Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Chemistry Class 12

Created by: Mohit Rajpoot

Class 12 : Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

The document Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes is a part of the Class 12 Course Chemistry Class 12.
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SOLVED SUBJECTIVE PROBLEMS

Example 1. A face-centred cubic solid of an element A has a largest sized guest atom B at the body centre octahedral hole if insertion of B doesn't affect the original unit cell dimension; determine the packing fraction of the solid.
Sol. In the given solid, there is one B and four A per unit cell.

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Also, under the condition of largest possible size of B, it will be in contact of A present at the face centres only and the following relationship will exist:
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes and 2(rA + rB) = a
solving,
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Now, packing fraction
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Example 2. An element A has BCC structure and another guest atom B, of largest possible size are present at each edge centres of unit cell of A but without disturbing the original unit cell dimension. Determining the void percentage of this solid. 

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Sol. Since B is closer to corner than to face centre, in close contact,
2(rA + rB) = a
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
In a unit cell, there are two A and 12 × = 3B
Packing fraction (f) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
⇒ Void space = 31.6%

Example 3. An element A has a BCC structure and another guest atoms B, of largest possible size, are present at the face centres, but without disturbing the unit cell dimension. Determine the packing fraction of this solid.

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Sol. In the above solid, face centres are closer to body centre, therefore, the relationship.
2(rA rB) = 0
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Now, packing fraction (f) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes = 0.684

Example 4. Calcium has face-centred cubic lattice and radius of calcium atom is 195.6 picometre. Determine the number of Ca atoms present on surfaces of a mm3 block of calcium metal assuming that atoms in the closest packing calcium metal assuming that atoms are in the closest packing. 
Sol. In FCC, the relation is

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
= 553.24 × 10-9 mm.
⇒ Area of a face of unit cell = a3 = 3.06 × 10-13 mm2
⇒ Surface area of metal block = 6 mm2
⇒ Total number of faces of unit cells present on surface
= Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes = 1.96 × 1013
Each face contributes two Ca atoms on surface as shown in diagram.
⇒ Total number of Ca atoms present on surface
= 2 × 1.96 × 1013 = 3.92 × 1013

Example 5. An atom crystallizes in hexagonal closed packed arrangement. Determine dimensions (radius and length) of a large cylindrical atom that can be accommodated in the centre of HCP, in terms of radius of host atom. 
Sol. The cylinder will pass through centre of middle layer and will lie between the face centres.

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Therefore,
Height of cylinder (h) = height of hexagon (h) = 2r
Since, in HCP : Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes,
where r = radius of atoms.
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
r = 1.266 r
Also, if R is the radius of cylinder, then in the case of closet contact :
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
⇒ R = 0.155 r,
⇒ h = 1.266 r

Example 6. A uniform cylindrical, polymer molecule crystallizes in body centred cubic array. Determine the packing fraction of this polymer in solid state assuming that molecules are in their closest contact. 
Sol. The arrangement of molecules can be represented as follows :

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Here,
 Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Packing fraction (f) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

= Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Example 7. Show the following arrays of atoms on a plane in an atomic FCC.

(A)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(B)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(C)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(D)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Sol. 
(A)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(B)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(C)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

(D)
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Example 8. Cesium bromide crystallizes in cubic arrangement like CsCl. Given the ionic radii of Cs  and Br- are 1.88 Å and 1.82 Å respectively, determine the packing fraction. 
Sol. Here Cs  ion is at body centre and Br- at the corners, therefore the relationship is
2(r++ r-) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes 
= 2 (1.88 1.82) Å,
⇒ a = 4.27 Å
Packing fraction (f) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev NotesPractice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Example 9. Titanium metal crystallizes in a BCC arrangement and radius of an atom is 142 picometre. Determine the atomic weight of titanium if the density is 16.6 g/cm3. Also determine the number of unit cells present in a 5 cm block of titanium metals. 
Sol. In BCC arrangement of atoms, the relationship between edge length radius of the atom is
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Density (r) = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Where, N = Number of atoms per unit cell,
M = Molar mass, NA = Avogadro's number
V = Volume of unit cell
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Here, the radius is taken in cm unit since, density is in g/cm3 unit.
Solving : M = 176. Also mass of 5 cm3 block = 5 × 16.6 = 83 g

⇒ Number of atoms is 5 cm3 block Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

⇒ Number of unit cells   Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes = Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes = 1.42 × 1023

Example 10. Copper metal crystallizes in face-centred cubic arrangement and surface of adjacent atoms along the edge of the unit cell are 106 picometres apart. Determine the density of metal. Atomic mass of copper metal is 63.5 u. 
Sol. According to the given information, a view of one face of the unit cell will be as shown below.
In FCC; 4√2a

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

The given distance is a - 2r = a - Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes

Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
⇒ a = 362 pm = 3.62 × 10-8 cm
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes = 8.89 g/cm3

 

Example 11. Potassium crystallizes in "body centred cubic" arrangement and the surfaces of the adjacent atoms along the edges of unit cells are 71.4 picometre apart. Determine the density of metal. Atomic weight of K = 39. ?
Sol. In BCC, the relationship between edge length and radius of atom is
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
Also, given a - 2r = 71.4 × 10-10 cm
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
 ⇒ 71.4 × 10-10 cm
⇒ a = 5.329 × 10-10 cm
Practice Questions(Solved) - Solid State, Class 12, Chemistry | EduRev Notes
= 0.855 g/cm3

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