The Solid State Quiz-3
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Q1.Which of the following is not a ferroelectric compound?:
Solution
(b) Factual statement
(b) Factual statement
Q2.The number of unit cells in 58.5 g of NaCl is nearly:
Solution
(c) 58.5 g NaCl=1 mol =6.02×10^23Na⨁Cl⊝ units One unit cell contains 4 Na⨁Cl⊝ units. Hence, number of the unit cell present = 6.02×10^23/4=1.5×10^23
(c) 58.5 g NaCl=1 mol =6.02×10^23Na⨁Cl⊝ units One unit cell contains 4 Na⨁Cl⊝ units. Hence, number of the unit cell present = 6.02×10^23/4=1.5×10^23
Q3. The density of an ionic compound (Mw=58.5) is 2.165 kg m-3 and the edge length of unit cell is 562 pm, then the closest distance between AB⊝ and Zeff of unit cell is
Solution
(a) ρ=Zeff*Mw/a3*NA ∴Zeff=4 (fcc-type structure) ∴For fcc, dA-B⊝=a2=5622=281 pm
(a) ρ=Zeff*Mw/a3*NA ∴Zeff=4 (fcc-type structure) ∴For fcc, dA-B⊝=a2=5622=281 pm
Q4. The interionic distance for cesium chloride crystal will be
Solution
(c) As CsCl is body-centered,d=sqrt(3)a/2
(c) As CsCl is body-centered,d=sqrt(3)a/2
Q5.The number of nearest neighbours and next nearest neighbours of an Na ion in a crystal of NaCl are, respectively,
Solution
(b) Na lies in OVs formed by Cl⊝ (Na touches six Cl⊝ ions) Na and Cl⊝ are not shown touching in the figure. Each atom shown is present at the face center of each cube Distance between two Na and Cl⊝=a2 Try to visualize two cubes exactly above these two cubes.Each atom shown is present at the edge center and body center of each cube Distance between two nearest Na=a2 Thus, the number of nearest neighbours of Na ion = 6 Cl⊝ ions The number of next nearest neighbours of Na ion = 12 Na ions Note: Next nearest neighbours are shown by number 1, 2, 3,…6 in Fig. b.Likeswise 6 next nearest neighbours of Na ions will be in the above two cube. Hence, total number of next nearest neightbours of Na ions=12 Na ions
(b) Na lies in OVs formed by Cl⊝ (Na touches six Cl⊝ ions) Na and Cl⊝ are not shown touching in the figure. Each atom shown is present at the face center of each cube Distance between two Na and Cl⊝=a2 Try to visualize two cubes exactly above these two cubes.Each atom shown is present at the edge center and body center of each cube Distance between two nearest Na=a2 Thus, the number of nearest neighbours of Na ion = 6 Cl⊝ ions The number of next nearest neighbours of Na ion = 12 Na ions Note: Next nearest neighbours are shown by number 1, 2, 3,…6 in Fig. b.Likeswise 6 next nearest neighbours of Na ions will be in the above two cube. Hence, total number of next nearest neightbours of Na ions=12 Na ions
Q6. A metal of density 7.5 × 103 kg m-3 has an fcc crystal structure with lattice parameter a = 400 pm. Calculate the number of unit cells present in 0.015 kg of the metal
Solution
(c) The volume available =0.0157.5×103 (Number of unit cells)400×10-123=0.0157.5×103 Number of unit cell =3.125×10^22
(c) The volume available =0.0157.5×103 (Number of unit cells)400×10-123=0.0157.5×103 Number of unit cell =3.125×10^22
Q7.If R is the radius of the octahedral voids and r is the radius of the atom in close packing, then r/R is equal to
Solution
(a) For OV, R/r i.e., r/r⊝or rvoid/r⊝=0.414 ∴r/R=1/0.414=2.41
(a) For OV, R/r i.e., r/r⊝or rvoid/r⊝=0.414 ∴r/R=1/0.414=2.41
Q8.In cubic ZnS (II-VI) compounds, if the radii of Zn and S atoms are 0.74 Å and 1.70 Å, the lattice parameter of cubic ZnS is
Solution
(b) rri.e.,rZnrS2-=0.74 Å1.70 Å=0.44 From radius ratio, it is expected that Zn2+ ion occupy OVs; however, the value of 0.44 is only slightly larger than rvoid/r=0.414 for OV. There is also some covalent character in the Zn2+-S2- interaction, which tends to shorten the interaction distance Note: Experimentally, it was found that Zn2+ ions occupy TVs ∴rZn2++rS2-=34a 0.74+1.70Å=34a⇒a=5.634 Å
(b) rri.e.,rZnrS2-=0.74 Å1.70 Å=0.44 From radius ratio, it is expected that Zn2+ ion occupy OVs; however, the value of 0.44 is only slightly larger than rvoid/r=0.414 for OV. There is also some covalent character in the Zn2+-S2- interaction, which tends to shorten the interaction distance Note: Experimentally, it was found that Zn2+ ions occupy TVs ∴rZn2++rS2-=34a 0.74+1.70Å=34a⇒a=5.634 Å
Q9.Which of the following is a ferroelectric compound?
Solution
(a) Factual statement
(a) Factual statement
Q10.Silver (atomic weight =108 g mol-1) has a density of 10.5 g cm-3. The number of silver atoms on a surface of area 10-12m2 can be expressed in scientific notation as y10x. The value of x is
Solution
(c) Volume of one mole of silver atoms =10810.5cm3/mol Volume of one silver atom =10810.516.022 ×1023 cm3 So, 43r3=10810.516.022 ×1023=1.708×10-23 r3=0.407×10-23cm3=0.407×10-29m3 Area of each silver atom, r2=0.407×10-29 m32/3 So, number of silver atoms in given area =10-120.407×10-29 m32/3=108π×2 =1.6×107=y10x So, x=7
(c) Volume of one mole of silver atoms =10810.5cm3/mol Volume of one silver atom =10810.516.022 ×1023 cm3 So, 43r3=10810.516.022 ×1023=1.708×10-23 r3=0.407×10-23cm3=0.407×10-29m3 Area of each silver atom, r2=0.407×10-29 m32/3 So, number of silver atoms in given area =10-120.407×10-29 m32/3=108π×2 =1.6×107=y10x So, x=7
