(s) -> Na+(g) + Cl-(g)}

, and calculate the lattice formation enthalpy. (c) The Solvay process achieves

\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2}

through the cycle:

\ce{CaCO3 ->[\Delta] A + B}

;

\ce{NaCl + NH3 + B + H2O -> C + D}

;

\ce{2C ->[\Delta] Na2CO3 + H2O + B}

;

\ce{A + H2O -> E}

;

\ce{E + 2D -> CaCl2 + 2H2O + 2NH3}

. Identify compounds A\–E.

Step 1 of 3: Unit cell and density

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N(\ce{Na+}) = N(\ce{Cl-}) = 4;\quad CN = 6{:}6;\quad a = 2(r_+ + r_-) = 0.566\ \text{nm};\quad \rho = \dfrac{4 \times 58.44}{(0.566 \times 10^{-9})^3 \times 6.022 \times 10^{23}} = 2.14 \times 10^{3}\ \text{kg m}^{-3}

Analysis

The fcc lattice gives 4 ions of each kind per cell with octahedral 6:6 coordination. Along the cell edge $\ce{Cl-}$ \– $\ce{Na+}$ \– $\ce{Cl-}$ touch, so $a = 2(0.102+0.181) = 0.566$ nm; the four formula units per cell then give $\rho = 2.14 \times 10^{3}$ kg m $^{-3}$ .

Common pitfall. The cell edge equals

2(r_+ + r_-)

, not

4r

of one ion: adjacent unlike ions touch along the edge.