Test: Packing Crystals (Old NCERT) - NEET MCQ

(d) CsCI has bcc structure.
Closest dista nce between Cs⁺ and Cl^- is the sum of radii of Cs⁺ and CI^- = 169 + 181 = 350 pm

Cubic diagonal = √3 a

Closest distance is one-half of the cubic diagonal

(d) AB₂ crystallises as cubic close packed (ccp).

Thus, A²⁺ is surrounded by eight B^- and B^- is surrounded by four A²⁺ .

Thus, coordinatio n number of A²⁺ = B

coordination number of B^- = 4

Number of atoms in tetrahedral in small cube of unit cell of atom B = 4

Thus, fraction occupied = 4/4 = 1 (100%)

(c)

AB = a, AC = a

BC² = AB² + AC² =2a²

Radius =r

then BC = CD + DE + EB = r + 2r + r = 4r 

∴ 16r² = 2a²

(c)

Tetrahedral holes = 8, Octahedral holes = 4 Cobalt atoms = 8/8 =1

Oxide ions = 4/2 =2

Thus, formula is CoO₂

(b) Unit cell of fee is divided into eight small cubes. Each cube has four atoms at alternate positions. When joined to each other, they make a regular tetrahedron.
Thus, number of tetrahedral void in small cube = 1 and in eight cubes, voids = 8.

Note  If number of atoms in a cube = N, then, number of octahedral voids = N and number of tetrahedral voids = 2N

(a) Edge length = a, Volume of one face = a³ Volume of six faces of a cube = 6a³ = V Radius = r

fee can be hexagonal closest packed (hep)

or cubic closest packed (ccp)

z = number of atoms in unit cell.

(c) Close packing in two-dimensions can be Type I A A A ...... ty p e The second row may be ^ placed in contact with the first one such that the spheres of A the second row are exactly . above those of the first row.
The spheres of the two rows A are aligned horizontally as well as vertically. In this arrangement, each sphere is in contact with four of its neighbours.
Thus, its coordination number = 4 and is called square close packing in two-dimensions.
Atom A (•) is in contact with four others.

Type II ABAB... type The second row may be placed above the first in a staggered manner such that its spheres fit in the depressions of the first row.
First row arrangement: A type Second row arrangement: 8 type Third row arrangement: A type and so on There is less free space and packing is more efficient than the square close packing. Each sphere is in contact with six of its neighbours and coordination number = 6 in two-dimensions. It is called two dimensional hexagonal close-packing. Voids formed in this case are triangular in shape. If the apex of the triangles downward in the first row, it is upward in the second row.

 (a)

In a bcc structure, the atoms tou ch each other along the body diagonal.

Radius of the atom = r

AB² = 8C² + AC² 
BC² = BD² + CD² = a² + a² = 2a² 

Volume of one atom = 

Volume of N₀ atoms in one mole of the metal

= 6.02 x 10²³ x 1.48 x 10^-23 cm³ = 8.93 cm³

(b)

Structure is cubic close packing.
Number of atoms of M = (from edge centre) +1 (from body centre) =2

Number of atoms of X =   (from corners of the cube) +  (from face centre)  =4

Thus, M : X = 2 : 4 = 1 : 2 Hence, the empirical formula of the compound is MX₂.

(a) Ratio of r+ /r_ will decid e the type of unit cell 

For AX    hence, AX has bcc structure.

Thus, octahedral or square planar structure.

(i)

Body-centred cubic Every cation (Cs⁺) is coordinated to eight anions (Br^- ) and every anion (Br^- ) is coordinated to eight cations (Cs⁺).
Thus, (i) -> (s)

(ii) Face-centred cubic Every cation (Na⁺) is coordinated to six anions (Cl^- ) and every anion (Cl^- ) is coordinated to six cations (Na⁺).
Thus, (ii) -> (q)

(iii) Face-centred cubic (tetrahedral void) Every cation (Zn²⁺) is coordinated to four anions (S^2-) and vice-versa.
Thus, (iii) —> (p)

(iv) Face-centred cubic (antifluorite) EveryCa²⁺ is coordinated to eight fluoride and every F^- is coordinated to fourCa2+ (to balance charge).

4Ca²⁺ + 8F^-
Thus, (iv) —> (r)

 (a,b,c)

(a) r = radius, a = edge length 

then for hcp, r  = 

Volume of one spherical atom 

Number of atoms in one unit cell = 4

∴ Volume of four atoms  

Volume of cube = a³

∴ Packing efficiency =

Thus, correct.

(b) In hep coordination number = 12, six at the diagonal and six at the edge length thus, correct.

(c) A tetrahedral void is formed when a sphere of a second layer is above the void of the first layer or vice-versa thus, correct.

(a, b)

(a) A tetrahedral void is formed when a sphere of a second layer is placed above triangular void in the first layer thus, correct.

(b) All the triangular voids are not covered by this arrangement thus, correct.

 (a, d)

Octahedral voids are formed in

(i) hexagonal close packing

(ii) face-centred cubic

(a)

Nearest neighbours are along the diagonal at the half distance apart, Diagonal AB = √3a

∴ Distance of the nearest neighbour

Next nearest neighbours are along the cell edge = 520 pm

K-atom at the centre is surrounded by eight K-atoms as nearest neighbours.

In bcc, coordination number is six as next nearest neighbours

(c)

Nearest neighbours are along the diagonal at the half distance apart, Diagonal AB = √3a

∴ Distance of the nearest neighbour

Next nearest neighbours are along the cell edge = 520 pm

K-atom at the centre is surrounded by eight K-atoms as nearest neighbours.

In bcc, coordination number is six as next nearest neighbours

(b)

Closest distance between two gold atoms is half of the diagonal.

Note There are total of 12 nearest neighbours (six along diagonal and six along the edges],

Number of atoms in tetrahedral in small cube of unit cell of B = 4
Thus, tetrahedral voids in unit cell of B = 8.

 Number of atoms of A = 2/3 of tetrahedral voids of B

Total number of atoms in lattice = 4 + 3 = 7

In fee or hep lattice, body centre is surrounded by six atoms on face centres. If these face centres are joined, one octahedron is formed.

Thus, number of octahedral void in the body centre of the cube = 1 There is also one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, three belonging to the same unit cell and three belonging to the two adjacent unit cells.

Each edge of the cube is shared between four adjacent unit cells,

thus, number of octahedral voids =12/4 =3

Thus, total octahedral voids =1+ 3 = 4

 Density (d) = 

where, M = Atomic mass = 48 g mol^-1

N₀ = Avogadro's number = 6.02 x 10²³ m ol^-1 !

z = Number of atoms in the unit cell a³ = Volume 

Note z = 4 indicates thatTi has fee structure.