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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.
| 1051. |
Which of the following statements is not ture ?A. Paramagnetic substances are weakly attracted by magnetic fieldB. Ferromagnetic substances cannot be magnetised permanentlyC. The domains in antiferromagnetic substances are oppositely oriented with respect to each otherD. Pairing of electrons cancels their magnetic moment in the diamagnetic substances. |
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Answer» Correct Answer - B Ferromagnetic substances can be magnetised permanently. |
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| 1052. |
If the number of close packed sphere be N, then the number of octahedral and tetrahedral voids generated respectively are A and B that refers toA. 2N, NB. 3N, NC. 3N, ND. N, 2N |
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Answer» Correct Answer - D Octahedral voids = number of lattice point 2 x octahedral voids - tetrahedral voids N, 2N are the voids . |
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| 1053. |
In a `C CP` lattice of `X and Y` atoms are present at the corners while `Y` atoms are at face centeres .Then the formula of the compound would be if one of the atoms from a corner is replaced by `Z` atoms (also monovalent)?A. `X_(7)Y_(2) Z_(2)`B. `X_(7)Y_(24) Z`C. `X_(24)Y_(7) Z`D. `XY_(24) Z` |
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Answer» Correct Answer - b `X_((7)/(8)) Y_(3) Z_((1)/(8)) rArr X_(7) Y_(24) Z` |
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| 1054. |
The number of close neighbours in a body-centred cubic unti cell of monoatomic substance is,A. 8B. 6C. 4D. 2 |
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Answer» Correct Answer - A bcc has a coordination number of 8. |
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| 1055. |
The arranegement `ABC, ABC, ABC …….` is referred asA. Cubic colose packingB. Hexagonal close packingC. Teragonal close packingD. Octabedral close packing |
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Answer» Correct Answer - a It represents ccp arrangement. |
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| 1056. |
The arranegement `ABC, ABC, ABC …….` is referred asA. octahedral close packingB. Hexagonal close packingC. Tetragonal close packingD. Cubic close packing. |
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Answer» Correct Answer - D It represents ccp arrangement. |
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| 1057. |
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` isA. `11.87 Å`B. `5.634 Å`C. `5.14 Å`D. `2.97 Å` |
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Answer» Correct Answer - B `(r_(o+))/(r_(ɵ)) = (i.e., (r_(Zn^(o+)))/(r_(S^(2-)))) = (1.74 Å)/(1.70 Å) = 0.44` From radius ratio, it is expected that `Zn^(2+)` ion occupy `OVs`, however, the value of `0.44` is only slightly larger than `r_("void")//r_(ɵ) = 0.414` for `OV`. There is alos some covalent character in the `Zn^(2+) -S^(2-)` interation, which tends to shortent the interatom distance. `:. (r_(Zn^(2+)) + r_(S^(2-))) = (sqrt(3))/(4)a` `(0.74 + 1.70)Å = (sqrt(3))/(4)a rArr a = 5.634 Å` |
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| 1058. |
What colour is observed when ZnO is heated ?A. YellowB. VioletC. GreenD. Blue |
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Answer» Correct Answer - B Upon heating white colour of ZnO changes to yellow |
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| 1059. |
In solid `NH_(3)` each `NH_(3)` molecule has six other `NH_(3)` molecules as nearest neighbouts. `DeltaH` of sublimation of `NH_(3)` at the melting point is `30.8 kJ mol^(-1)` and the estimated `DeltaH` of sublimation in the absence of `H-` bonding is `14.4kJ mol^(-1)`. What is the strength of a hydrogen bond in solid ammonia ?A. `6.5kJ mol^(-1)`B. `16.5kJ mol^(-1)`C. `5.5kJ mol^(-1)`D. `4.5kJ mol^(-1)` |
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Answer» Correct Answer - C Total strength of `H-` bond =`20.8-14.3` `=16.5kJ mol^(-1)` There are 6 nearest neighbours, but each `H-` bond involves 2 molecules. Thus, total` H-` bond `=(6)/(2)=3` `:. ` `H-` bond energy `=(16.5)/(3)=5.5kJ mol^(-1)` |
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| 1060. |
The melting point of `RbBr` is `682^(@)C`, while that of `NaF` is `988^(@)C` . The principla reason that melting point of `NaF` is much higher than that of `RbBr` is that `:`A. the two crystal are not isomorphousB. the molar mass of `NaF` is smaller thean that of `RbBr`C. the internuclear distance `r_(c)+r_(a)` is greater for `RbBr` than for `NaF`.D. the bond in `RbBr` has more covalent character than the bond in `NaF`. |
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Answer» Correct Answer - c This leads to stronger coulombic forces of attractions in `NaF`. |
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| 1061. |
The melting point of `RbBr` is `682^(@)C`, while that of `NaF` is `988^(@)C` . The principla reason that melting point of `NaF` is much higher than that of `RbBr` is that `:`A. the two crystals are not isomorphousB. the molar mass of NaF is smaller than that of RbBrC. the bond in RbBr has more covalent character than the bond in NaFD. the internuclear distance `(r_(c)+r_(a))` is greater for RbBr than for NaF. |
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Answer» Correct Answer - D Lattice energy is inversely proportional to inter ionic distance and so is the melting point of a crystalline solid. |
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| 1062. |
The ionic radii of `A^(+) and B^(-)` ions are ` 0.98 xx 10^(-10) m and 1.81 xx 10^(-10) m`. The coordination number of each ion in AB isA. 2B. 6C. 4D. 8 |
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Answer» Correct Answer - A Radius ratio , ` r_(+)/r_(-) = ( 0.98 xx 10^(-10))/( 1.81 xx 10^(-10)) = 0.541` It liess in the range 0.414 - 0.732. Hence, coordination number of each ion will be 6, as in the case of NaCl. |
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| 1063. |
Why does KCl turn violet on heating in the potassium vapours ? |
| Answer» Due to the formation of F-centres. | |
| 1064. |
Why is Frenkel defect not found in pure alkali metal halides ? |
| Answer» Because of high co-ordination number. | |
| 1065. |
Why is frenkel defect not found in pure alkali halides ? |
| Answer» This is due to the fact that alkali metal have large size which cannot fit into the intersitional sites. | |
| 1066. |
If the unit cell of a minearl has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahderal holes occupied by aluminium ions and n fraction of tetrahdral holes occupied by magnesium ions, m and n, respectively areA. `1/2 , 1/8`B. `1,1/4`C. `1/2, 1/2`D. `1/4,1/8` |
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Answer» Correct Answer - A In ccp lattice , Z =4 No. of O-atoms per unit cell = `4 ( O^(2-))` No.of octahedral voids = 4 and No. of tetrahedral voids = 8. As m fraction of octahedral voids is occupied by `Al^(3+)` ions. Therefore, `Al^(3+)` ions present = 4 m. similarly , ` Mg^(2+)` ions= 8 n. Hence, formula of the mineral is ` Al_(4m) Mg_(8n) O_(4)` . As total charge on the compound is zero, hence. 4 m ( +3) + 8 n ( +2) + 4( -2) =0 or 12 m + 16 n -8 =0 Substituting the given values of m and n m, equation is satisfied only when ` m = 1/2 and n = 1/8` ` ( as 12 xx 1/2 + 16 xx 1/8 -8 =0 )` |
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| 1067. |
Structure of a mixed oxide is cubic close-packed (c.c.p) the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. the formula of the oxide is :A. `A_(2)B_(3)O_(4)`B. `AB_(2)O_(2)`C. `ABO_(2)`D. `A_(2)BO_(2)` |
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Answer» Correct Answer - A Suppose number of ` O^(2-)` ions in close packing = n No. of octahedral voids = n No. of tetradral voids = 2n Divalent metal ions A present = `(2n)/4 = n/2 ` Monovalent metal ions B present =n Ratio `A :B : O^(2-) = n/2 : n :n = 1/2 :1:1 ` Hence, formula = ` AB_(2)O_(2)` |
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| 1068. |
which of the following represents correct order of conductivity in solids ?A. `K_("metals")gtgtK_("Insulators")ltK_("semiconductors")`B. `K_("metals")ltltK_("insulators")ltK_("semiconductors")`C. `K_("metals")ltK_("insulators")gtK_("semiconductors")=` zeroD. `K_("metals")ltK_("insulators")gtK_("semiconductors")cancel=` zero |
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Answer» Correct Answer - A Conductivity of metal, insulator and semiconductors can be represented in teht term of k. (Kappa) which depednds upon energygap beween valence band and conduction band. |
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| 1069. |
which of the following represents correct order of conductivity in solids ?A. `K_("metals")gtgtK_("insulators")ltK_("semiconductors")`B. `K_("metals")ltltK_("insulators")ltK_("semiconductors")`C. `K_("metals")~=K_("semiconductors")gtK_("insulators")="zero"`D. `K_("metals")ltK_("insulators")gtK_("semiconductors")ne"zero"` |
| Answer» Correct Answer - A | |
| 1070. |
Let `Mg TiO_(3)` exists in pervoskite structure. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the denstiy of unit cell if the radius of `Mg^(2+)` is `0.7 Å` and the corner ions are touching each other. [Given atomic mass of Mg = 24, Ti = 48] |
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Answer» Correct Answer - `~~65 g//cm^(3)` No. of `Mg^(2+)` per unit cell = 8 [At corners] `xx (1)/(8) = 1` No. of Ti per unit cell = 1 [body center ] `xx (1)/(1) = 1` No of O per unit cell = 6 [Face center] `xx (1)/(2) = 3` So formula `= MgTiO_(3)` Atom are removed along face diagonal No. of `Mg^(2+) = 6`[At corner] `xx (1)/(8) = (6)/(8) = (3)/(4)` No. of Ti per unit cell = 1 [Body center] `xx (1)/(1) = 1` No. of O per unit cell = 5[Face center] `xx (1)/(2) = (5)/(2)` So formula of compound `= Mg_((3)/(4)) TiO_((5)/(2))` Formular mass `= 24 xx (3)/(4) + 48 + 16 xx (5)/(2) = 18 + 48 + 40 = 106` amu As corner ion are touching so `= a = 2 r_(Mg^(2+)) = 2 xx 0.7 = 1.4 Å` `d = ("mass")/("Volume") = (106 xx 1.76 xx 10^(-24))/((1.4)^(3) xx 10^(-24)) g//cm^(3) = 64.5 g//cm^(3) ~~ 65 g//cm^(3)` |
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| 1071. |
Lithium crystallizes as body centered cubic crystals. If the length of the side of unit cell is 350=pm, the atomic radius of lithium is:A. 303.1pmB. 606.2pmC. 151.5pmD. 123.7pm |
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Answer» Correct Answer - C |
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| 1072. |
Lithium forms body centred cubic structrue. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will beA. 75 pmB. 300 pmC. 240 pmD. 152 pm |
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Answer» Correct Answer - D For BCC sturcture, `sqrt(3)a=4r` `r=(sqrt(3))/(4)a=(sqrt(3))/(4)xx351=152` pm |
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| 1073. |
Copper crystalline in a face centred cubic lattice with a unit cell length of `361p m` .What is the radius of copper atom in p m?A. `157`B. `181`C. `108`D. `128` |
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Answer» Correct Answer - 4 For a face-centred cubic unit cell, we have `4r=sqrt(2a)` or `r=sqrt(2)/(4)a` `=(1.414)/(4)(361p m)` `=127.6p m` |
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| 1074. |
Lithium metal crystallizes in a body centred cubic crystals. If the length of the side of the unit cell of lithium is `351p m` the atomic radius of the lithium will beA. `151.8p m`B. `75.5p m`C. `300.5p m`D. `240.8p m` |
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Answer» Correct Answer - 1 For a body centred cubic unit cell we have `4r=sqrt(3)a` or `r=(sqrt(3))/(4)` `=(1.732)/(4)(351p m)` `=151.98p m` |
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| 1075. |
Which of the following metals has different geometry compared to those of the others ?A. FeB. CoC. NiD. Cu |
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Answer» Correct Answer - A Fe possesses hcp arrangement while all others ccp arrangement |
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| 1076. |
Lithium forms body centred cube structure .The length of the side of its unirt cell is 351 pm Atomic radius of the lithium will beA. 75 pmB. 300 pmC. 240 pmD. 152 pm |
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Answer» Correct Answer - D For b.c.c structure, `sqrt(3)a=4r` `r=sqrt(3)/4a=sqrt(3)/4xx351=152"pm"` |
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| 1077. |
Which of the following statements is not correct ?A. The number of carbon atoms in the unit cell of diamond is 4B. The number of Bravais lattics in which a crystal can be catagorized is 14C. The fraction of the total volume occupied by the atoms in a primitive cell is `0.48`D. Molecular solids are generally volatile |
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Answer» Correct Answer - C In a primitive cell or simple cubic unit cell, the fraction of occupied space or packing efficiency is `0.52` and not `0.48` |
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| 1078. |
Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?A. 157B. 181C. 108D. 128 |
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Answer» Correct Answer - D Copper crystallises in a faces centred cubic lattice, Atomic radius `(r)=a/(2sqrt(2))` `=361/(2xx1.41)=127.6 approx 128 "pm"` |
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| 1079. |
select the correct statement (s)A. Co-ordinational no. of `Cs^(+)` and `Cl^(-)` are 8, 8 in CsCl crystalB. If radius ratio `(r_(c)//r_(a))lt0.225` then shap of compound must be linearC. If radius `(r_(c)//r_(a))` lies between 0.414 to 0.732 then shope of ionic compound may be square planner `(Ex. PtCl_(4)^(2-))`D. If radius ratio is less than than 0.155 then shape of compound is linaer |
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Answer» Correct Answer - A::C::D |
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| 1080. |
select the correct statement (s)A. CsCl change to NaCl structure on heatingB. NaCl changes to CsCl structure on applying pressureC. Co-ordination number decreses on applyping pressureD. Co-ordination number increses on heating |
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Answer» Correct Answer - A::B |
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| 1081. |
select the correct statement (s)A. A NaCl type AB crystal lattice can be interperted to be made up of two individual fcc type uint lattice of `A^(+)` and `B^(-)` fused togther is such a manner that the corner of one unit lattic becomes the edge centre of the otherB. In a fcc unit , cell the body centre is an octahedral voidC. In an scc lattice, there can be no octahedral voidD. In an scc lattice,the body centre is the octaedral |
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Answer» Correct Answer - A::B::C |
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| 1082. |
In a AB unit cell (Rock salt type) assuming `A^(+)` forming fcc : |
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Answer» Correct Answer - A::B::C |
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| 1083. |
For each of the following substances, identify the intermolecular force or forces that predominate. Using your knowledge of the relative strength of the vaious forces, rank the substance in order of their normal boiling points. `Al_(2)O_(3), F_(2), H_(2)O, Br_(2), IC l, NaCl`A. `F_(2) lt Br_(2) lt IC l`B. `H_(2)O lt NaCl lt Al_(2)O_(3)`C. `ICl lt H_(2)O`D. `H_(2)O lt ICl` |
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Answer» Correct Answer - A::B::C `A_(2)O_(3)-` ionic `Br_(2)`- vanderwaal `F_(2)-` vanderwaal ICl - dipole dipole `H_(2)O` - dipole - dipole (H - bonding) NaCl - ionic `F_(2) lt Br_(2) lt ICl lt H_(2)O lt NaCl lt Al_(2)O_(3)` |
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| 1084. |
Which one of the following crystal does not exhibit Frenkel defect?A. AgBrB. AgClC. CsClD. ZnS |
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Answer» Correct Answer - C |
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| 1085. |
In the unit cell of KCl (NaCl type), Cl^(-) ions constitute ccp and `K^(+)` ion fall into the octahedral holes. These holes are:A. One at the centre and 6 at the centres of the facesB. one at the centre and 12 at the centres of the edgesC. 8 at the centres of 8 small cubes forming the unit cellD. none of these |
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Answer» Correct Answer - B |
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| 1086. |
What fraction of the surface of a crystal of `Cd` at `T = 298 K` consists of vacancies? Assume that the energy needed to form a vacancy `= 0.5 Delta_("sub")H^(Θ)`. For `Cd(s), Delta_("sub")H^(Θ) = 112.0 kJ mol^(-1)`.A. `1.5 xx 10^(-10)`B. `3 xx 10^(-5)`C. `2 xx 10^(-6)`D. `3 xx 10^(-2)` |
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Answer» Correct Answer - A The number of vacancies (or Schottky defects ) (n) is given by. `n=Ne^(-E//2KT)` or `(n)/(N)=e^(-E//2KT)` But fraction of the surface of a crystal vacancy is given by `(n)/(N)=e^(-E//RT)` `="exp"[(-(0.5)(112.0xx10^(3)J"mol"^(-1)))/(8.314JK^(-1)"mol"^(-1)(298K))]=1.5xx10^(-10)` |
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| 1087. |
With which one of the following element silicon should be depend so as to give p- type semiconductor?A. AsB. SeC. BD. Ge |
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Answer» Correct Answer - c Silicon should be doped with group `13` element to give a p- type semiconductor |
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| 1088. |
What fraction of the surface of a crystal of `Cd` at `T = 298 K` consists of vacancies? Assume that the energy needed to form a vacancy `= 0.5 Delta_("sub")H^(Θ)`. For `Cd(s), Delta_("sub")H^(Θ) = 112.0 kJ mol^(-1)`. |
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Answer» The number of vacancies (or Schottky defects) `(n)` is given by. `n = Ne^(-E//2KT)` or `(n)/(N) = e^(-E//2KT)` But fraction of the surface of a crystal vacancy is given by `(n)/(N) = e^(E//2KT)` `= exp[(-(0.5)(112.0xx10^(3)Jmol^(-1)))/((8.314JK^(-1)mol^(-1))(298 K))] = 1.5 xx 10^(-10)` |
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| 1089. |
The correct statement in the following isA. The ionic crystal of `AgBr` has schottky defectB. The unit cell having crystal parameters, `a = b ne c, alpha = beta = 90^(@) , gamma = 120^(@)` is hexagonalC. In ions compound having frenkel defect the ratio `(gamma_(+))/(gamma_(-))` is highD. The coordination number of `Na^(+)` ion in `NaCI` is `4` |
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Answer» Correct Answer - b A crystal system is haxagojnal if the unit cells have `a= b!= c`axial ratio `alpha = beta = 90^(@) , gamma = 120^(@)` axial angles. |
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| 1090. |
The pyknometric density of `NaCl` crystal is `2.165 xx 10^(3 kg m^(-3)` while its `X`-ray density is `2.178 xx 10^(-3) kg m^(-3)`. The fraction of unoccupied sites in `NaCl` crystal is a. `5.96` b. `5.96 xx 10^(-2)` c. `5.96 xx 10^(-1)` d. `5.96 xx 10^(-3)`A. `5.99`B. `5.96 xx 10^(-2)`C. `5.96 xx 10^(-1)`D. `5.96 xx 10^(-3)` |
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Answer» Correct Answer - d Mole volume from pyknometric density `V_(p) = (M)/(2.165 xx 10^(3)) m^(3)` Molar volume from X-rays density `V_(x) = (M)/(2.178 xx 10^(3)) m^(3)` `:.` volume unoccupies`= (M)/(10^(3)) ((1)/(2.165)-(1)/(2.178))` `= (0.013 xx M xx 10^(-3))/(2.165 xx 2.178 )m^(3)` `:.` fraction of volume occupied `= (0.013 xx M xx 10^(3))/(2.165 xx 2.175) //(M xx 10^(-3))/(2.165)` `= 5.96 xx 10^(-3)` |
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| 1091. |
The pyknometric density of `NaCl` crystal is `2.165 xx 10^(3 kg m^(-3)` while its `X`-ray density is `2.178 xx 10^(-3) kg m^(-3)`. The fraction of unoccupied sites in `NaCl` crystal is a. `5.96` b. `5.96 xx 10^(-2)` c. `5.96 xx 10^(-1)` d. `5.96 xx 10^(-3)`A. `5.96 xx 10^(-1)`B. `5.96`C. `5.96 xx 10^(-2)`D. `5.96 xx 10^(-3)` |
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Answer» Correct Answer - d Difference `= 2.178 xx 10^(3) - 2.165 xx 10^(3) = 0.013 xx 10^(3)` Fraction unoccupied `= (0.013 xx 10^(3))/(2.173 xx 10^(3)) = 5.96 xx 10^(-3)` |
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| 1092. |
The pyknometric density of `NaCl` crystal is `2.165 xx 10^(3 kg m^(-3)` while its `X`-ray density is `2.178 xx 10^(-3) kg m^(-3)`. The fraction of unoccupied sites in `NaCl` crystal is a. `5.96` b. `5.96 xx 10^(-2)` c. `5.96 xx 10^(-1)` d. `5.96 xx 10^(-3)`A. 5.96B. `5.96xx10^(-2)`C. `5.96xx10^(-1)`D. `5.96xx10^(-3)` |
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Answer» Molar volume from pykonometic density = `("weight of mass")/("density")=("Weight" (W))/(2.165xx10^(3))m^(3)` Molar volume from X-ray density= `(W)/(2.178xx10^(3))m^(3)` Volume unoccupied `=(W)/(10^(3))((1)/(2.165)-(1)/(2.178))m^(3)=(0.013Wxx10^(-3))/(2.165xx2.178)` `therefore` Fraction unoccupied `=((0.013Wxx10^(-3))/(2.165xx2.178))//((Wxx10^(-3))/(2.165))=5.96xx10^(-3)` |
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| 1093. |
The pyknometric density of `NaCl` crystal is `2.165 xx 10^(3 kg m^(-3)` while its `X`-ray density is `2.178 xx 10^(-3) kg m^(-3)`. The fraction of unoccupied sites in `NaCl` crystal is a. `5.96` b. `5.96 xx 10^(-2)` c. `5.96 xx 10^(-1)` d. `5.96 xx 10^(-3)` |
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Answer» d. Molar volume from pyknometric density `= ("Weight or mass")/("Volume") = ("Weight" (W))/(2.165 xx 10^(3))m^(3)` Molar volume from `X`-ray density `= (W)/(2.178 xx 10^(3))m^(3)` Volume unoccupied `= (W)/(10^(3)) ((1)/(2.165) - (1)/(2.178))m^(3)` `= (0.013 W xx 10^(-3))/(2.165 xx 2.178)` `:.` Fraction unoccupied `= ((0.013 W xx 10^(-3))/(2.165 xx 2.178))//((W xx 10^(23))/(2.165))` `= 5.96 xx 10^(-3)` |
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| 1094. |
Stacking of square close packed layers give rise to:A. bcc structureB. fcc structureC. Simple cubic structureD. hcp structur |
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Answer» Correct Answer - C |
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| 1095. |
Which is / are correct statement ?A. Packing fraction in 2D-hcp is 0.785B. Packing fraction in AAA….. Is 0.52C. Packing fraction in ABAB…… is 0.74D. Packing fraction in ABCABC……. Is 0.26 |
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Answer» Correct Answer - A::B::C::D |
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| 1096. |
In cubic system how many atoms arrangement exsit in nature ? |
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Answer» Correct Answer - 3 |
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| 1097. |
(A): Thermodynamically all solids possess a tendency to acquire defects (R): During defects the entropy of the system increases in solids.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true and R is not the correct expalantion of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - 1 | |
| 1098. |
The ionic radii of `A^(+)` and `B^(-)` are `1.7 Å` and `1.8 Å` respectively . Find the co-ordination number of `A^(+)` |
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Answer» Correct Answer - 8 |
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| 1099. |
AB is an ionic. If the ratio of ionic radii of `A^(+)` and `B^(-)` is 0.52. what is the co-ordination number of `B^(-)`?A. 2B. 3C. 6D. 8 |
| Answer» Correct Answer - 3 | |
| 1100. |
Which is / are correct statement ?A. In simple cubic close packed arrangement no octahedral void is present at edge centre.B. In fcc unit cell octaherdral void and tetrahedral void are vacant .C. Packing fraction : simple cubic cell `lt` bcc unit cell `lt ` fcc unit cell.D. Size of void : cubic `gt` octachedral void ` gt` trtrahedral void. |
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Answer» Correct Answer - A::B::C::D |
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