Calculate number of atoms in FCC unit cell and density of NaCl crystal

Question

(a) Calculate the number of atoms per unit cell in an FCC crystal. (b) NaCl has an FCC lattice with edge length $a = 5.64$ A. Calculate the density of NaCl. Given: $M(\text{NaCl}) = 58.5$ g/mol, $N_A = 6.022 \times 10^{23}$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

In an FCC (face-centred cubic) unit cell:

Corner atoms: 8 corners, each shared by 8 unit cells = $8 \times \frac{1}{8} = 1$
Face-centre atoms: 6 faces, each shared by 2 unit cells = $6 \times \frac{1}{2} = 3$

\boxed{Z = 1 + 3 = 4 \text{ atoms per unit cell}}

NaCl has a rock-salt structure. The $\text{Cl}^-$ ions form an FCC lattice, and $\text{Na}^+$ ions occupy all octahedral voids (or vice versa).

$\text{Cl}^-$ per unit cell: 4 (FCC arrangement)
$\text{Na}^+$ per unit cell: 12 edges $\times \frac{1}{4}$ + 1 body centre = 3 + 1 = 4

So there are 4 NaCl formula units per unit cell ( $Z = 4$ ).

The density formula for a crystal:

\rho = \frac{Z \times M}{N_A \times a^3}

$a = 5.64$ A $= 5.64 \times 10^{-8}$ cm

a^3 = (5.64 \times 10^{-8})^3 = 179.4 \times 10^{-24} = 1.794 \times 10^{-22} \text{ cm}^3

\rho = \frac{4 \times 58.5}{6.022 \times 10^{23} \times 1.794 \times 10^{-22}}

\rho = \frac{234}{108.0} = \boxed{2.17 \text{ g/cm}^3}

This matches the experimental density of NaCl, confirming our structure.

Why This Works

The density formula connects the microscopic (unit cell, atomic mass) to the macroscopic (density). The numerator $ZM$ gives the total mass in one unit cell (in grams, when divided by $N_A$ ). The denominator $a^3$ gives the volume of the unit cell. Mass/volume = density.

The value $Z = 4$ for FCC is one of the most important numbers in solid-state chemistry. For comparison: simple cubic has $Z = 1$ , BCC has $Z = 2$ , and HCP also has $Z = 6$ (for the hexagonal cell) or effectively 2 per primitive cell.

Alternative Method

You can also calculate density using the packing fraction. FCC has 74% packing efficiency. If you know the atomic radius $r$ , use $a = 2\sqrt{2}r$ (FCC relation) and compute volume directly.

For JEE, the density formula $\rho = ZM/(N_A a^3)$ is used in at least 2-3 questions per session. Be careful with units: if $a$ is in cm, density comes in g/cm $^3$ . If $a$ is in pm, convert to cm first ( $1\text{ pm} = 10^{-10}\text{ cm}$ ). Unit conversion errors are the most common source of wrong answers.

Common Mistake

For NaCl, students often take $Z = 4$ and use the atomic mass of Na (23) or Cl (35.5) separately instead of the formula mass of NaCl (58.5). Since $Z = 4$ counts formula units, not individual atoms, you must use the formula mass. Alternatively, count Na and Cl atoms separately: density = $(4 \times 23 + 4 \times 35.5)/(N_A \times a^3)$ — same result.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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