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Perovskite Structure (CaTiO3)

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Perovskite (CaTiO3)
A cubic structure with Ca2+ at cube corners, Ti4+ at body center, and O2- at face centers in 12:6:2 coordination.
The ideal perovskite structure (ABO₃) is cubic with the large A cation (Ca2+) at cube corners coordinated by 12 oxygen atoms, the smaller B cation (Ti4+) at the body center in octahedral coordination with 6 oxygens, and oxide ions at face centers. The structure belongs to space group Pm3̄m with 1 formula unit per unit cell. The BO₆ octahedra share corners to form a 3D framework with A cations in the 12-coordinate cuboctahedral cavities. Most perovskites are distorted from ideal cubic symmetry due to size mismatch, tilting of octahedra, or off-center displacement of cations. This structure is enormously important, hosting ferroelectric (BaTiO₃), piezoelectric, superconducting (YBa₂Cu₃O₇), and colossal magnetoresistive materials.

Problem 1

Question: In the perovskite structure (CaTiO₃), where are the Ca²⁺, Ti⁴⁺, and O²⁻ ions located in the unit cell?

Solution

Answer:

Ca²⁺: At the 8 corners of the cube (8 × 1/8 = 1 per cell)

Ti⁴⁺: At the body center (1 per cell)

O²⁻: At the 6 face centers (6 × 1/2 = 3 per cell)

This gives the formula unit CaTiO₃.

Problem 2

Question: What is the coordination number of Ca²⁺ and Ti⁴⁺ in the perovskite structure?

Solution

Answer:

Coordination number of Ca²⁺ (A-site) = 12 (cuboctahedral coordination by 12 O²⁻)

Coordination number of Ti⁴⁺ (B-site) = 6 (octahedral coordination by 6 O²⁻)

The coordination ratio is 12:6

Problem 3

Question: Calculate the tolerance factor (t) for CaTiO₃ if r(Ca²⁺) = 1.00 Å, r(Ti⁴⁺) = 0.61 Å, and r(O²⁻) = 1.40 Å.

[Tolerance factor: t = (r_A + r_O)/[√2(r_B + r_O)]]

Solution

Calculation:

t = (r_Ca + r_O)/[√2(r_Ti + r_O)]

t = (1.00 + 1.40)/[√2(0.61 + 1.40)]

t = 2.40/[1.414 × 2.01]

t = 2.40/2.84

t = 0.845

Since 0.75 < t < 1.0, the perovskite structure is stable (ideal cubic perovskite has t ≈ 1).

Numerical Problem

Question: In the ideal perovskite structure CaTiO₃, Ca2+ ions are at cube corners, Ti4+ is at the body center, and O2- ions are at face centers.

(a) How many Ca2+ ions are effectively present per unit cell?

(b) How many Ti4+ ions are effectively present per unit cell?

(c) How many O2- ions are effectively present per unit cell?

(d) Write the ratio Ca : Ti : O based on your answers. Does it match the formula CaTiO₃?

Solution

(a) Number of Ca2+ ions per unit cell:

Ca2+ ions are at the 8 corners of the cube

Each corner atom is shared by 8 unit cells

Contribution from corners = 8 × (1/8) = 1 Ca2+ ion

(b) Number of Ti4+ ions per unit cell:

Ti4+ ion is at the body center

Body center atom belongs entirely to one unit cell

Number of Ti4+ ions = 1 Ti4+ ion

(c) Number of O2- ions per unit cell:

O2- ions are at the 6 face centers

Each face-centered atom is shared by 2 unit cells

Contribution from face centers = 6 × (1/2) = 3 O2- ions

(d) Ratio verification:

From our calculations:

• Ca : Ti : O = 1 : 1 : 3

This matches the formula CaTiO₃ ✓

Summary: The unit cell contains exactly 1 formula unit of CaTiO₃, making perovskite one of the simplest cubic structures to understand.

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