Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm^-3. The value of  Avogadro's number based on this data will be -

(Atomic mass of Fe = 56.0 u)

1. 6.04 x 10²³ mol^-1

2. 12.08 x 10²⁴ mol^-1

3. 6 x 10²² mol^-1

4. 6 x 10^-23 mol^-1

Niobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm^-3, calculate atomic radius of niobium, given its atomic mass 93u.

1.  149 cm

2.  143 $\mathrm{\rho m}$

3.  133 $\mathrm{\rho m}$

4.  143 cm

A compound forms hcp structure. The total number of voids in 0.5 mol will be- 

1. 9.033$\times$10²³

2. 6.011$\times$10²³

3. 5.023$\times$10²³

4. 7.033$\times$10²³

Assertion : An important feature of Fluorite structures is that cations being large in size occupy FCC

lattice points whereas anions occupy all the tetrahedral voids giving the formula unit $A{B}_2$(A: cation B: anion).

Reason : There are 6 cations and 12 anions per FCC unit cell of the Fluorite structure.

Assertion : In NaCl crystal each Na ion is touching 6 $C{l}^{‐}$ ions but these $C{l}^{‐}$ ions do not touch each other.

Reason : The radius ratio roe $r_{Na}/r_{Cl-}$ is greater than 0.414, required for exact fitting.

Assertion : The ratio of $Z{n}^{2+ }$and F- per unit cell of ZnS and $Ca{F}_2$ is 1:2. 

Reason : In ZnS $Z{n}^{2+}$ ions occupy alternate tetrahedral voids and in $Ca{F}_2$ F- ions occupy all the tetrahedral voids of FCC unit cell.

Sodium metal crystallizes in bcc lattice with cell edge $=4.29 \stackrel{0}{A}$. The radius of sodium atom will be-

(1) 1.50 $\stackrel{0}{A}$                                                 

(2) 1.86  $\stackrel{0}{A}$

(3) 2.80 $\stackrel{0}{A}$                                               

(4) None of these 

The space lattice of graphite is

1. Cubic                                     

2. Tetragonal

3. Rhombic                                 

4. Hexagonal

The edge length of the cube is 400pm. Its body diagonal would be-

(1) 500 pm                                         

(2) 693 pm

(3) 600 pm                                           

(4) 566pm

A metallic element exists as a cubic lattice. Each edge of the unit cell is 2.88 $\stackrel{0}{A}$. The density of the metal is 7.20 g $c{m}^{-3}$. How many unit cell will be present in 100 g of the metal -

(1) $6.85\times {10}^2$                             

(2) $5.82\times {10}^{23}$

(3) $4.37\times {10}^5$                             

(4) $2.12\times {10}^6$