Practice "Short Questions Answers (2 Marks) " of Chapter 1 => Solid State for CBSE NCERT.Q-1.What are F- centers ? why are the solids containing the F- centers are
paramagnetic?
Ans-1. =>The free electrons trapped in the anion vacancies are termed as the Fcenters.
=> The solids containing the F-centres are paramagnetic because the electrons
occupying the vacant sites are unpaired.
Q-2. A unit cell consists of a cube in which there are anions at each corner and one at the center of the unit cell. The cations are the center of the each face.
how many
A) cations and
B)anions make up the unit cell?
C)What is the simplest formula of the compound?
A-2. A)The cation at the center of each face is shared by two unit cells.
Hence no. of cations= 6 X ½ = 3
B) The anion at each corner is shared by 8 unit cells . the anion at the center is not shared by any other unit cell.
Hence no. of anions = 8 X 1/8 + 1 = 2
C) Since there are 3 cations and 2 anions the simplest formula of the compond is
A3B2.
Q-3.Exess of potassium in the KCl makes the crystal appears violet. Explain why?
Ans-3. => When KCl is heated in an atmosphere of K metal vapour , the metal K deposits on the surface of the KCl crystal.
=> The chloride ions diffuse into the surface and combine with K atoms. The electrons produced by the ionization of the K Atoms then diffuse into the crystals and are then trapped in the anion vacancies called F- centers.
=> The excess of the K in KCl makes the crystal appear violet.
Q-4. The two ions A+ and B- have radii 88 and 200 pm respectively. In the close
packed crystal of compound AB, predict the coordination number of A.
Ans- 4. I n the close-packed arrangement, A+ will be in the intersites of the close –packed arrangement of B
Radius of cation = 88 pm
Radius of anion =200 pm
Radius ratio = r+/r- = 88pm/200pm = 0.44
Since the radius ratio lies between 0.414—0.732, the ion A+ MUST OCCUPY THE OCTAHEDRAL SITE. Hence the coordination no. of A+ is 6.
Q-5. A solid AB has NaCl structure . if the radius of the cation A+ is 140 pm
calculate the maximum possible values of the radius of anion B-.
Ans-5. Radius ratio = r+/r- = 0.414
OR 140/ r B- = O.414
SO r B- = 140/0.414
= 338 pm.
Q-6. Copper crystallises in face-centred cubic lattice and has a density of 8.930 g
mol-3at 293 K. Calculate the edge length of unit cell. [At. mass of Cu = 63.5
a.m.u, Avogadro's constant NA = 6.02 x 1023].
Ans-6. Density = Mass of unit cell/Volume of unit cell
8.93 = (4 x 63.5)/(a3 x 6.02 x 1023).
a3 = 47.24 x 10-24
a = 3.6 x 10-8 cm. = 360 pm.
Q7. The edge length of NaCl unit cell is 500 pm. What is the density of NaCl in
g/cm3?
[NA = 6.02 x 1023, Na = 23.0, Cl = 35.5 a.m.u.]
Ans-7. d = (4 x 58.5)/[6.02 x 1023 x (500 x 10-10)3]
= 3.12 g/cm3.
Q.-8. Addition of CdCl2 to the crystal of AgCl will produce schottky
defect, but the same is not produced when NaCl crystal are added,
Why ?
Ans-8. => The replacement of one Ag+ ion with Cd+ ion necessitates the
removal of other Ag+ ion from the lattice to maintain the electrical
neutrality of the crystal.
=> But in case of NaCl, Na+ and Ag+ both are monovalent.
Q.9. Ferromagnetic and Ferrimagnetic substances become para magnetic
upon heating . Why ?
Ans-9. => The temperature at which they are changed into paramagnetic is
called curie temperature.
=> This is because the realignment of electrons spin or their magnetic moments which are now oriented in one particular direction.
Q.10. Explain the term 'dislocations' in relation to the crystal.
Ans-10. The defects which result from improper orientation of planes with
respect to one-another in the crystal are called dislocations.
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