The radius of a calcium ion is $$94$$ pm and of the oxide ion is $$146$$ pm. The possible crystal structure of calcium oxide will be

The type of void that a cation occupies in an ionic solid is predicted from the radius-ratio rule.

Define the radius ratio as
$$\text{radius ratio } = \frac{r_{+}}{r_{-}}$$
where $$r_{+}$$ is the radius of the cation and $$r_{-}$$ is the radius of the anion.

For calcium oxide we are given
$$r_{+} \,(Ca^{2+}) = 94 \text{ pm}, \quad r_{-} \,(O^{2-}) = 146 \text{ pm}$$

Calculate the radius ratio:
$$\frac{r_{+}}{r_{-}} = \frac{94}{146} \approx 0.644$$

Now recall the standard radius-ratio ranges:

• $$0.225 \lt \frac{r_{+}}{r_{-}} \lt 0.414$$  → tetrahedral void, coordination number (CN) = 4
• $$0.414 \lt \frac{r_{+}}{r_{-}} \lt 0.732$$  → octahedral void, CN = 6
• $$0.732 \lt \frac{r_{+}}{r_{-}} \lt 1.000$$  → cubic void, CN = 8

The computed value $$0.644$$ lies in the interval $$0.414 \lt 0.644 \lt 0.732$$, corresponding to an octahedral void.

Therefore each $$Ca^{2+}$$ is surrounded by six $$O^{2-}$$ ions (and vice versa), giving the NaCl-type (rock-salt) lattice where the coordination number is 6 : 6, i.e., octahedral coordination.

Option C which is: octahedral.