Statement 1: CsCl has body-centred cubic arrangement.

Statement 2: CsCl has one Cs⁺ ion and 8 Cl^- ions in its unit cell.

$\mathrm{CsCl} \mathrm{follows}$$8 : 8 \mathrm{type} \mathrm{of} \mathrm{arrangement}$.

That means, $$\begin{gathered} {\text{Co.No. of Cs}}^{+ }\mathrm{ion} = 8; \mathrm{and} {\mathrm{Cs}}^{+}\mathrm{ion} \mathrm{present} \mathrm{in} \mathrm{cubic} \mathrm{voids} \mathrm{or} \mathrm{body} \mathrm{centre} \\ \mathrm{Co}.\mathrm{No}. \mathrm{of} {\mathrm{Cl}}^{-}\mathrm{ion} = 8; \mathrm{and} \mathrm{ions} \mathrm{follows} \mathrm{simple} \mathrm{cubic} \mathrm{arrangement}. \mathrm{They} \mathrm{occupy} \mathrm{corners} \mathrm{of} \mathrm{the} \mathrm{cube}. \end{gathered}$$

$$\begin{gathered} \mathrm{Number} \mathrm{of} {\mathrm{Cs}}^{+}\mathrm{ions} \mathrm{per} \mathrm{unit} \mathrm{cell} = 1\times 1 =1 \\ \mathrm{Number} \mathrm{of} {\mathrm{Cl}}^{- }\mathrm{ions} \mathrm{per} \mathrm{unit} \mathrm{cell} =8\times \frac{1}{8}=1 \end{gathered}$$

Therefore, CsCl has one $C{s}^{+}$ ion and one $C{l}^{-}$ ion in its unit cell.