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

(A) 20/3 $\pi $r^{3}

(B) 24/3 $\pi $r^{3}

(C) 12/3 $\pi $r^{3}

(D) 16/3 $\pi $r^{3}

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(D) Volume of atoms in cell = 4/3 $\pi $r^{3} x n (n =4 for fcc)

= 4/3 x $\pi $r^{3} x 4 = 16/3 $\pi $r^{3}

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The arrangement of X^{- }ions around A^{+} ion in solid AX is given in the figure (not drawn to scale).

If the radius of X^{- } is 250 pm,the radius of A+ is:

(a)104 pm

(b)125 pm

(c)183 pm

(d) 57 pm

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(a) The figure represents an octahedral void

r_{A}^{+}/ r_{X}^{-}=0.414

r_{A}^{+}=0.414x250=104 pm

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Select the correct statement (s)-

(a) The C.N. of cation occupying a tetrahedral hole is 4.

(b) The C.N. of cation occupying a octahedral hole is 6.

(c) In schottky defects, density of the lattice decreases,

(A) a,b

(B) b,c

(C) a,b,c

(D) a,c

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(C). Since tetrahedral holes are surrounded by 4 nearest neighbours. So, the C.N. of cation occupying tetrahedral hole is 4. Since octahedral hole is surrounded by six nearest neighbours. So, C.N. of cation occupying octahedral is 6. In schottky a pair of anion and cation leaves the lattice.So, density of lattice decreases.

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Among the following types of voids. which one is the largest void:-

(A) Triangular system

(B) Tetragonal system

(C) Monoclinic system

(D) Octahedral

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(D) The vacant spaces between the spheres in closed packed structure is called void. The voids are of two types, tetrahedral voids and octahedral voids. Also radius of tetrahedral voids and octahedral voids are r_{void} = 0.225 x r_{sphere} and r_{void} = 0.411 x r_{sphere} respectively. Thus, octahedral void is

larger than tetrahedral void.

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The arrangement ABC ABC.. is referred to as.

(a) octahedral close packing

(b) hexagonal clase packing

(c) tetrahedral close packing

(d) cubic close packing

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(d) It represents ccp arrangement.

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The ratio of cations to anion in a closed pack tetrahedral is:

(a) 0.414

(b) 0.225

(c) 0.02

(d) none of these

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(b) r^{+}/ r^{-} for tetrahedral void=0.225-0.414;

r^{+}/ r^{-} for triangular=0.155-0.225

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For a solid with the following structure, the co-ordination number of the point B is:

(a) 3

(b) 4

(c) 5

(d)6

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(d) It is evident from figure that B occupies tetrahedral voids and thus, co-ordination number is six.

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A crystal formula AB_{3} has A ions at the cube corners and B ions at the edge centres. The coordination number of A and B respectively

(A) 6 and 6

(B) 2 and 6

(C) 6 and 2

(D) 8 and 8

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(c) Co-ordination of cation/Co-ordination no of anion

= charge of cation/charge of anion = 3/1 =3

Now Co-ordination no. of A is as shown so of B is 2.

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Compute the percentage void space per unit volume of unit cell in zinc fluoride structure.

1. 15.03%

2. 22.18%

3. 18.23%

4. 25.07%

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(D) Anions occupy fcc positions and half of the tetrahedral holes are occupied by cations. Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres per unit volume of the unit cell is

For tetrahedral holes, r_{c}/r_{a} = 0.225

= {1+(0.225)^{3}} = 0.7493

Void volume = 1-0.7493 = 0.2507/unit volume of unit cell.

% Void space= 25.07%

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8 : 8 co-ordination of CsCl is found to change into 6 : 6

co-ordination on:

(a) applying pressure

(b) increasing temperature

(c) both (a) and (b)

(d) none of these

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(b) High temperature changes 8 : 8 co-ordination to 6 : 6 whereas high pressure changes 6 : 6

co-ordination to 8 : 8.

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