Interstitial Voids

Q: What are interstitial voids in solids?

Interstitial voids are the empty spaces between the spheres of a packed crystal. In close packing there are two kinds: tetrahedral voids, each surrounded by 4 spheres, and octahedral voids, each surrounded by 6 spheres. Smaller atoms or ions often occupy these voids.

Q: How many tetrahedral and octahedral voids are there in a ccp structure?

For N spheres in cubic or hexagonal close packing there are N octahedral voids and 2N tetrahedral voids. An fcc unit cell with 4 atoms therefore has 4 octahedral and 8 tetrahedral voids; an hcp cell with 6 atoms has 6 and 12.

Q: What is the radius ratio of a tetrahedral void?

The largest sphere that fits in a tetrahedral void has radius 0.225 times the radius of the surrounding spheres. It follows from placing the four spheres on alternate corners of a cube: they touch along face diagonals, and the void centre is half a body diagonal from each.

Q: Where are octahedral voids located in an fcc unit cell?

In an fcc unit cell the octahedral voids are at the body centre, which belongs wholly to the cell, and at the midpoints of the 12 edges, each shared by 4 cells. That gives 1 + 3 = 4 octahedral voids per cell.

Q: Which void is bigger, tetrahedral or octahedral?

The octahedral void is bigger. Its limiting radius ratio is 0.414 compared with 0.225 for the tetrahedral void, so it can hold a sphere about 1.84 times larger in radius. Tetrahedral voids are, however, twice as numerous.

Q: How do you find the formula of a compound from void occupancy?

Take N ions forming the close-packed lattice. There are N octahedral and 2N tetrahedral voids. Multiply each by the fraction occupied to get the number of the other ions, write the ratio in whole numbers and check that the charges balance.

Q: Are voids and radius ratio in the JEE Main syllabus?

No. Solid state, including voids and radius ratio, was removed from JEE Main and NEET from 2024. JEE Advanced 2026 still lists close packing, nearest neighbours, ionic radii and radius ratio, so voids remain important there.

Q: What questions on voids are asked in JEE Advanced?

Typical questions ask for void counts in fcc or hcp, the ratio of tetrahedral to octahedral void sizes, the largest atom that fits in a void, distances between voids, and formulas of compounds when ions fill a fraction of the voids.

ChemistrySolid StateFor JEE aspirants

Interstitial voids are the empty spaces left between spheres even in the tightest packing. A close-packed crystal has two kinds: small tetrahedral voids, each surrounded by 4 spheres, and larger octahedral voids, each surrounded by 6. For $N$ spheres in ccp or hcp there are $N$ octahedral and $2 N$ tetrahedral voids. This page derives the limiting radius ratios (0.155, 0.225, 0.414, 0.732), locates every void in the fcc and hcp cells, and uses void counts to write formulas. Solid state is now in the JEE Advanced syllabus only.

On this page1What a void is2Triangular void3Tetrahedral void4Octahedral void5Cubic void6Voids between layers7Voids in fcc and hcp8Formulas from voids

Key Formulas - Quick Reference

★ Must learn Limiting radius ratio $r / R$ : triangular $0.155$ , tetrahedral $0.225$ , octahedral $0.414$ , cubic $0.732$ .
★ Must learn For $N$ close-packed spheres: octahedral voids $= N$ , tetrahedral voids $= 2 N$ .
★ Must learn fcc cell ( $Z = 4$ ): 8 tetrahedral voids (minicube centres), 4 octahedral voids (body centre + 12 edge centres × $\frac{1}{4}$ ).
hcp cell ( $Z = 6$ ): 12 tetrahedral and 6 octahedral voids.
★ Must learn In fcc: corner to tetrahedral void $= \frac{3}{4} a$ ; nearest T-T $= \frac{a}{2}$ ; nearest O-O $= \frac{a}{2}$ ; along an edge $2 R + 2 r_{oc t} = a$ .
Void size order: triangular $<$ tetrahedral $<$ octahedral $<$ cubic (coordination 3, 4, 6, 8).
Largest atom in a void of an fcc metal: $r = 0.225 R$ (tetrahedral), $0.414 R$ (octahedral), with $R = \frac{2}{4} a$ .

1. What Is an Interstitial Void?

Even the best packing of spheres fills only 74% of space. The remaining 26% is not one big gap; it is split into many small cavities between the spheres, called interstitial voids (or holes, or interstices). In ionic compounds and alloys, smaller particles sit in these voids, so knowing their size, number and position explains most crystal structures.

The size of a void is the radius

r

of the largest sphere that fits into it without pushing the surrounding spheres (radius

R

) apart. It is quoted as the limiting radius ratio

r / R

, which depends only on how many spheres surround the void.

2. Triangular Void (2D, CN 3)

In a single close-packed layer, three touching spheres leave a small hollow between them. The centres form an equilateral triangle of side $2 R$ ; the void centre is the centroid.

Figure 1: The triangular void, the hollow between three touching spheres in one layer. From

cos 3 0^{\circ} = \frac{R}{R + r}

, the largest sphere that fits has

r = 0.155 R

cos 3 0^{\circ} = \frac{R}{R + r} \Rightarrow R + r = \frac{2 R}{3} \Rightarrow \frac{r}{R} = \frac{2}{3} - 1 = 0.155

3. Tetrahedral Void (CN 4)

When a sphere of the second layer sits on a triangular hollow of the first, the 4 spheres enclose a tetrahedral void: the lines joining their centres form a regular tetrahedron. The void is the smaller of the two 3D voids in close packing.

Figure 2: A tetrahedral void is surrounded by 4 spheres at the corners of a tetrahedron, which fit exactly on alternate corners of a cube. Face diagonal

2 x = 2 R

and half body diagonal

\frac{3}{2} x = R + r

give

r = 0.225 R

Place the four spheres on alternate corners of a cube of edge $x$ . Neighbouring spheres lie along a face diagonal and touch, so $2 x = 2 R$ . The void is at the cube centre, half a body diagonal from each sphere:

R + r = \frac{3}{2} x = \frac{3}{2} \cdot 2 R = \frac{3}{2} R \Rightarrow \frac{r}{R} = \frac{3}{2} - 1 = 0.225

The same result follows from the tetrahedral angle. The angle between two sphere-void lines is

10 9^{\circ} 2 8^{'}

; half of it is

5 4^{\circ} 4 4^{'}

, and

sin 5 4^{\circ} 4 4^{'} = \frac{R}{R + r}

gives

r / R = 0.225

4. Octahedral Void (CN 6)

An octahedral void is enclosed by 6 spheres: 4 in a square and 1 each above and below the square. In close packing it forms where a triangular hollow of one layer lies directly over a hollow of the next, pointing the other way (3 spheres below, 3 above).

Figure 3: An octahedral void is surrounded by 6 spheres (a square of 4, plus one above and one below). Across the square,

2 R + 2 r = 2 (2 R)

, so

r = 0.414 R

In the square of 4 touching spheres (side $2 R$ ), the diagonal passes through two sphere centres and the void:

2 R + 2 r = 2 (2 R) \Rightarrow \frac{r}{R} = 2 - 1 = 0.414

5. Cubic Void (CN 8)

A cubic void lies at the centre of 8 spheres at the corners of a cube. It does not occur in close packing; it appears in simple cubic arrangements, for example of $C l^{-}$ ions in CsCl.

Figure 4: The cubic void at the centre of a simple cube of 8 touching spheres. With

a = 2 R

and

3 a = 2 R + 2 r

, the void fits

r = 0.732 R

. It is not found in close packing, only in simple cubic arrays.

a = 2 R, 3 a = 2 R + 2 r \Rightarrow \frac{r}{R} = 3 - 1 = 0.732

Figure 5: The more spheres surround a void, the bigger it is. Order of size: triangular

<

tetrahedral

<

octahedral

<

cubic.

Void	Surrounding spheres (CN)	Shape	Limiting $r / R$	Found in
Triangular	3	equilateral triangle	$\frac{2}{3} - 1 = 0.155$	planar sites, e.g. B in $B_{2} O_{3}$
Tetrahedral	4	tetrahedron	$\frac{3}{2} - 1 = 0.225$	ccp and hcp (2 per sphere)
Octahedral	6	octahedron	$2 - 1 = 0.414$	ccp and hcp (1 per sphere)
Cubic	8	cube	$3 - 1 = 0.732$	simple cubic arrays (CsCl)

Exam Trick Memorise the four ratios as 0.155, 0.225, 0.414, 0.732: each is (a simple surd) minus 1.

\frac{2}{3} = 1.155

1.5 = 1.225

2 = 1.414

3 = 1.732

. Drop the 1 and you have the ratio.

Key idea

Void size grows with coordination: CN 3, 4, 6, 8 → r/R 0.155, 0.225, 0.414, 0.732.

Quick Recall: tap to check

Which is bigger, a tetrahedral or an octahedral void?

Octahedral (

r = 0.414 R

) is almost twice the tetrahedral (

0.225 R

): ratio

0.543

Why is there no cubic void in ccp?

In ccp every void is surrounded by 4 or 6 spheres; a cube of 8 spheres around one hole occurs only in simple cubic stacking.

Derive the octahedral ratio in one line.

Square diagonal:

2 R + 2 r = 22 R

, so

r / R = 2 - 1

6. Voids Between Two Close-Packed Layers

Look down on layer A with layer B on top. Every position in the second layer falls into one of two situations, which is why exactly two kinds of voids exist in close packing.

Figure 6: Where layer B covers a hollow of A, one B sphere and three A spheres enclose a tetrahedral void (T). Where a hollow of B lies over a hollow of A, three spheres above and three below enclose an octahedral void (O).

Tetrahedral void (T): a sphere of B sits over a hollow of A. One B sphere + three A spheres. There is a second set too: above each A sphere, under a hollow of B (one A sphere + three B spheres).
Octahedral void (O): a hollow of B lies over a hollow of A, the two triangles pointing opposite ways. Three A spheres + three B spheres.

Counting through a large crystal gives a simple rule: each sphere is associated with one octahedral void and two tetrahedral voids.

For

N

spheres in a close-packed structure (ccp or hcp): number of octahedral voids

= N

; number of tetrahedral voids

= 2 N

. So 1 mol of atoms has

N_{A}

octahedral and

2 N_{A}

tetrahedral voids.

Key idea

One sphere → one octahedral + two tetrahedral voids, in both hcp and ccp.

7. Locating Voids in the fcc and hcp Cells

7.1 Tetrahedral voids in fcc (ccp)

Divide the fcc cell into 8 minicubes of edge $a /2$ . Each minicube has 4 of its alternate corners occupied (one cube corner and three face centres), exactly the tetrahedral arrangement of Figure 2, so its centre is a tetrahedral void.

Figure 7: Split the fcc cell into 8 minicubes of edge

a /2

: each minicube centre is a tetrahedral void (its 4 surrounding atoms are one corner and three face centres). So there are 8 per cell, twice the number of atoms.

8 tetrahedral voids per cell, all inside: $8 = 2 Z$ .
They lie on the 4 body diagonals, 2 per diagonal, each at $\frac{3}{4} a$ from the nearest corner.
Two voids on the same diagonal are $\frac{3}{2} a$ apart; the closest pair of tetrahedral voids (in adjacent minicubes) is $\frac{a}{2}$ apart.

7.2 Octahedral voids in fcc (ccp)

Figure 8: Octahedral voids in fcc sit at the body centre (surrounded by the 6 face-centre atoms) and at the 12 edge centres (each shared by 4 cells):

1 + 12 \times \frac{1}{4} = 4

, equal to the number of atoms.

The body centre is surrounded by the 6 face-centre atoms: 1 octahedral void, wholly inside.
Each of the 12 edge centres is surrounded by 2 corner atoms and 4 face-centre atoms (of the 4 cells sharing that edge): $12 \times \frac{1}{4} = 3$ .
Total $= 1 + 3 = 4 = Z$ . Nearest O-O distance $= \frac{a}{2}$ ; tetrahedral to octahedral void $= \frac{3}{4} a$ .

7.3 Voids in hcp

Figure 9: In hcp (

Z = 6

), tetrahedral voids lie in pairs along vertical lines through the atoms: 8 inside plus 12 on the vertical edges shared by 3 prisms, total 12. The 6 octahedral voids sit over the open c hollows, wholly inside.

Unit cell	$Z$	Tetrahedral voids $= 2 Z$	Octahedral voids $= Z$
ccp (fcc)	4	8	4
hcp	6	12 (8 inside + 12 × $\frac{1}{3}$ )	6 (all inside)

Tetrahedral void

CN 4, $r = 0.225 R$ . $2 N$ per $N$ spheres. In fcc: minicube centres (8, inside). Smaller, so it takes small cations such as $Z n^{2 +}$ in ZnS.

Octahedral void

CN 6, $r = 0.414 R$ . $N$ per $N$ spheres. In fcc: body centre and edge centres (4). Larger, so it takes cations such as $N a^{+}$ in NaCl.

7.4 How big an atom fits?

In an fcc metal, $R = \frac{2}{4} a$ . The largest atom that fits without distortion has $r = 0.414 R$ in an octahedral void and $0.225 R$ in a tetrahedral void. Along a cube edge the geometry is direct: two corner atoms and the edge-centre void lie in a line, so $2 R + 2 r = a$ .

JEE Advanced Interstitial alloys and compounds use exactly these holes. Small atoms (H, B, C, N) enter the octahedral voids of transition-metal lattices to form hard, high-melting interstitial compounds such as TiC,

M n_{4} N

F e_{3} H

and steel (C in the octahedral voids of fcc iron, austenite). Because the atoms fit into existing holes, the metal keeps its lattice but becomes harder and less ductile. A carbon atom (radius about 77 pm) is larger than the octahedral hole of fcc iron (about 52 pm), which is why only a few per cent of carbon dissolves and the lattice is strained.

Key idea

fcc: 8 T at minicube centres, 4 O at body + edge centres. hcp: 12 T, 6 O. Always T = 2Z, O = Z.

Quick Recall: tap to check

Where are the octahedral voids in an fcc cell?

At the body centre (1) and the 12 edge centres (each

\frac{1}{4}

): 4 in all.

Distance of a tetrahedral void from the nearest corner in fcc?

\frac{3}{4} a

, a quarter of the body diagonal.

How many voids of each kind in 0.2 mol of a ccp metal?

Octahedral

0.2 N_{A} = 1.20 \times 1 0^{23}

; tetrahedral

2.41 \times 1 0^{23}

8. Writing Formulas from Void Occupancy

In many ionic solids the larger ions (usually anions) form the close-packed lattice and the smaller ions fill a fraction of the voids. Take $N$ lattice ions, multiply the void counts ( $N$ octahedral, $2 N$ tetrahedral) by the fractions filled, and write the ratio.

Figure 10: Every void-occupancy formula question follows these steps. The charge check catches most slips.

Oxide ions in ccp: take $N$ $O^{2 -}$ .
Half the octahedral voids hold $A^{3 +}$ : $A = \frac{1}{2} \times N = \frac{N}{2}$ .
One eighth of the tetrahedral voids hold $B^{2 +}$ : $B = \frac{1}{8} \times 2 N = \frac{N}{4}$ .
Ratio $A : B : O = \frac{1}{2} : \frac{1}{4} : 1 = 2 : 1 : 4$ , giving $A_{2} B O_{4}$ .
Charge check: $2 (+ 3) + (+ 2) + 4 (- 2) = 0$ . This is the spinel structure ( $MgA l_{2} O_{4}$ ).

Exam Trick Octahedral = one each, tetrahedral = two each. If a question says 'all tetrahedral voids', the small ion is twice as many as the lattice ion (fluorite-type

Ca F_{2}

with roles swapped, or

N a_{2} O

); 'all octahedral voids' gives a 1 : 1 compound (NaCl type).

Key idea

Formula = (fraction × void count) : lattice ions, with O = N and T = 2N; always finish with a charge check.

Quick Recall: tap to check

Anions ccp, cations in half the tetrahedral voids. Formula?

Cations

= \frac{1}{2} \times 2 N = N

: 1 : 1, as in zinc blende (ZnS).

Oxide ccp,

M^{2 +}

in all octahedral voids. Formula?

MO (rock-salt type, for example MgO).

What does the charge check confirm?

That the chosen fractions give a neutral formula; a mismatch means a slip in the void counts.

8.1 The whole concept at a glance

Figure 11: Interstitial voids at a glance: size, coordination, count and where they sit.

9. Solved Examples

Solved Example 1

In a face-centred cubic arrangement of metallic atoms, what is the ratio of the sizes of tetrahedral and octahedral voids?
(A) 0.543
(B) 0.732
(C) 0.414
(D) 0.637

Solution:

Answer: (A).

\frac{r _{t e t r ah e d r a l}}{r _{oc t ah e d r a l}} = \frac{0.225 R}{0.414 R} = 0.543

Solved Example 2

The numbers of tetrahedral and octahedral holes in a ccp array of 100 atoms are respectively
(A) 200 and 100
(B) 100 and 200
(C) 200 and 200
(D) 100 and 100

Solution:

Answer: (A). $2 N = 200$ tetrahedral and $N = 100$ octahedral voids.

Solved Example 3

Copper has a face-centred cubic structure with edge length 3.61 Å. What is the size of the largest atom that could fit into the interstices of the copper lattice without distorting it?

Solution:

Figure 12: Along a cube edge, two corner atoms and the octahedral void at the edge centre lie in a line, so

2 R + 2 r = a

. For copper this gives

r = 0.529

Å, the largest atom that fits without distortion.

The largest interstice is the octahedral void. Atoms touch along the face diagonal, so $R = \frac{2}{4} a$ .

r = 0.414 R = 0.414 \times \frac{2 \times 3.61}{4} = 0.414 \times 1.276 = 0.53 \overset{˚}{A}

Check with the edge: $r = \frac{a}{2} - R = 1.805 - 1.276 = 0.529$ Å.

Solved Example 4

In a close-packed structure of mixed oxides, the lattice is made of oxide ions; one half of the octahedral voids are occupied by trivalent cations

A^{3 +}

and one eighth of the tetrahedral voids by divalent cations

B^{2 +}

. Derive the formula of the oxide.

Solution:

Per oxide ion: 1 octahedral void and 2 tetrahedral voids.

$A^{3 +}$ $= 1 \times \frac{1}{2} = \frac{1}{2}$ ; $B^{2 +}$ $= 2 \times \frac{1}{8} = \frac{1}{4}$ .

Formula $A_{1/2} B_{1/4} O$ = $A_{2} B O_{4}$ . Charge: $+ 6 + 2 - 8 = 0$ .

Solved Example 5

A compound is formed by elements X and Y. Atoms of Y form ccp and atoms of X occupy one third of the tetrahedral voids. What is the formula?

Solution:

For $N$ Y atoms: tetrahedral voids $= 2 N$ ; X $= \frac{1}{3} \times 2 N = \frac{2 N}{3}$ .

X : Y $= \frac{2}{3} : 1 = 2 : 3$ , so $X_{2} Y_{3}$ .

Solved Example 6

Atoms of element B form hcp and atoms of A occupy two thirds of the tetrahedral voids. The formula is
(A)

A_{2} B_{3}

(B)

A_{4} B_{3}

(C)

A_{3} B_{4}

(D) AB

Solution:

Answer: (B). A $= \frac{2}{3} \times 2 N = \frac{4 N}{3}$ ; A : B $= 4 : 3$ , giving $A_{4} B_{3}$ . Void counts per atom are the same in hcp and ccp.

Solved Example 7

Nickel is fcc with a = 352 pm. What is the radius of the largest atom that fits in a tetrahedral void of nickel?

Solution:

$R = \frac{2}{4} \times 352 = 124.5$ pm, so $r = 0.225 \times 124.5 = 28.0$ pm.

Only very small atoms such as H fit here without distortion.

Solved Example 8

In an fcc cell of edge a, the shortest distance between two tetrahedral voids is
(A)

\frac{3}{4} a

(B)

\frac{a}{2}

(C)

\frac{a}{2}

(D)

\frac{3}{2} a

Solution:

Answer: (B). Tetrahedral voids are at the centres of the 8 minicubes of edge $\frac{a}{2}$ , so neighbouring voids are $\frac{a}{2}$ apart. $\frac{3}{2} a$ is the separation of the two voids on one body diagonal, and $\frac{3}{4} a$ is corner-to-void.

Solved Example 9

Oxide ions form a ccp lattice and

A l^{3 +}

ions occupy two thirds of the octahedral voids. Find the formula and confirm it is neutral.

Solution:

For $N$ oxide ions: $A l^{3 +}$ $= \frac{2}{3} N$ .

Al : O $= \frac{2}{3} : 1 = 2 : 3$ , so $A l_{2} O_{3}$ . Charge: $2 (+ 3) + 3 (- 2) = 0$ . This is the corundum structure (strictly, oxide ions in hcp).

Practice Questions

Derive the limiting radius ratio for a triangular void.Answer: $cos 3 0^{\circ} = \frac{R}{R + r}$ gives $0.155$ .
How many tetrahedral and octahedral voids are in 0.5 mol of a ccp solid?Answer: Tetrahedral $6.022 \times 1 0^{23}$ ; octahedral $3.011 \times 1 0^{23}$ .
N forms ccp and M occupies one third of the octahedral voids. Formula?Answer: $M N_{3}$ .
Aluminium (fcc) has atomic radius 143 pm. Radius of the largest atom for its octahedral voids?Answer: $0.414 \times 143 = 59.2$ pm.
In an fcc cell with a = 400 pm, how far is a tetrahedral void from the nearest corner?Answer: $\frac{3}{4} \times 400 = 173$ pm.
How many octahedral and tetrahedral voids does one hcp unit cell contain?Answer: 6 octahedral, 12 tetrahedral.
Anions form ccp; cations fill all tetrahedral voids. What is the cation : anion ratio and an example?Answer: 2 : 1, antifluorite, for example $N a_{2} O$ .

Common Mistakes to Avoid

Watch out

Saying tetrahedral voids are larger because there are more of them. They are more numerous ( $2 N$ ) but smaller ( $0.225 R$ against $0.414 R$ ).
Using $N$ tetrahedral voids for $N$ spheres. It is $2 N$ tetrahedral and $N$ octahedral.
Placing the tetrahedral voids of fcc at the body centre. The body centre is an octahedral void; tetrahedral voids are at the minicube centres.
Counting edge-centre octahedral voids as whole voids. Each is shared by 4 cells: $12 \times \frac{1}{4} = 3$ .
Mixing up the corner-to-void distance $\frac{3}{4} a$ with the void-to-void distance $\frac{a}{2}$ .
Expecting a cubic void (0.732) in ccp or hcp. It occurs only in simple cubic arrays.
Forgetting that $r / R$ is the largest sphere that fits without distortion; a bigger atom can enter a void but pushes the lattice apart.
Skipping the charge check after writing a formula from void occupancy.

Frequently Asked Questions

What are interstitial voids in solids?

Interstitial voids are the empty spaces between the spheres of a packed crystal. In close packing there are two kinds: tetrahedral voids, each surrounded by 4 spheres, and octahedral voids, each surrounded by 6 spheres. Smaller atoms or ions often occupy these voids.

How many tetrahedral and octahedral voids are there in a ccp structure?

For N spheres in cubic or hexagonal close packing there are N octahedral voids and 2N tetrahedral voids. An fcc unit cell with 4 atoms therefore has 4 octahedral and 8 tetrahedral voids; an hcp cell with 6 atoms has 6 and 12.

What is the radius ratio of a tetrahedral void?

The largest sphere that fits in a tetrahedral void has radius 0.225 times the radius of the surrounding spheres. It follows from placing the four spheres on alternate corners of a cube: they touch along face diagonals, and the void centre is half a body diagonal from each.

Where are octahedral voids located in an fcc unit cell?

In an fcc unit cell the octahedral voids are at the body centre, which belongs wholly to the cell, and at the midpoints of the 12 edges, each shared by 4 cells. That gives 1 + 3 = 4 octahedral voids per cell.

Which void is bigger, tetrahedral or octahedral?

The octahedral void is bigger. Its limiting radius ratio is 0.414 compared with 0.225 for the tetrahedral void, so it can hold a sphere about 1.84 times larger in radius. Tetrahedral voids are, however, twice as numerous.

How do you find the formula of a compound from void occupancy?

Take N ions forming the close-packed lattice. There are N octahedral and 2N tetrahedral voids. Multiply each by the fraction occupied to get the number of the other ions, write the ratio in whole numbers and check that the charges balance.

Are voids and radius ratio in the JEE Main syllabus?

No. Solid state, including voids and radius ratio, was removed from JEE Main and NEET from 2024. JEE Advanced 2026 still lists close packing, nearest neighbours, ionic radii and radius ratio, so voids remain important there.

What questions on voids are asked in JEE Advanced?

Typical questions ask for void counts in fcc or hcp, the ratio of tetrahedral to octahedral void sizes, the largest atom that fits in a void, distances between voids, and formulas of compounds when ions fill a fraction of the voids.

Previous year questions on Interstitial Voids

2 questions from past papers, each with a step-by-step solution.

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