Solid State MCQ Questions & Answers in Physical Chemistry | Chemistry
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151.
A solid has a $$'bcc'$$ structure. If the distance of nearest approach between two atoms is $$1.73\mathop {\text{A}}\limits^{\text{o}} ,$$ the edge length of the cell is
A
314.20$$\,pm$$
B
1.41$$\,pm$$
C
200$$\,pm$$
D
216$$\,pm$$
Answer :
200$$\,pm$$
For $$bcc$$ structure
$$d = \frac{{\sqrt 3 a}}{2}$$
where $$d =$$ distance between two atoms $$a =$$ edge length
$$\eqalign{
& 1.73 = \frac{{\sqrt 3 }}{2}a \cr
& a = \frac{{2 \times 1.73}}{{\sqrt 3 }} \cr
& = 2\mathop {\text{A}}\limits^{\text{o}} = 200pm \cr} $$
152.
Which one of the following statements about packing in solids is incorrect ?
A
Coordination number in $$bcc$$ mode of packing is 8
B
Coordination number in $$hcp$$ mode of packing is 12
C
Void space in $$hcp$$ mode of packing is 32%
D
Void space is $$ccp$$ mode of packing is 26%
Answer :
Void space in $$hcp$$ mode of packing is 32%
The $$hcp$$ arrangement of atoms occupies 74% of the available space and thus has 26% vacant space.
153.
A crystal is formed by two elements $$X$$ and $$Y$$ in cubic structure. $$X$$ atoms are at the corners of a cube while $$Y$$ atoms are at the face centre. The formula of the compound will be
A
$$XY$$
B
$$X{Y_2}$$
C
$${X_2}{Y_3}$$
D
$$X{Y_3}$$
Answer :
$$X{Y_3}$$
No. of $$X$$ atoms ( at the corners ) $$ = \frac{1}{8} \times 8 = 1$$
No. of $$Y$$ atoms $$\left( {fcc} \right) = 6 \times \frac{1}{2} = 3$$
Hence, the formula is $$X{Y_3}.$$
154.
Pure silicon and germanium behave as
A
conductors
B
semiconductors
C
insulators
D
piezoelectric crystals
Answer :
insulators
Pure $$Si$$ and $$Ge$$ are insulators because of the absence of any hole or free electron. They show semiconductor
behaviour only after doping.
155.
Which of the following is not a crystalline solid ?
A
$$KCl$$
B
$$CsCl$$
C
$$Glass$$
D
$$Rhombic$$ $$S$$
Answer :
$$Glass$$
Glass is an amorphous solid.
156.
The $${r_ + }/{r_ - }$$ ratio of $$ZnS$$ is 0.402. Pick out the
incorrect statements from the following?
A
$$ZnS$$ is 4 : 4 coordination compound.
B
$$ZnS$$ does not crystallize in rock salt type lattice because $${r_ + }/{r_ - }$$ is too small to avoid overlapping of $${S^{2 - }}\,ions.$$
C
$$Z{n^{2 + }}\,ion$$ is too small to fit precisely into the octahedral voids of $${S^{2 - }}\,ions.$$
D
$$Z{n^{2 + }}\,ion$$ is too large to fit into the octahedral voids of $${S^{2 - }}\,ions.$$
Answer :
$$Z{n^{2 + }}\,ion$$ is too large to fit into the octahedral voids of $${S^{2 - }}\,ions.$$
$$Z{n^{2 + }}\,ion$$ is too small to fit into the octahedral voids of $${S^{2 - }}\,ions.$$
157.
A metal has a $$fcc$$ lattice. The edge length of the unit cell is $$404\,pm.$$ The density of the metal is $$2.72\,g\,c{m^{ - 3}}.$$ The molar mass of the metal is : ( $${N_A}$$ Avogadro’s constant $$ = 6.02 \times {10^{23}}\,mo{l^{ - 1}}$$ )
A
$$30\,g\,mo{l^{ - 1}}$$
B
$$27\,g\,mo{l^{ - 1}}$$
C
$$20\,g\,mo{l^{ - 1}}$$
D
$$40\,g\,mo{l^{ - 1}}$$
Answer :
$$27\,g\,mo{l^{ - 1}}$$
Density is given by
$$d = \frac{{Z \times M}}{{{N_A}{a^3}}};$$ where $$Z = $$ number of formula units present in unit cell, which is 4 for $$fcc$$ $$a=$$ edge length of unit cell. $$M=$$ Molecular mass
$$\eqalign{
& 2.72 = \frac{{4 \times M}}{{6.02 \times {{10}^{23}} \times {{\left( {404 \times {{10}^{ - 10}}} \right)}^3}}}\,\,\left( {\because \,\,1pm = {{10}^{ - 10}}cm} \right) \cr
& M = \frac{{2.72 \times 6.02 \times {{\left( {404} \right)}^3}}}{{4 \times {{10}^7}}} \cr
& = 26.99 = 27\,gm\,\,mol{e^{ - 1}} \cr} $$
158.
Which of the following is true about the charge acquired by $$p$$ - type semiconductors ?
A
Positive
B
Neutral
C
Negative
D
Depends on concentration of $$p$$ impurity
Answer :
Neutral
Both $$p$$ - type and $$n$$ - type semiconductors are electrically neutral because each atom of impurity added is neutral in itself containing equal number of protons and electrons.
159.
Which of the following is not true about the ionic solids ?
A
Bigger ions form the close packed structure.
B
Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size.
C
Occupation of all the voids is not necessary.
D
The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.
Answer :
The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids.
The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ion present at the lattice points. As we know the radii of octahedral or tetrahedral void is related to radii of atom $$(r)$$ as radius of octahedral void $$ = 0.414\,R$$
and radius of tetrahedral void $$ = 0.225\,R$$
where, $$R =$$ radius of the sphere in close packing.
160.
The empty space in the body centred cubic lattice is
A
68%
B
52.4%
C
47.6%
D
32%
Answer :
32%
Packing fraction of $$bcc$$ = 68%
Empty space = 100 - 68 = 32%