InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Iodine molecules are held in the crystals lattice by ……..A. London forcesB. dipole-dipole interractionsC. covalent bondsD. coulombic forces |
|
Answer» Correct Answer - C |
|
| 2. |
Iodine molecules are held in the crystal lattice by:A. London forcesB. Dipole -dippole interactionsC. Covvalent bondsD. Coulombic forces |
|
Answer» Correct Answer - A Iodine molecules are a class of non -polar molecular solid in which constituents molecules are held together by London or dispersion forces , these solid are doft and nono- conductor of electricity. |
|
| 3. |
Which of the following compounds are not isomorphous? a. Copper sulphate and zine sulphate b. Zinc sulphate and manganeous sulphate c. Calcium carbonte and ferrous sulphate d. Zine sulphate and ferrous sulphate |
|
Answer» `(a, c)` a. `CuSO_(4)*5H_(2)O` and `ZnSO_(4)*7H_(2)O` are not ismorphous. Different moles of `H_(2)O`. b. `ZnSO_(4)*7H_(2)O` and `MnSO_(4)*7H_(2)O` are isomorphous with `7H_(2)O` in both compounds. c. `CaCO_(3)` and `FeSO_(4)*7H_(2)O` are not isomorphous, since the moles of `H_(2)O` are not same. d. `ZnSO_(4)*7H_(2)O` and `FeSO_(4)*7H_(2)O` are isomorphous with same number of moles `H_(2)O`. |
|
| 4. |
How many unit cells are present in a cube-shaped crystal of NaCl of mass 1.00g?A. `2.57xx10^(21)` unit cellsB. `5.14xx10^(21)` unit cellsC. `1.28xx10^(21)` unit cellsD. `1.71xx10^(21)` unit cells |
|
Answer» Correct Answer - 1 `58.5grArr"N"` formula units `1grarr(N)/(58.5)` formula units (b) 4 formula units `rarr` 1 unit cell `(N)/(58.5)rarr(N)/(58.xx4)1"unit cell"=(6xx10^(23))/(234)` |
|
| 5. |
Which of the following statements is/are correct?A. Dislocation of ion from lattice site to intersitial site is called Frenkel defect.B. Missing of +ve and -ve ions from their respective postion producing a parir of holes is called Sschorttky defect.C. The presence of ions in the vacant intersitial sites along with lattice point is called interstital defect.D. Non-stoichiometric NaCl is yellow solid. |
| Answer» Non-stoichionetric NaCl has F-centers due to anion vacancy defect. | |
| 6. |
which of the following is not the characteristic of ionic solids?A. Very low value of electrical conductivity in the molten stateB. Brittle natureC. Very strong forces of interactionsD. Anisotropic nature |
|
Answer» Correct Answer - A Ionic solids are hard and brittle because of the presence of very strong forces of attraction. Because of highly ordered arrangement of particles, these are anisotropic in nature. In molten state or in aqueous solution, these are good conductor of electricity because of the presence of free ions. |
|
| 7. |
For which of the following cases, answer is `4`A. Coordination number of `Zn^(2+)` in `Z` inc blendeB. Number of body diagonal planes in a cubeC. Formula units in rock salt structureD. Formula units in `CsCl` structure |
|
Answer» Correct Answer - A::C a. `CN` of `Zn^(2+)` ion in `ZnS = 4`. b. Number of body diagonal planes in a cube `= 6`. c. Formula unit is rock salt structure `= 4`. d. Formula unit in `CsCl` structure `= 1`. |
|
| 8. |
The existence of a substance in more the one solid modifications is known as//or .Any compound having more than two cystal structures is calledA. IsomorphismB. Polymorp hismC. NeomorphismD. Crystallomorphism |
|
Answer» Correct Answer - 2 Different crystal structures of same solid substance are polymorphs |
|
| 9. |
Ionic solids are characterised byA. Good conductivity in solid stateB. High vapour pressureC. Low melting pointD. Solubility in polar solvents |
|
Answer» Correct Answer - 4 Ionic substances are soluble in polar solvents |
|
| 10. |
Aluminium metal has a density of `2.72 g cm^(-3)` and crystallizes in a cubic lattice with an edge of `404 pm`. Which of the following is correct?A. It forms bcc unit cellB. It forms fcc unit cellC. Its `CN = 8`D. Its `CN` is `6` |
|
Answer» Correct Answer - B `rho = Z_(eff xx Aw of Al)/(N_(a) xx a^(3)), Z_eff = ?` `2.72 g cm^(-3) = (Z_(eff) xx 27)/(6.023 xx 10^(23) xx (404 xx 10^(-10))^(3)` `:. Z_(eff) = 4` Hence, it forms `fcc` structure with `CN = 12`. |
|
| 11. |
An octahedron hasA. `8` cornersB. `8` facesC. `8` edgesD. `8` edges |
|
Answer» Correct Answer - B::D An octahedron has `8` faces and `12` edges. |
|
| 12. |
For the spinel structure `(MgAl_(2)O_(4))`, the correct statement is/areA. 50% Ovs are occupied by ions.B. `Al^(3+)` is equally distributed in TVs and Ovs.C. Oxide ions occupy ccp lattice.D. 12.5% TVs are occupied by ions. |
|
Answer» For spinel structure `O^(2-)` ions form fcc arrangement. `therefore ` Number of `O^(2-)` ions =4. Number of TV=8, Number of OV=4 Number of `Mg^(2+)` ions `=(1)/(8)xxTV =(1)/(8)xx8=1` Number of `Al^(3+)` ions `=(1)/(2)xxOV=(1)/(2)xx4=2` So, 50%OVs are occuped and 1/8th , i.e., 12.5% TVs are occupied. |
|
| 13. |
Aluminium metal has a density of `2.72 g cm^(-3)` and crystallizes in a cubic lattice with an edge of `404` pm. Which is//are correct?A. It forms an fcc unit cell.B. It forms a bcc unit cell.C. Its coordination number is `8`.D. Its coordination number is `12`. |
|
Answer» Correct Answer - A::D Use any relation: `rho=(Z_(eff)xxAw)/(N_(A)xxa^(3))` or `rho=(Z_(eff)xxAwxx1.67xx10^(-24)g)/(a^(3))` `:. 2.72 g cm^(-3) = (Z_(eff)xxAwxx1.67xx10^(-24)g)/((404xx10^(-10))^(3)cm^(3))` `:. Z_(eff) approx = 4` `a. Z_(eff) = 4`, means fcc structure. d. `CN` of fcc structure `= 12` |
|
| 14. |
In the mineral, spinel, having the formula ` MgAl_(2)O_(4)` oxide ions ar arranged , in the cubic close packing, ` Mg^(2+)` ions occupy the tetrahedrel voids while `Al^(3+)` ions occupy the octahedral voids. (i) What precnetage of tetrahedral voids is occupied by ` Mg^(2+) ` ions ? (ii) What precentage of octahedral voids is occupied by ` Al^(3+)` ions ? |
|
Answer» According to the formula ` MgAl_(2)O_(4)` , if there are 4 oxide ions, there will be ` 1 Mg^(2+)` ions and `2Al^(3+) ` ions. But if the ` 4O^(2-)` ions are in cp arrangement, there will be 4 octahedral voids and 8 tetrahedral voids. Thus ` 1 Mg^(2+)` ions is presnet in one of the 8 tetrahedral voids. %e of tetrahedral voids occupied by ` Mg^(2+) = 1/8 xx 100 = 12.5%` Similarly ` 2Al^(3+)` ions are presnet in two octahedral voids out of 4 available . % of octahedral voids occupied by ` Al^(3+) = 2/4 xx 100 xx 50 %` |
|
| 15. |
For the spinel structure `(MgAl_(2)O_(4))`, the correct statement is//areA. `50% OV_(s)` are occupied by ions.B. `Al^(3+)` is equally distributed in `TVs` and `OVs`.C. Oxide ions occupy ccp lattice.D. `12.5% TVs` are occupied by ions. |
|
Answer» Correct Answer - A::C::D For spinel structure, `O^(2-)` ions from fcc arrangement. `:. `Number of `O^(2-)` ions `= 4`. Number of `TV = 8`, Number of `OV = 4` Number of `Mg^(2+)` ions `= 1/8 = TV = 1/8 xx 8 = 1` Number of `Al^(3+)` ions `= 1/2 xx OV = 1/2 xx 4 = 2` So, `50% OVs` are occupied and `1//8` lth, ie., `12.5% TV` occupied. |
|
| 16. |
Which of the following statements is/are correct?A. Dislocation of ion from lattice site to interstitial site is called Frenkel defect.B. Missing of `+ve` and `-ve` ions from their respective position producing a pair of holes is called Schottky defect.C. The presence of ions in the vacent interstital sites along with lattice point is called interrstital defect.D. Non-stoichiometric `NaCl` is yellow solid. |
| Answer» Correct Answer - A::B::C::D | |
| 17. |
If the radius of anion is `0.20 nm`, the maximum radius of cation which can be filleed in respective voids is correctly matched inA. `r_(o+) = 0.0828 nm` for tetrahedral voidB. `r_(o+) = 0.045 nm` for triangular voidC. `r_(o+) = 0.1464 nm` for tetrahedral voidD. None of the above. |
|
Answer» Correct Answer - A::B::C `r_(o+) // r_(ɵ) = 0.732, 0414`, and `0.225` maximum for `OV, TV`. and angular triangular defect. |
|
| 18. |
If radius of anion is `0.20nm`, the maximum radius of cations which can be filled in respective voids are correctly matched in `:`A. `r^(+)=0.045nm` triangular voidB. `r^(+)=0.0828nm` for tetrahedral voidC. `r^(+)=0.1464nm` for octahedral voidsD. none of these |
|
Answer» Correct Answer - a,b,c `(r^(+))/(r^(-))` maximum for triangular, tetrahedral and octahedral are `0.225,0.414` and `0.732` respectively. |
|
| 19. |
A spinel is an important class of oxide consisting two types of metal ions withs the oxide ions arranged in ccp layers . The normal spinel has one -eight of the tetrahendral holes occupied by one type of metal ions and one- half of the octaherdral holes occupied by another type of metal ion. Such a spine is formed by `Mg^(2+), Al^(3+) and O^(2-)` . The netutrality of the crystal is benig maintained. The fromule of the spinel is :A. `mg_(2)AlO_(2)`B. `MgAl_(2)O_(4)`C. `Mg_(3)Al_(3)O_(6)`D. none of the above |
| Answer» Correct Answer - B | |
| 20. |
A solid made up of ions of `A` and `B` possess edge length of unit cell of `0.5664nm` has four formula units. Among the two ions, the smaller one occupy the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass `9.712xx10^(-23)g.` The solide `AB` is supposed to show `:`A. SchottKy defectB. Freenkel defectC. non - stoichimetric defectD. none of the above |
| Answer» Correct Answer - A | |
| 21. |
A solid made up of ions of `A` and `B` possess edge length of unit cell of `0.5664nm` has four formula units. Among the two ions, the smaller one occupy the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass `9.712xx10^(-23)g.` The structural arrangement of `AB` is `:`A. cubicB. octahedralC. tetrahedralD. triangular |
|
Answer» Correct Answer - b `(r_(A^(+)))/(r_(B^(-)))=(0.826)/(1.994)=0.414,` `i.e.,` octahedral arrangement |
|
| 22. |
A solid made up of ions of `A` and `B` possess edge length of unit cell of `0.5664nm` has four formula units. Among the two ions, the smaller one occupy the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass `9.712xx10^(-23)g.` Which statement is wrong about solid is `:`A. Cation fills the voidB. The solid has octahedral voids filled by `A^(+)` ionsC. Anion fills the voidD. Each anion has six octahedral holes |
|
Answer» Correct Answer - c Anions being larger forms space lattice. |
|
| 23. |
A solid made up of ions of `A` and `B` possess edge length of unit cell of `0.5664nm` has four formula units. Among the two ions, the smaller one occupy the interstitial void and the larger ions occupy the space lattice with ccp type of arrangement. One molecule of solid has mass `9.712xx10^(-23)g.` The solide `AB` is supposed to show `:`A. Schottky defectB. Frenkel defectC. non`-` stoichiometric defectD. none of these |
|
Answer» Correct Answer - A `AB` shows Schottky defect due to missing of one `A^(+)` and one `B^(-)` ion like `NaCl`. |
|
| 24. |
What makes a glass different from a soild such as quartz ? Under what conditons could quartz be converted into glass ? |
| Answer» Glass is an amorphous solid in which the constituent particels `(SiO_(4)` tetrahdral) have only a short range order and there is no long range order. In quartz, the constituent particles ` (SiO_(4)` tetrahdra) have short range as will as long range order. on melting quartz and then cooling i repidly , it is converted into glass. | |
| 25. |
CsCl has cubic sructare, its density is `3.99 g cm^(-3) ` . What is the distance between ` Cs^(+) and Cl^(-)` ions ? ( At. Mass of Cs= 133) |
|
Answer» CsCl has BCC struture. It has noe formula unit in the unit cell . So Z=1 ` p = ( Z xx M)/ (a^(3) xx N_(0) ) or a^(3) = (Z xx M)/( p xx N_(0)) = (1 xx (133 + 35.5) "g mol" ^(-1))/ (3.99 " g cm"^(-3) xx 6.02 xx 10^(23) " mol" ^(-1)) = 70.15 xx 10^(-24) cm^(3)` ` a = ( 70. 15)^(1//3) xx 10 ^(-8) cm^(3) = ( 70.15) ^(1//3) xx 10^(2) " pm" " " ( 1"pm" =10 ^(-10)" cm")` ` = 4.124 xx 10^(2) " pm" = 412.4 "pm"` For BCC structure, Interionic distance (d) ` = (sqrt3a)/2= 1.732/2 xx 412.4 = 357 ` pm (see solved sample problem 2, page 1/34) ( To solve ` (70.15)^(1//3) ` , put ` x = ( 70.15)^(1//3)` , Then ` log x = 1/3 log 70.15 = 1/3 xx 1.8460 = 0.6153` x = Antilog 0.6153 = 4.124 |
|
| 26. |
The `CsCl` has cubic structure of `Cl^(-)` ions in which `Cs^(+)` in which `Cs^(+)` ion is present in the body centre of the cube. It density is `3.99g cm^(-3)`. The radius of `Cs^(+)` ion if the radius of `Cl^(-)` ion is `180 p m` is `:`A. `180.6p m`B. `276.8p mC. `176.8p m`D. `280.6 p m` |
|
Answer» Correct Answer - c The unit cell of `CsCl` has cubic arrangement of `Cl^(-)` ions and `Cs^(+)` ion is present in the body centre of the cube. Therefore, the unit cell contains one `Cs^(+)` and one `Cl^(-)` ion, `i.e.,z=1` Molar mass of `CsCl=168.5g mol^(-1)` Now Density`rho=(zxxm)/(a^(3)xxN_(0))` or `a^(3)=(1xx168.5)/(3.99xx6.023xx10^(23))` `=7.02xx10^(-23)cm^(3)` `r_(Cs^(+))+r_(Cl^(-))=356.8p m` `r_(Cs^(+))=356.8-180=176.8 p m` |
|
| 27. |
The `CsCl` has cubic structure of `Cl^(-)` ions in which `Cs^(+)` in which `Cs^(+)` ion is present in the body centre of the cube. It density is `3.99g cm^(-3)`. Then length of the edge of unit cell is `:`A. `312p m`B. `412p m`C. `436 p m`D. `536p m` |
|
Answer» Correct Answer - b The unit cell of `CsCl` has cubic arrangement of `Cl^(-)` ions and `Cs^(+)` ion is present in the body centre of the cube. Therefore, the unit cell contains one `Cs^(+)` and one `Cl^(-)` ion, `i.e.,z=1` Molar mass of `CsCl=168.5g mol^(-1)` Now Density`rho=(zxxm)/(a^(3)xxN_(0))` or `a^(3)=(1xx168.5)/(3.99xx6.023xx10^(23))` `=7.02xx10^(-23)cm^(3)` Let the length of edge of unit cell `=a` `:.a^(3)=7.02xx10^(-23)cm^(-3)=70.2xx10^(-24)cm^(-3)` or `a=4.12xx10^(-8)cm=412p m` |
|
| 28. |
`CsCl` has cubic structure. Its density is `3.99 g cm^(-3)`. What is the distance between `Cs^(o+)` and `Cl^(Θ)` ions? (Atomic mass of `Cs = 133`) |
|
Answer» `CsCl` has `bc c` structure , so `Z_(eff) = 1` `rho = (Z_(eff) xx Mw)/(rho^(3) xx 10^(-30) xx N_(A))` or `a^(3) = (Z_(eff) xx Mw)/(a^(3) xx 10^(-30) xx N_(A))` `= (1 xx (133 + 35.5))/(3.99 xx 10^(-30) xx 6.02 xx 10^(23))` `= 70.15 xx 10^(6)` `a = (70.15)^(1//3) xx 10^(2)` `= 4.12 xx 10^(2) "pm" = 412` pm Interionic distance `= (sqrt3a)/(2)` `= (1.732)/(2) xx 412` `=356.8` |
|
| 29. |
In an `LiI` crystal, `I^(ɵ)` ions from a cubical close-packed arrangement, a nd `Li^(o+)` ion occupy octahedral holes. What is the relationship between the edge length of the unit cells and radii of the `I^(ɵ)` ions if `a = 60` "pm". |
|
Answer» Cubical close packing imlied that the unit cell of the `I^(ɵ)` ion is fcc, and for that (from Table `1.12` and Fig. `1.67`) `r_(o+) + r_(ɵ) = a//2 = (600)/(2) = 300` pm From `OV`, `(r_(o+))/(r_(ɵ)) = 0.414` `r_(o+) = 0.414 xx r_(ɵ)` `= 0.414 xx 300 = 124.2` pm |
|
| 30. |
The maximum ra dius of sphere that can be fitted in the octahedral hole of cubical closed packing of sphere of raius `r` isA. `0.155 r_(A)`B. `0.125 r_(A)`C. `0.414 r_(A)`D. `0.732 r_(A)` |
|
Answer» Correct Answer - c For octahedral void `(r_(B))/(r_(A)) = 0.414` or, `r_(B) = 0.414 r_(A)` |
|
| 31. |
The maximum radius of sphere that can be fitted in the octahedral hole of cubical closed packing of sphere of radius I isA. 0.732IB. 0.414 IC. 0.225ID. 0.155I |
|
Answer» Correct Answer - B r=0.414R. |
|
| 32. |
The unit cube length for `LiCl` (`NaCl` structure) is `5.14 Å`. Assuming anion-anion contact, calculate the ionic radius for chloride ion. |
|
Answer» Interionic distance of `LiCl` `= (5.14)/(2) = 2.57 Å` `:. BC = sqrt(AB^(2) + AC^(2)) = sqrt((2.57)^(2) + (2.57)^(2)) = 3.63` Radius of `Cl^(Θ)` ion `= (1)/(2) xx 3.63 = 1.81 Å` |
|
| 33. |
KCI cyrstallizes in the saem type of lattice as does NaCl. Given that `(r_(Na^(o+))/(r_(Cl_(-)))=0.5` and `(r_(Na^(o+)))/(r_(K^(o+)))=0.7` Calculate the ratio of side of the unit cell for KCl to that for NaCl.A. 0.6B. 1.1C. 2.2D. 3.3 |
|
Answer» NaCl crystallizes in the face-centered cubic unit cell, such that `r_(Na^(o+))+r_(Cl^(-))=a//2` where a is the edge length of unit cell. Now since `r_(Na^(o+))//r_(Cl^(-))=0.5` and `r_(Na^(o+))//r_(K^(o+))=0.7` we will have `(r_(Na^(o+))+r_(Cl^(-)))/(r_(Cl^(-)))=(r_(K^(o+)))/(r_(Na^(o+))//0.5)=(0.5)/(r_(Na^(o+))//r_(K^(o+)))(0.5)/(0.7)` [Add 1 to both sides] or `(r_(K^(o+)))/(r_(Na^(o+))+r_(Cl^(-)))=(1.2)/(0.7)xx(1)/(1.5)` or `(a_(KCl)//2)/(a_(NaCl)//2)=(1.2)/(0.7xx1.5)` or `(a_(KCl))/(a_(NaCl))=(1.2)/(1.05)=1.143`. |
|
| 34. |
`LiI` occurs as cubical closed packing. If the edge lenth of unit cell is `624` pm, determine the ionic radii of `Li^(o+)` and `I^(Θ)` inos. |
|
Answer» The cubical closed packing has a face-centred cubic unit cell. `I^(Θ)` occupy the corners and the face centres. These ions touch each other along the face diagonal of the cube. Hence `4r_(I^(Θ)) = sqrt2a` `r_(I^(Θ)) = (a)/(sqrt2a)= (624 "pm")/(2(1.414))= 220.65` pm Now along the edge , we will have `I^(Θ) Li^(o+) I^(Θ)` arrangement, where `I^(Θ)` ions are at the corners and `Li^(o+)` ions at the centre of the edge (octahedral void). Since in cloest packing they touch each other, we will have `2r_(I^(Θ)) + 2r_(Li^(o+)) = a` or `r_(Li^(o+))= (a)/(2) - r_(I^(Θ)) = (624 "pm")/(2) - 220.65 pm = 91.35` pm |
|
| 35. |
Casium may be considered to form interpentrating simple primitice cubic crystal. The edge length of unit cell is `412` pm. Determine a. The density of `CsCl`. b. The inoic radius of `Cs^(o+)` if the ionic radius of `Cl^(Θ)` is `181` pm. Given: `Aw (Cs) = 133 g mol^(-1)` |
|
Answer» The interpenetrating simple primitive cubic crystals mean the unit cell is body-centred where `Cl^(Θ)` ions occupy corners and `Cs^(o+)` ion and one `Cl^(Θ)` ion (or one molecule of `CsCl`) per unit cell. From the expression, `rho = (Z_(eff) xx Aw)/(a^(3) xx N_(A))` `= (1)/((412 xx 10^(-12) m)^(3)) ((168.5 g mol^(-1))/(6.023 xx 10^(23) mol^(-1)))` `= 0.4 xx 10^(7) g m^(-3) -= 4.0 g cm^(-3)` Now since `Cs^(o+)` ions touch the two chlorides along the cross diagonal of the cube, we will have `2r_(c) + 2r_(a) = sqrt3a` or `r_(c) = ((sqrt3)/(2)) a - r_(a) = ((sqrt3)/(2)) (412 "pm") - (181 "pm")` `= 175.8` pm |
|
| 36. |
`KCI` crystallizes in the same type of lacttice as does `NaCl`. Given that `(r_(Na^(o+)))/(r_(Cl^(Θ))) = 0.5` and `(r_(Na^(o+)))/(r_(K^(o+))) = 0.7` Calculate (a) the ratio of side of the unit cell for `KCl` to that for `NaCl`, and (b) the ratio of density of `NaCl` to that `KCl`. |
|
Answer» `NaCl` crystallizes in the face-centred cubic unit cell, such that `r_(Na^(o+)) + r_(Cl^(Θ)) = a//2` where `a` is the edge length of unit cell. Now since `r_(Na^(o+)//r_(Cl^(Θ)) = 0.5` and `r_(Na^(o+))//r_(K^(o+)) = 0.7`, we will have `r_(Na^(o+) + r_(Cl^(Θ)))/(r_(Cl^(Θ))) = 1.5, [r_(Na^(o+))/r_(Cl^(Θ)) + 1= 0.5 + 1 ]` and `r_(Na^(o+))/r_(Cl^(Θ)) = r_(K^(o+))/(r_(Cl^(Θ))//0.5)= (0.5)/(r_(Na^(o+))//r_(K^(o+))) = (0.5)/(0.7)` [Add `1` to both sides] or ` (r_(K^(o+)) + r_(Cl^(Θ)))/(r_(Na^(o+)) + r_(Cl^(Θ))) = (1.2)/(0.7) xx (1)/(1.5)` or `or (a_(KCl)//2)/(a_(NaCl)//2) = (1.2)/(0.7 xx 1.5)` or `(a_(KCl))/(a_(NaCl)) = (1.2)/(1.05 )= 1.143` Now since `rho = (Z_(eff))/(a^(3)) ((Mw)/(N_(A)))` We will have `rho_(NaCl)/(rho_(KCl)) = ((a_(KCl))/(a_(NaCl)))^(3) (Mw_(NaCl))/(Mw_(KCl))` `= (1.143)^(3) ((58.5)/(74.5)) = 1.172` |
|
| 37. |
A compound made up of elements A and B crystallises in the cubic structure . Atoms A are present at the corners and alternate face centres whereas atoms B are present at the edge centres as well as body centre . What is the formula of the compound ?A. `A_2B`B. ABC. `AB_2`D. `A_2B_2` |
|
Answer» Calculation of number of atoms of A in unit cell Contribution from alternate faces `=2xx1/2=1` Contribution from alternates faces `=2xx1/2=1` `therefore` Total number of atoms of A in unit cell = 1+1=2 Calculation of number of atoms of B in unit cell Contribution from edge centre `=12xx1/4=3` Contribution from body centre `=1xx1=1` `therefore` Total number of atoms of B in unit cell = 3+1=4 Formula of the compound is `A_2B_4" or " AB_2` |
|
| 38. |
An element crystallises in `f.c.c.` lattice having edge length `400 p m`. Calculate the maximum diameter, which can be placed in interstitial sites without disturbing the structure. |
|
Answer» For a fcc structure, edge length (a) is related to the radius (r) `r=(a)/(2sqrt2)=((400"pm")/(2sqrt2))=141.42"pm"` In octahedral (fcc) void R = 0.414 r `=0.414xx141.42` Diameter `=2R=2xx0.414xx141.42=117.096"pm."` |
|
| 39. |
KCl crystallizes int the same type of lattic as done NaCl . Given that `r_(Na^(+))//r_(Cl^(-)) = 0.50` and `r_(Na^(+))//r_(K^(+)) = 0.70` , Calcualte the ratio of the side of the unit cell for KCl to that for NaCl: |
|
Answer» `(r_(Na^(+)))/(r_(Cl^(-)))=0.55, (r_(Na^(+)))/(r_(Cl^(-)))+1=1.55` `(r_(K^(+)))/(r_(Cl^(-)))=0.74, (r_(K^(+)))/(r_(Cl^(-)))+1=1.74` Dividing (ii) by (i), `(r_(K^(+))+r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=(1.74)/(1.55)=1.122` |
|
| 40. |
Look ato the sodium choride structure in the figure and use it to calculate `(r^(+))/(r^(-))` |
|
Answer» Along an edge, `2r^(+)+2r^(-)=a` `r^(+)=((a)/(2)-r^(-))` Along a diagonal, `4r^(-)=asqrt2` `therefore" "a=(4r^(-))/(sqrt2)=2sqrt2r^(-)` `therefore" "r^(+)=(2sqrt2r^(-))/(2)-r^(-)` `r^(+)=r^(-)[sqrt2-1]` `(r^(+))/(r^(-))=sqrt2-1=1.414-1=0.414.` |
|
| 41. |
In the laboratory , sodium choride is made by burning the sodium in the atmosphere of cholrine which is yellow in colour .The cause of yellow colour isA. Presence of `Na^(+)` ions in the crystal latticeB. Presence of `Cl^(-)` ions in the crystal latticeC. Presence of electron in the crystal latticeD. Presence of face centered cubic crystal lattice |
|
Answer» Correct Answer - c Yellow colour on heating `NaCI` in the presence of Na is due to presence of electron in ation vacancies (f-centres). |
|
| 42. |
What type of crystal defect is indicated in the diagram given below `{:(Na^(o+),Cl^(ɵ),Na^(o+),Cl^(ɵ),Na^(o+),Cl^(ɵ)),(Cl^(ɵ),square,Cl^(ɵ),Na^(o+), square, Cl^(ɵ)),(Na^(o+),Cl^(ɵ),square,Cl^(ɵ),Na^(o+),Cl^(ɵ)),(Cl^(ɵ),Na^(o+),Cl^(ɵ),Na^(o+),square,Na^(o+)):}`A. Frenkle defectB. Schottky defectC. Interstitial defectD. Frenkle defect and Schottky defect |
|
Answer» Correct Answer - b Equal no , of `Na^(+) `and `CI^(-)` is missing completely `rArr ` schattky defect. |
|
| 43. |
What type of crystal defect is indicated in the diagram given below `{:(Na^(o+),Cl^(ɵ),Na^(o+),Cl^(ɵ),Na^(o+),Cl^(ɵ)),(Cl^(ɵ),square,Cl^(ɵ),Na^(o+), square, Cl^(ɵ)),(Na^(o+),Cl^(ɵ),square,Cl^(ɵ),Na^(o+),Cl^(ɵ)),(Cl^(ɵ),Na^(o+),Cl^(ɵ),Na^(o+),square,Na^(o+)):}`A. Frenkel and Schottky defectsB. Schottky defectC. Interstitial defectD. Frenkel defect |
|
Answer» Correct Answer - B These defect are produced when one `+ve` and one `-ve` ion are missing from their respective position removed. |
|
| 44. |
How many unit cell are present in a cubic-shaped ideal crystal of `NaCl` of mass `1.0 g`?A. `1.28xx10^(21)` unit cellsB. `1.71xx10^(21)` unit cellsC. `2.57xx10^(21)` unit cellsD. `5.14xx10^(21)`unit cells |
|
Answer» Correct Answer - C `n=4` for `f.c.c.,` Also no. of atoms in `1 g NaCl=(6.-23xx10^(23))/(58.5)` No. of unit cell present in `1g NaCl` `=(6.023xx10^(23))/(58.5xx4)` `=2.57xx10^(21)` unit cell |
|
| 45. |
How many unit cell are present in a cubic-shaped ideal crystal of `NaCl` of mass `1.0 g`?A. `1.28 xx 10^(21)`B. `1.71 xx 10^(21)`C. `2.57 xx 10^(21)`D. `5.14 xx 10^(21)` |
|
Answer» Correct Answer - C For fcc, `Z_(eff) = 4//"unit cell"` `Mw` of `NaCl = 58.5 g mol^(-1)` Number of atoms in `1.0 g NaCl = (6 xx 10^(23))/(58.5)` Number of unit cells in `0.1 g NaCl` `= (6 xx 10^(23))/(58.5 xx 4) = 2.57 xx 10^(21) "unit cells"` |
|
| 46. |
The edge of unit of `FC C Xe` crystal is 620 pm .The radius of Xe atom isA. `189.37` pmB. `209.87` pmC. `219.25` pmD. `235.16` pm |
|
Answer» Correct Answer - C `r=(a)/(2sqrt(2))=(620)/(2sqrt(2))=219.25` pm |
|
| 47. |
How many unit cell are present in a cubic-shaped ideal crystal of `NaCl` of mass `1.0 g`?A. `2.57xx10^(21)` unit cellsB. `5.14xx10^(21)` unit cellsC. `1.28xx10^(21)` unit cellsD. `1.71xx10^(21)` unit cells |
|
Answer» Correct Answer - A `(1)/(58.5)xx6.023xx10^(23)=1.029xx10^(22)` A unit cell contains `4Na^(+)` ion and `4Cl^(-)` ions `therefore` Unit cell `=(1.029xx10^(22))/(4)=2.57xx10^(21)` unit cell. |
|
| 48. |
The edge of unit of `FC C Xe` crystal is 620 pm .The radius of Xe atom isA. 219.25pmB. 235.16pmC. 189.37pmD. 209.87pm |
|
Answer» Correct Answer - A `r=(a)/(2sqrt(2)):r=(620)/(2sqrt(2))`=219.25 pm. |
|
| 49. |
The interionic distance for cesium chloride crystal will beA. `a`B. `a//2`C. `sqrt3a//2`D. `2a//sqrt3` |
|
Answer» Correct Answer - C As `CsCl` is body-centred, `d = sqrt(3)a//2`. |
|
| 50. |
The interionic distance for cesium chloride crystal will beA. aB. `(a)/(2)`C. `sqrt(3a)/(2)`D. `(2a)/sqrt(3)` |
|
Answer» Correct Answer - C Ad CsCl is body-centred, `d=sqrt(3)a//2`. |
|