Animated Solution for Chemistry - States of Matter: A crystal is made up of metal ions $M_1$ and $M_2$ and oxide ions. Oxide ions form a ccp lattice structure. The cation $M_1$ occupies $50\%$ of octahedral voids and the cation $M_2$ occupies $12.5\%$ of tetrahedral voids of oxide lattice. The oxidation numbers of $M_1$ and $M_2$ are, respectively

Welcome to a fascinating puzzle from the world of solid-state chemistry! In this problem, we are tasked with finding the oxidation states of two different metal cations, $M_1$ and $M_2$, embedded within a crystal lattice formed by oxide ions. This requires a beautiful blend of spatial geometry and fundamental chemical principles.

The foundation of our crystal is the cubic close-packed (ccp) lattice formed by the oxide ions ($O^{2-}$). The first crucial piece of information we need is the effective number of atoms per unit cell, denoted by $Z$. For a ccp lattice (which is geometrically identical to a face-centered cubic or fcc lattice), $Z=4$. This means there are effectively 4 oxide ions in one unit cell.

Now, how does this relate to the empty spaces, or 'voids', within the lattice? The geometry of close packing dictates a strict relationship: Number of Octahedral Voids (OV) is exactly equal to $Z$. So, we have $4$ octahedral voids. Number of Tetrahedral Voids (TV) is exactly double that, $2Z$. So, we have $8$ tetrahedral voids.

With our voids mapped out, let's place our metal cations according to the problem's constraints: Cation $M_1$ occupies $50\%$ of the octahedral voids. Since there are 4 octahedral voids in total, $50\%$ of 4 gives us exactly $2$ ions of $M_1$ per unit cell. Cation $M_2$ occupies $12.5\%$ of the tetrahedral voids. $12.5\%$ is equivalent to the fraction $\frac{1}{8}$. Since there are 8 tetrahedral voids, $\frac{1}{8}$ of 8 gives us exactly $1$ ion of $M_2$ per unit cell.

By combining the number of each type of ion present in the unit cell, we can construct the empirical formula of the crystal. We have: $2$ ions of $M_1$ $1$ ion of $M_2$ * $4$ ions of $O^{2-}$

This gives us the empirical formula: $(M_1{)}_2(M_2{)}_1{O}_4$.

Here is where the chemistry magic happens. Any stable ionic crystal must be electrically neutral overall. This means the total positive charge contributed by the metal cations must perfectly balance the total negative charge contributed by the oxide anions.

Let's set up an algebraic equation. Let the oxidation state (charge) of $M_1$ be $x$, and the oxidation state of $M_2$ be $y$.

The total positive charge is $2(x)+1(y)$. The total negative charge from the four oxide ions is $4\times (-2)=-8$.

Setting the sum of all charges to zero, we get:

We now have a linear equation with two variables: $2x+y=8$. To find the specific values of $x$ and $y$, we can test the pairs of oxidation numbers provided in the options:

(a) $+2,+4$: Substituting $x=2$ and $y=4$, we get $2(2)+4=4+4=8$. This matches perfectly! (b) $+1,+3$: Substituting $x=1$ and $y=3$, we get $2(1) + 3 = 5 eq 8$. (c) $+3,+1$: Substituting $x=3$ and $y=1$, we get $2(3) + 1 = 7 eq 8$. (d) $+4,+2$: Substituting $x=4$ and $y=2$, we get $2(4) + 2 = 10 eq 8$.

Therefore, the correct oxidation numbers for $M_1$ and $M_2$ are $+2$ and $+4$, respectively. This problem beautifully demonstrates how macroscopic properties like electrical neutrality are deeply rooted in the microscopic geometric arrangement of atoms!