The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 Å which crystallizes in cubic lattice. If the density is 2 g/cc then find the radius of metal atom.(NA = 6 × 10²³).Give the answer in pm.

217 pm
210 pm
220 pm
205 pm

Solution

In the Bragg’s equation for diffraction of X-rays, n represents for

quantum number
an integer
Avogadro’s numbers
moles

Solution

Bragg’s equation is n? = 2dsin? where n is a positive integer i.e., 1, 2, 3, 4 etc.
which stands for serial order of diffracted beams

Potassium fluoride has NaCl type structure. What is the distance between K⁺ and F^– ions if cell edge is ‘a’ cm.

2a cm
a/2 cm
4a cm
a/4 cm

Solution

Distance between K⁺ and F^- =\(\frac{1}{2}\) × length of the edge

The pyknometric density of sodium chloride crystal is 2.165 × 10³kg m^-3 while its X-ray density is 2.178 × 10³ kg m^-3.The fraction of unoccupied sites in sodium chloride crystalis

5.96 ×10^-3
5.96 × 10⁴
5.96 ×10^-2
5.96 × 10^-1

Solution

CsBr crystallises in a body centered cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs =133 and that o fBr = 80 amu and Avogadro number being 6.02 × 10²³mol^-1, the density of CsBr is

0.425 g/cm³
8.5 g/cm³
4.25 g/cm³
82.5 g/cm³

Solution

For a cubic geometry the limiting \(\frac{r^{+}}{r^{-}}\) is :

0.414
0.155
> 0.731
0.731

The second order Bragg diffraction of X-rays with = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60º. The distance between the scattering planes in the crystal is

0.575 Å
1.00 Å
2.00 Å
1.15 Å

Solution

The limiting radius ratio for tetrahedral shape is:

0 to 0.155
0.225 to 0.414
0.155 to 0.225
0.414 to 0.732

Solution

For tetrahedral shape radius ratio is 0.225 – 0.414

In the calcium fluoride structure, the coordination number of the cation and anion are respectively

6, 6
8, 4
4, 4
4, 8

Solution

In CaF2, Ca²⁺ ions have ccp arrangement and F^- ions occupy the tetrahedral voids.In ccp arrangement total no. of Ca²⁺ ions is 4 and no. of tetrahedral voids is always 2n i.e. 8. Hence for every Ca²⁺ there are 8F^- ions as C.N, for F^- there are 4 Ca²⁺ ions.

Edge length of a cube is 400 pm. Its body diagonal would be

500 pm
566 pm
600 pm
693 pm

Solution

Body diagonal (d) of a cubic crystal of edge length (a)is given by,
d= a\(\sqrt{3}\)
putting a = 400 pm, we get
d= \(\sqrt{3}\) × 400 pm = 692.8pm ≈ 693 pm

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