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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.
| 1201. |
Xenon crystallizes in the face-centred cubic lattice and the edge of the unit cell is `620` pm. What is the nearest neighbour distance and what is the redius of xenon atom? |
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Answer» Here `a = 620 "pm", d = ?, r = ?` For the face-centred cubic lattice, `d = (a)/(sqrt2) = (620 "pm")/(1.414) = 438.5 "pm"` `r = d//2 = 438.5//2 = 219.25`. Pm |
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| 1202. |
The total number of octahedral void (s) per atom present in a cubic close packed structure isA. 1B. 2C. 4D. 8 |
| Answer» Correct Answer - 3 | |
| 1203. |
`NH_(4)Cl` crystallises in a body centred cubic lattice with a unit cell distance of 387 pm. Calculate (a) the distance between the oppositely charged ions in the lattice. (b) the radius of `NH_(4)^(+)` ion if that of `Cl^(-)` ion is 181 pm. |
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Answer» Correct Answer - (a) 335.15 pm, (b) 154.15 pm (a) In a body cented cubic lattice Charged ions each other along the cross-diagonal of the cube. Therefore `2r^(+)+2r^(-)=3sqrta," "r^(+)+r^(-)=(sqrt3)/(2)a," "r^(+)+r^(-)=(sqrt3)/(2)("387pm")=335.15"pm"` (b) Now, `r^(+)(NH_(4)^(+))=335.15-r^(-)(Cl^(-))=335.15-r^(-)(181)=154.15"pm."` |
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| 1204. |
Lithium borohydride crystallizes in an orthorhombic system with 4 molecule per unit cell. The unit cell dimensions are `a=6.8Å,b=4.4Å` and `c=7.2Å`. If the molar mass is `21.76`, calculate density of crystal. |
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Answer» Correct Answer - `0.6709g cm^(-3);` |
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| 1205. |
Most crystals show good cleavage because their atoms ions or molecules areA. weakly bonded togetherB. strongly bonded togetherC. spherically symmetricalD. arranged in planes |
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Answer» Correct Answer - 4 A soild is said to be crystalline if its various constituent particle (i.e, atoms,ions, or molecules) are arranged in a defintie geometric pattern in three dimensional space so that there is short range as well as long range order of constituent particles, Due to well defined ordered structure in three dimensions, crystals give a clean cleavage on being cut with a sharp knife. |
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| 1206. |
When electrons are trapped into the crystalline anion vacancy the defect is known asA. schottky defectB. Stoichiometric defectC. Frenkel defectD. F-centres |
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Answer» Correct Answer - 4 The lattice sites containing the electrons trapped in the anion vacancies are called `F-`centres because they are responsible for imparting colour to the crystals (`F=`Farbe which is a German word for colour). |
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| 1207. |
For orthorhombic system axial ratios are `a nebnec` and the axial angle areA. `alpha=beta=gamma ne90^(@)`B. `alpha=beta=gamma=90^(@)`C. `alpha=gamma =90^(@),beta ne90^(@)`D. `alpha ne beta ne gamma ne 90^(@)` |
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Answer» Correct Answer - 2 In orthorhombic crystal system axial distances (or edge lengths) are releated as `a ne b ne c` and axial angle are releated as `alpha=beta=gamma=90^(@)` |
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| 1208. |
The metal calcium crystallises in fcc unit cell with a = 600 pm. Calculate the density of the metal if it contains `0.2%` Schottky defects.A. `2.05g//cc`B. `5g//cc`C. `20g//cc`D. `50g//cc` |
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Answer» Correct Answer - 1 Due to schottky defects no. of atoms decreases `z=4-4xx(0.2)/(100)=3.992, rho=(z.m)/(N_(0)a^(3))` |
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| 1209. |
In the fluorite structure the coordination number of `Ca^(2+)` ion isA. `4`B. `6`C. `8`D. `3` |
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Answer» Correct Answer - 3 In a fluorite structure `(CaF_(2))` every `Ca^(2+)` ion is surrounded by `8` fluoride ion trapped in tetrahedral voids.Thus the `CN` of `Ca^(2+)` ions is eight . |
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| 1210. |
Mention one property which is caused due to the presence of F-centre in a solid . |
| Answer» F-centre is responsible for the colour and paramagnetic behaviour of the solid. | |
| 1211. |
Niobium crystallizes in body-centred cubic structure. If the density is `8.55 g cm^(-3)`, calculate the atomic radius of niobium using its atomic mass `93 u`.A. 143pmB. 331pmC. 190pmD. 300pm |
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Answer» Correct Answer - 1 `a^(3)=3.31xx10^(-24)cc` `a=3.31xx10^(-8)cm=331pm` `r=(sqrt(3)a)/(4)=143pm` |
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| 1212. |
The density of copper metal is ` 8.95 " g cm"^(-3)` . If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body- centred cubic or a face- centred cubic? ( Given At. Mass of Cu= 63.54 g ` " mol"^(-1) and N_(A) = 6.02 xx 10^(23) " mol"^(-1))` |
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Answer» If copper atom were simple cubic, a =2 r=2 ` xx` 127.8 pm = 256. 6 pm ` Z = 1 : P = ( Z xx M)/( a^(3) xx N_(0)) = ( 1xx 63.54 " g mol"^(-1))/ (( 255.6 xx 10^(-10) "cm") ^(3) xx ( 6.02 xx 10^(23) mol^(-1)) ) = 6.34 " g cm" ^(-3)` Actual ` p = 8.95 " g cm"^(-3)` Hence, copper atom is not simple cubic. If copper atom were body - centred. ` a = (4r)/(sqrt3) = ( 4xx 127.8)/(1.732) " pm" = 295.15` pm ` Z=2 , p = ( Z xx M)/( a^(3) xx N_(0)) = ( 2 xx 63.54 " g mol"^(-1))/ ( ( 295. 15 xx 10^(-10) "cm") xx 6.02 xx 10^(23) "mol" ^(-1)) = 8.21 " g cm"^(-3)` Hence, copper atom is not body- centred. If copper atom were face - centred. `a= 2sqrt2r = 2xx 1.414 xx 127.8 "pm" = 361 . 4"pm"` `Z=4 , p = ( Z xx M)/(a^(3)xxN_(0)) = ( 4xx63.54" g mol"^(-1))/((361.4xx10^(-10) "cm") xx 6.02 xx 10^(23) "mol" ^(-1))= 8.94 " g cm" ^(-3)` Hence, copper is face - centred cubic. |
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| 1213. |
Niobium crystallize in a body centred cubic structure. If density is ` 8.55 " g cm"^(-3)` , calculate atomic radius of niobium, given that its atomic mass is 93n. |
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Answer» ` a^(3) = ( M xx Z)/(p xx N_(0) xx 10^(-30)) = ( 93 " g mol"^(-1) xxx2)/( 8.55 " g cm"^(-3) xx 6.02 xx 10^(23) "mol"^(-1) xx 10^(-30))= 3.61 xx 10^(7) = 36.1 xx 10^(7) = 36.1 xx 10^(6)` ` a = (36.1)^(1//3) xx 10^(2) "pm" = 3.304 xx 10^(2) "pm" = 330.4 "pm"` ` [x = (36.1)^(1//3) , log x = 1/3 log 36.1 = 1/3 xx 1.5575 = 0.519 or x = " antilog" 0.519 = 3.304`) For body - centred cubic , ` r= sqrt3/4 or = 0.433 a,= 0.433 xx 330.3 "pm" = 143.1 "pm" ` |
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| 1214. |
The density of `KBr` is `2.75 g cm^(-3)`. The length of the unit cell is `654` pm. Atomic mass of `K = 39, Br = 80`. Then what is true about the predicted nature of the solid?A. Face centred cubicB. simple cubic systemC. body centred cubic systemD. none of these |
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Answer» Correct Answer - A `Z=(d xxa^(3)xxN_(0)xx10^(-30))/(M)" "(becausea=cm)` `=(2.85xx(654)^(3)xx(6.02xx10^(23))xx10^(-30))/(39+80)=3.9~~4` Hence, it is face-centred. |
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| 1215. |
The density of KBr is ` 2.75 " g cm^(-3)` , The length of edge of the unit cell is 654 pm. Predict, the type of cubic lattice to which unit cell of KBr belongs ` ( N_(0) = 6.023 xx 10^(23) " mol" ^(-1) ` , At mass : K = 29 , Br = 80) |
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Answer» For cubic crystals ` p = ( Z xx M)/( a^(3) xxN_(0))` ` or Z = ( p xx a^(3) xx N_(0))/ M = (( 2.75 "g cm" ^(-3)) ( 654xx 10^(-10) "cm")^(3) xx ( 6.023 xx 10^(23) "mol"^(-1)))/ (( 39+ 80) "g mol" ^(-1)) = 3.89=4` Thus, there are four formula units of KBr, present per unit cell. Hence, it has face- centred cubic lattic. |
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| 1216. |
The size of the crystals reflect the conditions of growth, rather than the internal constitution of the crystal. NaCl crystals grown at the bottom of a beaker are squre plates whose thickness is never greater than half-their width. Thus the growth rate in either of the two horizontal orientations is twice that in the vertical direction. The angle between between two characteristic faces of octahedra and cube of NaCl crystal are `theta_(1)`and `theta_(2)` then correct relation isA. `theta_(1) gt theta_(2)`B. `theta_(1) lt theta_(2)`C. `theta_(1) = theta_(2)`D. `theta_(1) = 2theta_(2)` |
| Answer» Correct Answer - C | |
| 1217. |
The size of the crystals reflect the conditions of growth, rather than the internal constitution of the crystal. NaCl crystals grown at the bottom of a beaker are squre plates whose thickness is never greater than half-their width. Thus the growth rate in either of the two horizontal orientations is twice that in the vertical direction. The appearance of NaCl crystal grown, suspended in a solution of Urea isA. CubeB. OctahedraC. LinearD. Square |
| Answer» Correct Answer - B | |
| 1218. |
A match box exbibitsA. cubic geometryB. monoclinic geometryC. orthorhombic geometryD. tetragonal geometry |
| Answer» Correct Answer - 3 | |
| 1219. |
A match box exbibitsA. Cabic geometryB. Orthormabic geometryC. Monoclinetry geometryD. Tetragonal geometry |
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Answer» Correct Answer - b Orihorhombic gromatry has `a!= b!= c and a = beta = gamma = 90^(@)`.The sharp of match abeys this geometry. |
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| 1220. |
The intermetallic compound `LiAg` crystallizes in cubic lattice in which both lithium and silver have coordination number of `8`. The crystal class isA. simple cubicB. body centred cubicC. face-centred cubicD. none of these |
| Answer» Correct Answer - 2 | |
| 1221. |
The characteristic features of solids areA. Definite shapeB. Definite sizeC. Definite shape and sizeD. Definite shape, size and rigidity. |
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Answer» Correct Answer - D Solids have definite shape,size and rigidity. |
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| 1222. |
The solid `NaCI` is a bad conductoe of electricty sinceA. in solid `NaCI` there are no ionsB. solid `NaCI` is covalentC. in solid `NaCI` there is no velocity ofionsD. in solid `NaCI` there are no electrons |
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Answer» Correct Answer - c Solid `NaCI` is a had conductor of electricity become into are not free to move. |
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| 1223. |
In a tetragonal crystalA. `a = b = c, alpha = beta = 90^(@) != gamma`B. `alpha = beta = gamma = 90^(@), a = b != c`C. `alpha = beta = gamma = 90^(@), a != b != c`D. `alpha = beta = 90^(@), gamma = 120^(@), a = b != c` |
| Answer» Correct Answer - B | |
| 1224. |
The solid `NaCI` is a bad conductoe of electricty sinceA. in solid `NaCI` there are no ionsB. solid `NaCI` is covalentC. in solid `NaCI` there is no velocity of ionsD. in solid `NaCI` there are no electrons |
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Answer» Correct Answer - c Solid `NaCI` is a had conductor of electricity become into are not free to move |
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| 1225. |
Which of the following is a ferroelectric compound?A. `BaTiO_(3)`B. `K_(4)[Fe(CN)_(6)]`C. `Pb_(2)O_(3)`D. None of these |
| Answer» Correct Answer - A | |
| 1226. |
Which of the following is not a ferroelectric compound?A. Rochelle saltB. `K_(4)[Fe(CN)_(6)]`C. `BaTiO_(3)`D. `KH_(2)PO_(4)` |
| Answer» Correct Answer - B | |
| 1227. |
Which of the following is a ferroelectric compound?A. `BaTiO_(3)`B. `K_(4)[Fe(CN)_(6)]`C. `Pb_(2)O_(3)`D. `PbZrO_(3)` |
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Answer» Correct Answer - A `BaTiO_(3)` is a ferroelectric compound. |
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| 1228. |
Which of the following is a ferroelectric compound?A. `BaTiO_(3)`B. `K_(6)[Fe(CN)_(6)]`C. `Pb_(2)O_(3)`D. `PbZrO_(3)` |
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Answer» Correct Answer - a `BaTiO_(3)` is a ferroelectic compound |
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| 1229. |
The three states of matter are solid, liquid and gas .Which of the following statement is/are about them?A. Gases and liquid have viscosity as a common propertyB. The molecules in all the three states posses random transtational motionC. Gases cannot be converted into solids without passing through the liquid phaseD. Solids and liquid have vapour presure as a common property |
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Answer» Correct Answer - a Both gases and liquids posses fluidity and hence viscosity of molecules in the solid state does not have transtational motion. |
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| 1230. |
Iron exhibits bcc structure at room temperatutre .Above `900^(@)C` .it transitions to fcc structure .The ratio of density of ions at room temperature to that at `900^(@)C`(assuming molar mass and atomic , radii of ion remain s constant with temperature ) isA. `sqrt(3)/(sqrt(2))`B. `(4sqrt(3))/(3(sqrt(2))`C. `(3sqrt(3))/(4(sqrt(2))`D. `(1)/(2)` |
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Answer» Correct Answer - c In fcc atoms along face diagram touch each other i.e,`sqrt(2)a = 4r` In fcc atoms along body diagram touch each other i.e,`sqrt(3)a = 4r` BCC FCC `4r = sqrt(3) a 4r = sqrt(2) a` `4r = sqrt(3)a 4r = sqrt(2)a` `a = (4r)/(sqrt(3)) a= (4r)/(sqrt(2))` `(d_(BC C))/(d_(FC C))= ((Z_(BC C) xx M)/(N_(A)a^(3)))/((Z_(FC C) xx M)/(N_(Aa^(3)))) = ((Na((4r)/(sqrt(3)))^(3))/(4xxM))/(N_(A) xx((4r)/(sqrt(3)))^(3)) = (3)/(4) sqrt((3)/(2))` |
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| 1231. |
Iron exhibits `bcc` structure at roomj temperature. Above `9000^(@)C`, it transformers to `fcc` structure. The ratio of density of iron at room temperature to that at `900^(@)C` (assuming molar mass and atomic radius of iron remains constant with temperature) isA. `sqrt(3)/sqrt(2)`B. `(4sqrt(3))/(3sqrt(2))`C. `(3sqrt(3))/(4sqrt(2))`D. `(1)/(2)` |
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Answer» Correct Answer - C BCC FCC `4r=sqrt(3)a 4r=sqrt(2)a` `a=(4r)/sqrt(3) a=(4r)/sqrt(2)` `(d_(BC C))/(d_(FC C))=((Z_(BC C)xxM)/(N_(A)a^(3)))/((Z_(FC C)xxM)/(N_(A)a^(3)))=((2xxM)/(N_(A)((4r)/(sqrt(3)))^(3)))/((4xxM)/(N_(A)xx((4r)/(sqrt(2)))^(3)))=(3)/(4)sqrt((3)/(2))` |
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| 1232. |
Calculate the number `(n)` of atoms contained within `(a)` cubic cell, `(b)` a body `-` centred cubic cell, `(c)` a face `-` centred cubic cell. |
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Answer» Correct Answer - `(a)`1, `(b)`2,(c) 4;` |
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| 1233. |
The number of atoms contained in one face -centred cubic unit cell of monoatomic substance is :A. 1B. 2C. 4D. 6 |
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Answer» Correct Answer - C The number of atoms present in sc,fcc and bcc unit cell are 1,4,2 respectively. |
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| 1234. |
if a metal has a bcc crystal structure, the coordination number is 8,because `:`A. each atom touches four atoms in the layer above it, four in the layer below it and none in its own layerB. each other touches four atoms in the layer above it. four in the layer below it and one in its own layerC. two atoms touch four atoms in the layer above them, four in the layer below them, and none in their own layer.D. each atom touches eight atoms in the layer above it, eight in the layer below it and none in its own layer |
| Answer» Correct Answer - A | |
| 1235. |
Which of the following is not a crystal system?A. CubicB. HexagonalC. TriclinicD. Orthorhombic. |
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Answer» Correct Answer - C The most unsymmetrical system is triclinic. `(a ne b ne c, alpha ne beta ne gamma= 90^(@))`. |
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| 1236. |
In a monoclinic unit cell the relation of sides and angles are respectivelyA. `a ne b ne c, alpha = gamma = 90^(@), beta ne 90^(@)`.B. `a = b = c, alpha = beta = gamma = 90^(@)`C. `a = b = c, alpha = beta = 90^(@), gamma =120^(@)`D. `a ne b = c, alpha = beta = gamma = 120^(@)` |
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Answer» Correct Answer - A For monoclinic crystal, `a ne b ne c, alpha = gamma = 90^(@), beta ne 90^(@)`. |
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| 1237. |
Tetragonal crystal system has the unit cell dimensions:A. `a = b = c and alpha = beta = gamma = 90^(@)`B. `a = b != c and alpha = beta = gamma = 90^(@)`C. `a != b != c and alpha = beta = gamma = 90^(@)`D. `a = b != c and alpha = beta = 90^(@), gamma = 120^(@)` |
| Answer» Correct Answer - B | |
| 1238. |
A two dimensional lattice of closest packed identical circles , indicating a suitable unit cell is represent as follows then the fractional void area is A. `0.93`B. `0.093`C. `0.17`D. `0.017` |
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Answer» Correct Answer - B Let r be the radius of a circle. The unit cell shown in the figure is composed of two equilateral triangles, ABC and BCD, each with side 2r and s=32r, the area of the unit cell is `2(3r,r^(3))^(1//2)=2r^(2)sqrt(3)`. The unit cell contains `(1)/(6)+(1)/(3)+(1)/(6)+(1)/(3)=1` circle of area `pir^(2)`. The fractional void area is `(2r^(2)sqrt(3)-pir^(2))/(2r^(2)sqrt(3))=0.093` |
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| 1239. |
In a tetragonal crystalA. a=b=c, `alpha =beta =90^(@) ne Y`B. ` alpha=beta = gamma = 90^(@) , a = b ne =0`C. ` alpha = beta = gamma =90^(@) , a ne b ne c`D. `alpha=beta = 90^(@) , gamma = 120^(@) , a=b ne c` |
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Answer» Correct Answer - C |
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| 1240. |
Byavais lattices are ofA. 10 typesB. 8 typesC. 7 typesD. 14 types |
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Answer» Correct Answer - D There are 14 Bravais lattices . |
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| 1241. |
An examplle of a face centred cubic lattice isA. ZineB. SodiumC. copperD. Caesium choride |
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Answer» Correct Answer - D |
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| 1242. |
Given below are dimension lattices with nicely shaded regions . You just have to find the total number of particles in the regions . A. `7(1)/(2)`B. `8(1)/(2)`C. `9(1)/(2)`D. 4 |
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Answer» Correct Answer - A `{:((a),(b),"Total"),(1//4,1,2),(1//2,1//4,2),(1//6,1//2,2),(1//6,-,1//2),(1//3,1//6,1):}` |
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| 1243. |
An example of a face centred cubic lattice isA. ZincB. SodiumC. CopperD. Caesium chloride |
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Answer» Correct Answer - C Copper has F.C.C. structure |
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| 1244. |
The most unsysmmetrical and symmeterical systems are, respectively:A. Tertragonal, CubicB. triclinic. CubicC. Rhombohedral, HexagonalD. Orthohombic, Cubic |
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Answer» Correct Answer - B |
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| 1245. |
In graphics carbon atoms are joined togather due toA. Covalent bondingB. van der waals forcesC. Metallic bondingD. lonic bonding |
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Answer» Correct Answer - a Graphite is `sp^(2)` hybridised and a covalent crystal. |
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| 1246. |
In X-ray diffraction experiment at which one of the following path difference between the two waves , destructive interference is observed (`gamma`=wavelength of x-rays)A. `lambda`B. `2lamdba`C. `3lambda`D. `1.5lambda` |
| Answer» Correct Answer - 4 | |
| 1247. |
Cesium bromide crystallizes in the cubic system. Its unit cell has a `Cs^(+)` ion at the body centred and a `Br` ion at each corner. Its density is `4.44g cm^(-3)`. Determine the length of the unit cell edge. |
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Answer» Correct Answer - `4.30Å;` |
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| 1248. |
Cesium chloried on heating to `760 K` changes inA. CsCl (g)B. NaCl structureC. antifluorite structureD. ZnS structure |
| Answer» Correct Answer - B | |
| 1249. |
Solid NaCl is a bad conductor of electricity becauseA. In solid NaCl there are no ionsB. Solid NaCl is covalentC. In solid NaCl there is no velocity of ionsD. In solid NaCl there are no electrons. |
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Answer» Correct Answer - C Solid NaCl is a bad conductor of electricity because ions are not free to move. |
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| 1250. |
Define the term doping. Pure silicon is an insulator. Silicon doped with phosphorus is a semiconductor. Silicon doped with gallium is also a semi-conductor. What is difference between the two types of semi-conductors ? |
| Answer» Silicon-doped with phosphorus will form n-type semi-conductor while silicon doped with gallium will form p-type semi-conductor. | |