Question 3 of 8: Question III — Crystal Structure II: BaTiO₃ Perovskite (20 marks)

Question III — Crystal Structure II: BaTiO₃ Perovskite (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

III.1 — Unit cell of BaTiO₃ (Ba at corners)

In the standard perovskite setting, Ba²⁺ occupies the 8 cube corners (each shared 1/8, giving 1 Ba/cell), Ti⁴⁺ occupies the body centre (coordination 6, octahedrally surrounded by O²⁻, giving 1 Ti/cell), and O²⁻ occupies the 6 face centres (each shared 1/2, giving 3 O/cell). Formula check: Ba₁Ti₁O₃ — matches BaTiO₃. The O²⁻ face-centre positions alone do form an FCC sub-lattice, as stated in the question; adding the corner Ba²⁺ and body-centre Ti⁴⁺ on top of that FCC anion array is exactly the perovskite structure.

III.2 — Lattice parameter from density

Given. $A_{Ba}=137$, $A_{Ti}=48$, $A_O=16$ g/mol; $\rho=6\ \text{g/cm}^3$; $Z=1$ formula unit per cell.

Find. Lattice parameter $a$.

Approach. $\rho=ZM/(a^3N_A)$, so $a=\left(ZM/(\rho N_A)\right)^{1/3}$, with $M$ the BaTiO₃ formula weight.

1. Formula weight. $$M=A_{Ba}+A_{Ti}+3A_O=137+48+3(16)=233\ \text{g/mol}$$
2. Solve for $a$. $$a^3=\frac{ZM}{\rho N_A}=\frac{(1)(233)}{(6)(6.023\times10^{23})}=6.448\times10^{-23}\ \text{cm}^3$$ $$a=(6.448\times10^{-23})^{1/3}=\boxed{4.01\times10^{-8}\ \text{cm}=0.401\ \text{nm}}$$

0.401 nm is close to the accepted room-temperature (cubic, above the 130°C Curie point) BaTiO₃ lattice parameter of 0.403 nm — a sound check on the given density.

III.3 — Alternate origin: Ba at body centre

Approach. Shifting the coordinate origin by $(\tfrac12,\tfrac12,\tfrac12)$ (i.e. re-choosing which lattice point is labelled "corner") moves every ion by that same vector, mod 1, without changing the physical crystal at all.

1. Ba²⁺: was at the corners $(0,0,0)$; shifting by $(\tfrac12,\tfrac12,\tfrac12)$ sends it to $\boxed{(\tfrac12,\tfrac12,\tfrac12)\text{, the body centre}}$.
2. Ti⁴⁺: was at the body centre $(\tfrac12,\tfrac12,\tfrac12)$; shifting sends it to $(1,1,1)\equiv(0,0,0)$, i.e. $\boxed{\text{the 8 cube corners}}$.
3. O²⁻: was at the 6 face centres, e.g. $(\tfrac12,\tfrac12,0)$; shifting sends it to $(1,1,\tfrac12)\equiv(0,0,\tfrac12)$ — the centres of the 12 cube edges. Formula check: 8 Ti$\times\tfrac18$ + 1 Ba + 12 O$\times\tfrac14$ = 1 Ti + 1 Ba + 3 O, unchanged, $\boxed{\text{O}^{2-}\text{ at the 12 edge centres}}$.