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Published on: 02/06/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
Write a note on Frenkel defect.
2.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. calculate the edge length of unit cell.
3.
What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure?
4.
Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?
5.
KF crystallizes in fcc structure like sodium chloride. Calculate the distance between K+ and F− in KF. (given : density of KF is 248 g cm-3)
1.
(i) Frenkel defect arises due to the dislocation of ions from its crystal lattice.
(ii) The ion which is missing from the lattice point occupies an interstitial position.
(iii) This defect is shown by ionic solids in which cation and anion differ in size.
(iv) Unlike Schottky defect, this defect does not affect the density of the crystal.
For example AgBr, in this case, small Ag+ ion leaves its normal site and occupies an interstitial position.
2.
For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4} \)
\(\therefore a =\frac{4 r}{\sqrt{2}} \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm
3.
1. The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. The coordination number of atoms in a bcc structure is '8'.
4.
Number of X type atoms in the unit cell \(=8 \times \frac{1}{8}=1\)
Number of Y type atoms in the unit cell \(=1 \times \frac{1}{1}=1\)
Hence the formula is XY (or) X1Y1
5.
\(\text { Density }(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\mathrm{n}=4, \mathrm{M}=\text { Molar mass of } \mathrm{KF}=58.1 \mathrm{~g} / \mathrm{mol} \)
\(\rho=2.48 \mathrm{~g} \mathrm{~cm}^{-3} \)
\(\mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} \)
\(a^{3} =\frac{n M}{\rho N_{A}}=\frac{4 \times 58.1}{2.48 \times 6.023 \times 10^{23}} \)
\(a^{3} =15.55 \times 10^{-23} \)
\(a^{3} =0.1555 \times 10^{-21} \)
\(a =\sqrt[3]{0.1555 \times 10^{-21}} \)
\(a =0.5375 \times 10^{-7} \mathrm{~cm}=5.375 \times 10^{-8} \mathrm{~cm}=537.5 \mathrm{pm} \)
\(d =\frac{a}{\sqrt{2}}(\text { for fcc }) [\therefore r = \frac{a\sqrt{2}}{4}]\)
\(=\frac{537.5}{1.414}=380.13 \mathrm{pm}\)
\(\therefore\) The distance between K+ and F- in KF = 380.13 pm
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