Analyze Cubic Unit Cell Geometry for Binary Compound

To determine the correct statement(s), we analyze the given cubic unit cell structure of the binary compound step-by-step:

1. Determination of the Empirical Formula (Option A)

• Position of Cations ($\text{M}$): The cations ($\text{M}$, represented by solid black circles) are located at the midpoints of the 4 vertical edges of the unit cell. Since each edge center is shared equally among 4 adjacent unit cells: $\text{Number of M ions per unit cell} = 4 \times \frac{1}{4} = 1$

• Position of Anions ($\text{X}$): The anions ($\text{X}$, represented by open circles) are located at the midpoints of 4 horizontal edges (2 on the top face and 2 on the bottom face). Since each edge center is shared equally among 4 adjacent unit cells: $\text{Number of X ions per unit cell} = 4 \times \frac{1}{4} = 1$

• Empirical Formula: The stoichiometric ratio of $\text{M} : \text{X} = 1 : 1$. Therefore, the empirical formula of the compound is $\text{MX}$.

Hence, Option (A) is correct.

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2. Analysis of Coordination Geometries (Option B)

Let the cubic unit cell have edge length $a$ with a corner at the origin $(0,0,0)$.

• Coordination Geometry of Cation M: Consider a cation $\text{M}$ located at the vertical edge center $(0, 0, a/2)$. Its four nearest neighbor $\text{X}$ anions are located at: $\left(0, \frac{a}{2}, 0\right), \quad \left(0, -\frac{a}{2}, 0\right), \quad \left(\frac{a}{2}, 0, a\right), \quad \left(-\frac{a}{2}, 0, a\right)$ Two of these neighbors lie in the $yz$-plane at $z=0$, and the other two lie in the $xz$-plane at $z=a$. These four neighbors form a tetrahedral geometry around cation $\text{M}$.

• Coordination Geometry of Anion X: Consider an anion $\text{X}$ located at the top-edge center $\left(\frac{a}{2}, 0, a\right)$. Its four nearest neighbor $\text{M}$ cations are located at: $\left(0, 0, \frac{a}{2}\right), \quad \left(a, 0, \frac{a}{2}\right), \quad \left(0, 0, \frac{3a}{2}\right), \quad \left(a, 0, \frac{3a}{2}\right)$ All four neighboring cations lie entirely in the single plane $y = 0$, forming a square planar geometry around anion $\text{X}$.

Since cation $\text{M}$ has tetrahedral coordination geometry while anion $\text{X}$ has square planar coordination geometry, they possess distinct coordination geometries.

Hence, Option (B) is correct.

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3. Ratio of M-X Bond Length to Edge Length (Option C)

The bond length between cation $\text{M}$ at $(0, 0, a/2)$ and anion $\text{X}$ at $\left(\frac{a}{2}, 0, a\right)$ is given by the distance formula: $d_{\text{M-X}} = \sqrt{\left(\frac{a}{2} - 0\right)^2 + (0 - 0)^2 + \left(a - \frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \frac{a}{\sqrt{2}}$

Thus, the ratio of the $\text{M-X}$ bond length to the unit cell edge length $a$ is: $\frac{d_{\text{M-X}}}{a} = \frac{1}{\sqrt{2}} \approx 0.707$

Since $\frac{1}{\sqrt{2}} \approx 0.707 \neq 0.866$ (where $0.866 = \frac{\sqrt{3}}{2}$), Option (C) is incorrect.

Hence, Option (C) is incorrect.

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4. Ratio of Ionic Radii (Option D)

For the cation $\text{M}$, which occupies a site with tetrahedral coordination (coordination number 4), the ideal limiting radius ratio $r_{\text{M}} / r_{\text{X}}$ for structural stability is: $\frac{r_{\text{M}}}{r_{\text{X}}} = \sqrt{\frac{3}{2}} - 1 \approx 0.225$

The value $0.414$ corresponds to octahedral or square planar coordination.

Hence, Option (D) is incorrect.

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Conclusion

The correct statements are A and B.