Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length Lx, body-centred

Q: Why is the packing efficiency of FCC higher than BCC?

In an FCC lattice, atoms are packed as closely as geometrically possible (Cubic Close Packing), occupying 74% of the space. BCC is slightly more open, occupying only 68%, because the atoms at the corners are not touching each other directly.

Q: How do we know the relation between edge length and radius for BCC?

In a BCC unit cell, the atoms touch along the body diagonal. The length of the body diagonal is $\sqrt{3}L$, and it contains two full atomic radii (the center atom) plus two half radii (the corner atoms), equaling $4r$. So, $\sqrt{3}L = 4r$.

Q: Can we assume the atomic radii are the same for all three metals?

Absolutely not! The problem explicitly gives relations between $r_x$, $r_y$, and $r_z$. You must use these specific ratios to compare their edge lengths.

Q: Why did the molar masses cancel out so neatly in the density ratio?

The problem was designed that way! By giving $M_z = \frac{3}{2} M_y$ and $M_z = 3 M_x$, we can deduce that $3 M_x = \frac{3}{2} M_y$, which simplifies to $M_y = 2 M_x$. When taking the ratio $d_x/d_y$, the $M_x$ terms perfectly cancel out.

Q: Is it necessary to calculate the exact decimal values for the edge lengths?

Not strictly necessary, but estimating them (like $\sqrt{2} \approx 1.414$ and $\sqrt{3} \approx 1.732$) makes it much easier to confidently compare their relative sizes without getting lost in complex fractions.

Question

Atoms of metals x, y, and z form face-centred cubic (fcc) unit cell of edge length $L_x$, body-centred cubic (bcc) unit cell of edge length $L_y$, and simple cubic unit cell of edge length $L_z$, respectively.
If $r_z = \frac{\sqrt{3}}{2} r_y$; $r_y = \frac{8}{\sqrt{3}} r_x$; $M_z = \frac{3}{2} M_y$ and $M_z = 3 M_x$, then the correct statement (s) is (are)
[Given : $M_x$, $M_y$, and $M_z$ are molar masses of metals x, y, and z, respectively.
$r_x$, $r_y$, and $r_z$ are atomic radii of metals x, y, and z, respectively.]

Options:

(A) Packing efficiency of unit cell of x > Packing efficiency of unit cell of y > Packing efficiency of unit cell of z

(B) $L_y > L_z$

(C) $L_x > L_y$

(D) Density of x > Density of y

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to a beautiful problem from the Solid State chapter that tests your spatial reasoning and algebraic stamina. We are introduced to three distinct metals: **x**, **y**, and **z**. Each of these metals crystallizes in a different lattice structure. Metal **x** forms a Face-Centered Cubic (FCC) lattice, metal **y** forms a Body-Centered Cubic (BCC) lattice, and metal **z** forms a Simple Cubic (SC) lattice.

Before we even look at the complex mathematical relations given in the problem, we can immediately tackle the first option. Packing efficiency is a fundamental property of these crystal lattices. It tells us what percentage of the total unit cell volume is actually occupied by the atoms.

For an FCC lattice, the atoms are packed as closely as geometrically possible, yielding a packing efficiency of about $74\%$. A BCC lattice is slightly more open, with an efficiency of $68\%$. Finally, a Simple Cubic lattice is the least efficient, occupying only about $52.4\%$ of the space. Therefore, the packing efficiency follows the order: $x > y > z$. This makes **Option (A) absolutely correct**.

### The Geometry of Edge Lengths

Now, let's dive into the geometry. The problem provides specific relationships between the atomic radii of these metals:
$r_z = \frac{\sqrt{3}}{2} r_y$
$r_y = \frac{8}{\sqrt{3}} r_x$

To compare their unit cell edge lengths ($L_x$, $L_y$, $L_z$), we must express all radii in terms of a single variable. Let's use $r_x$ as our base. By substituting the second equation into the first, we get:
$r_z = \frac{\sqrt{3}}{2} \left( \frac{8}{\sqrt{3}} r_x ight) = 4 r_x$

Next, we recall the standard geometric formulas relating the edge length $L$ to the atomic radius $r$ for each lattice type:
For FCC (metal x): $L_x = 2\sqrt{2} r_x \approx 2.828 r_x$
For BCC (metal y): $L_y = \frac{4}{\sqrt{3}} r_y$
For SC (metal z): $L_z = 2 r_z$

Let's substitute our radius relations into these edge length formulas to see how big these unit cells really are:
$L_y = \frac{4}{\sqrt{3}} \left( \frac{8}{\sqrt{3}} r_x ight) = \frac{32}{3} r_x \approx 10.67 r_x$
$L_z = 2 (4 r_x) = 8 r_x$

Comparing these values, it is glaringly obvious that $10.67 r_x > 8 r_x > 2.828 r_x$. Therefore, the order of edge lengths is **$L_y > L_z > L_x$**. This confirms that **Option (B) is correct**, and Option (C) is incorrect.

### The Density Showdown

Finally, we need to compare the densities of metal x and metal y. The density $d$ of a unit cell is given by the master equation:
$$d = \frac{Z \cdot M}{N_A \cdot L^3}$$
Where $Z$ is the effective number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro's number, and $L^3$ is the volume of the unit cell.

For metal x (FCC), $Z_x = 4$. For metal y (BCC), $Z_y = 2$. 
We are also given relations for their molar masses: $M_z = \frac{3}{2} M_y$ and $M_z = 3 M_x$. By equating these two expressions for $M_z$, we find a direct relationship between the masses of x and y:
$\frac{3}{2} M_y = 3 M_x \implies M_y = 2 M_x$

Now, let's set up the ratio of their densities:
$$\frac{d_x}{d_y} = \frac{Z_x M_x / L_x^3}{Z_y M_y / L_y^3} = \frac{4 M_x}{2 (2 M_x)} 	imes \left( \frac{L_y}{L_x} ight)^3$$

Notice how beautifully the mass terms cancel out! The ratio simplifies entirely to the cube of the ratio of their edge lengths:
$$\frac{d_x}{d_y} = 1 	imes \left( \frac{32/3 r_x}{2\sqrt{2} r_x} ight)^3 = \left( \frac{16}{3\sqrt{2}} ight)^3$$

Let's evaluate the term inside the parenthesis. $3\sqrt{2}$ is approximately $3 	imes 1.414 = 4.242$. So, we have $\frac{16}{4.242}$, which is clearly much greater than 1 (it's roughly 3.77). Cubing a number greater than 1 yields an even larger number.

Since $\frac{d_x}{d_y} > 1$, it strictly implies that **$d_x > d_y$**. Thus, **Option (D) is correct**.

This problem is a masterclass in keeping your variables organized and trusting the algebra to simplify beautifully at the end!

Select Answer:

Visualized Solution

$C r y s t a l S t r u c t u r es o f x, y, an d z$

$P a c k in g E f f i c i e n cy C o m p a r i so n$

$R e l a t in g A t o mi c R a d ii$

$E d g e L e n g t h s (L) v s R a d ii (r)$

$E x p r ess in g L in t er m s o f r_{x}$

$C o m p a r in g E d g e L e n g t h s$

$D e n s i t y o f U ni t C e l l$

$C a l c u l a t in g D e n s i t y R a t i o (d_{x} / d_{y})$

$C o n c l u s i o n f or D e n s i t y$

$S u mma r y & E x t e n s i o n s$

The Sigma Insight: Solid State

Analyzing the Setup

The Geometry of Edge Lengths

The Density Showdown

Similar Questions

Total volume of atoms present in a face-centred cubic unit cell of a metal is ( $r$ is atomic radius)

Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?

The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively.

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ' $a$ ', the closest approach between two atoms in metallic crystal will be

At $10 0^{\circ} C$ , copper (Cu) has FCC unit cell structure with cell edge length of $x \overset{˚}{A}$ . What is the approximate density of Cu (in $g cm^{- 3}$ ) at this temperature? [Atomic mass of Cu $= 63.55 u$ ]

The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s).

The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is (are)

The density (in g cm $^{- 3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ______. Use: Atomic mass of metal = 105.6 amu and Avogadro's constant = $6 \times 1 0^{23} mol^{- 1}$

Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

Copper crystallises in fcc with a unit cell length of $361 pm$ . What is the radius of copper atom?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Crystal Structures of x, y, and z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Packing Efficiency Comparison

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating Atomic Radii

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Edge Lengths (L) vs Radii (r)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expressing L in terms of rx​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Edge Lengths

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Density of Unit Cell

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Density Ratio (dx​/dy​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion for Density

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Summary & Extensions