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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.
151. |
Classify the following solids in different categories based on the nature of intermolecular forces operating in them : Potassium sulphate , tin , benzene , urea , ammonia , water , zinc sulphide , graphite , rubidium , argon , silicon carbide . |
Answer» Ionic solids Potassium sulphate , zinc sulphide (as they have ionic bonds) Covalent solids Graphite , silicon carbide (as they are covalent giant molecules ) . Molecular solids Benzene , urea , ammonia , water , argon (as they have covalent bond) . Metallic solids Rubidium , tin ( as these are metals ) . |
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152. |
(a) Based on the nature of intermolecular forces, classify the following solids . silicon carbide, Argon (b) ZnO turns yellow on heating , why ? (c ) what is meant by groups 12-16 compounds . Give an example. |
Answer» Correct Answer - (a) `C_(6) H_(6) ` = Non - polar molecular solid , Ag = Metallic solid (b) N/A (C ) N/A |
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153. |
The correct statement in the following isA. The ionic crstal of AgBr has Schottky defectB. The unit cell having crystal parameters, `a=b nec,alpha=beta=90^(@),gamma=120^(@)` is hexagonalC. In ionic compounds having Frenkel defect the ratio `(gamma_(+))/(gamma_(-))` is highD. The coordination number of `Na^(+)` ion in NACl is 4. |
Answer» Correct Answer - B A crystal system is hexagonal if its unit cell having `a=bnec` axial ratio and `alpha=beta=90^(@),gamma=120^(@)` axial angles. |
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154. |
In a crystalline solid, having formula `AB_(2)O_(4)` oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids. (a) What percentage of the tetrahedral voids is occupied by A? (b) What percentage of the octahedral voids is occupied by B ?A. 0.5B. 0.25C. 0.75D. `12.5%` |
Answer» Correct Answer - D In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion. Since the formula shows the presenece of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four . Out of the eight tetrahedral voids one is occupied by A. `:.` Percentage of tetrahedral voids occupied `=(1)/(8)xx100=12.5%` |
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155. |
How many molecules are there in the unit cell of sodium chioride?A. `2`B. `4`C. `6`D. `8` |
Answer» Correct Answer - b In `NaCI` (rock salty) Number of `Na^(+)` ions `= 12` (at edge centers ) `xx (1)/(4) xx 1` (at body centers ) `xx 1 = 4 ` Number of `CI^(-)` ions `= 8` (at centers )`xx (1)/(8) xx 6` (at face centers ) `xx (1)/(4)` Thus `4` formula units per unit cell |
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156. |
Nicker crystallizes in an fcc unit cell with a cell-edge length of 0.3524nm. Calcualte the radius of the nickel atomA. 0.1624nmB. 0.1246nmC. 0.2164nmD. 0.1426nm |
Answer» Correct Answer - B For an fcc lattice, `d_("face")=asqrt(2)` or `4r_(Ni)=asqrt(2)` `r_(Ni)=(asqrt(2))/(4)=sqrt(2)/(4)xx0.3524=0.1246` nm |
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157. |
ZnO crystals on heating acquires the formuls `Zn(1+x)O`. Or There is an increase in conductivity when silicon doped with phosphorus. Give reason. |
Answer» On heating, ZnO loses oxygen `O^(2-)` ions and changes to `Zn^(2+)` ions. Due to the presence of excess `Zn^(2+)` ions, `Zn(1+x)O` is formed. Or As as result of doping of group 14 element with group 15 element, the extra electrons available are responsible for increase in electrical conductivity. n-p type semi-conductors are formed as a result of this. |
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158. |
In which of the following crystal system F.C.C unit cells exists?A. Cubic, hexagonalB. Tetragonal, orthorhombicC. Orthorhobic , cubicD. Triclinic, monoclinic |
Answer» Correct Answer - 3 | |
159. |
Limiting radius ratio `((r_(+))/(r_(-)))` for co-ordination number six (octahedral arrangement ) isA. 0.155-0.225B. 0.225-0.414C. 0.414-0.732D. 0.732-1.000 |
Answer» Correct Answer - C For a octahedral arrangement co-ordination number is 6 and radius ratio `(r_(+)//r_(-))` is 0.414-0.732. |
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160. |
An `AB_(2)` type structure is found inA. `NaCI`B. `Al_(2)O_(3)`C. `CaF_(2)`D. `N_(2)O` |
Answer» Correct Answer - c `AB_(2)` type of terthedral is present in `CaF_(2)` `:. AB_(2) hArr A^(2++) " " 2B^(-), CaF_(2) hArr Ca^(2+)+2F^(-)` |
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161. |
How many molecules are there in the unit cell of sodium chioride?A. 2B. 4C. 6D. 8 |
Answer» Correct Answer - B In NaCl (rock salt): Number of `Na^(+)` ions =12 (at edge centers) `xx(1)/(4)+1` (at body centre)`xx1=4`. Number of `Cl^(-)` ions =8 (at corners)`xx(1)/(8)+6` (at face centre)`xx(1)/(2)=4`. Thus 4 formula units per unit cell, ex`-NaNO_(3),CaSO_(4)`, calcite `CaCO_(3),HgS` |
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162. |
Which of the following systems is not correctly characeriseed?A. cubic a=b=c,`alpha=beta=gamma=90^(@)`B. tetragonal , a=b `cancel=c,alpha=beta=gamma=90^(@)`C. orthorhombic, `acancel=bcancel=c,alpha=beta=gamma=90^(@)`D. rhombohedral, `a=bcancel=c,alpha=beta=gamma=90^(@)` |
Answer» Correct Answer - 4 | |
163. |
Pure silicon doped with phosphorus is:A. AmorphousB. n-type semi conductorC. p-type semi-conductorD. insulator |
Answer» Correct Answer - B | |
164. |
Out of seven crystal systems how many have body centred unit cell?A. 4B. 3C. 2D. 7 |
Answer» Correct Answer - 2 | |
165. |
If `Z` is the number of atoms in the unit cell that represent the closed packing sequence `---ABCAB---` the number of terrahedral in the unit cell is equal toA. `Z`B. `2Z`C. `Z//2`D. `Z//4` |
Answer» Correct Answer - b Number of terthedral voids in the unit cell `= 2 xx` number of atoms `= 2Z` |
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166. |
At the limiting value of radius ration `(r^+)/(r^-)`A. Force of attraction are larger than the forces of repulsionB. Forces of attraction are smaller than the forces of repulsionC. Forces of attraction and repulsion are just equalD. None of these |
Answer» Correct Answer - C At the limiting value of radius ratio `(r^+/r^-)`, the forces of attraction and repulsion are just equal . |
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167. |
The limiting radius ratios of the complexes `[Ni(CN)_(4)]^(2-)` and `[NiCl_(4)]^(2-)` are respectivelyA. `0.225-0.414, 0.225-0.414`B. `0.414-0.732,0.414-0.732`C. `0.225-0.414,0.414-0.732`D. `0.414-0.732,0.225-0.414` |
Answer» Correct Answer - A `[Ni(CN)_(4)]^(2-)` has square planar geometry `[NiCl_(4)]^(2-)` has tetrahedral geometry They have same limiting radius ratio |
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168. |
At the limiting value of radius ratio `r_(+)//r_(-)`A. Forces of attraction are larger than the forces of repulsionB. Forces of attraction are smaller than the forces of repulsionC. Forces of attraction andrepulsion are just equalD. None of these |
Answer» Correct Answer - C At the limiting value of `r^(+)//r^(-)`, forces of attraction and repulsion are equal. |
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169. |
When potassium chloride is dissolved in waterA. Face centred cubicB. Body centred cubicC. Simple cubicD. Simple tetragonal. |
Answer» Correct Answer - A Face centred cubic lattice found in KCL and NaCl. |
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170. |
Glass is a:A. Superccoled liquidB. Crystalline solidC. Liquid crystalD. none of these |
Answer» Correct Answer - 1 | |
171. |
The unit cell with crystallographic dimensions, `a ne b ne c, alpha =gamma=90^(@) and beta ne 90^(@)` is :A. Tetragonal systemB. orthorhombic systemC. monoclinic systemD. Triclinic system |
Answer» Correct Answer - B `a ne b ne c, alpha = beta = gamma = 90^(@) ` represents orthorhombic system. |
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172. |
How many unit cells are possible for the crystallographic dimensions as `a cancel=bcancel=c alpha=gamma=90^(@) alphacacel=beta`A. 2B. 1C. 4D. 3 |
Answer» Correct Answer - 2 | |
173. |
The unit cell with crystallographic dimensions, `a ne b ne c, alpha =gamma=90^(@) and beta ne 90^(@)` is :A. monoclinicB. tetragonalC. triclinicD. orthorhombic |
Answer» Correct Answer - A | |
174. |
Example of unit cell with crystallographic dimension `a != b != c, alpha = gamma = 90^(@) `isA. calciteB. graphiteC. rhombic sulphurD. monoclinic sulphur |
Answer» Correct Answer - d Monoclinic sulphur is an example of monoclinic crystal system |
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175. |
How many types of Bravais lattices can occur in crystalline solids?A. `14`B. `7`C. `10`D. `11` |
Answer» Correct Answer - 1 There are seven crystal derived form seven poimitive unit cell. Some crystal systems, may have one or more types of lattices depending the number of lattice points. If there are lattice points only at the eight corners of a unit cell ,the result is a simple or primitive (P) lattice. A unit cell which has lattice points at the eight cornors and the six face centres creates a face centered `(F)` lattice. A unit cell which has eight lattice points at the corners and two more at the centres of a pair of any two opposite faces created an endcenred `(E)` . If a unit cell has eight lattice points at the cornors and one at the body centre,the result is a body centred `(B)` lattice. The unit of the type `F,E` and `B` are called non promitive cells . Based on the presence of lattice points in the seven crystal systems. There are fourteen Bravais lattices : orthorhombic (4), cubic (3), tetragonal (2), monoclinic (2), rhombohedral (1), hexagonal (1), and triclinic (1). |
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176. |
Argon crystallizes in FCC arrangement and density of solid and liquid argon is 3/7 and 3gm/cc respectively. Find percentage of empty space in liquid Ar.A. 17B. 34C. 68D. 82 |
Answer» Correct Answer - B First method Let the volume of sloid Ar=100mL. Mass of solid Ar= `Vxxrho=100xx1.59gcm^(-3)=159g` Volume of liquid Ar= `("Mass")/(rho)=(159g)/(1.42gcm^(-3))=112mL` Since Ar crystallizes in fcc-type lattice, the packing fraction (PF)=0.74 Actual volume occupie by Ar=PF xx Volume of solid Ar=`0.74xx100=74mL` % Empty space in liquid Ar= `((112-74))/(112)=33.92%` |
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177. |
Calculate the void space in closest packing of `n` spheres of radius 1 unit, `n` spheres of radius `0.414` units, and `2n` spheres of radius `0.225` units.A. 19B. 18C. 91D. 40 |
Answer» Correct Answer - A Since the packing is closest , therefore it is in fcc-type situation `Z_(eff)=4`/unit cell. |
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178. |
The limiting radius ratio of the complex `[Ni(CN)_(4)]^(2-)` isA. 0 to 0.155B. 0.115-0.225C. 0.225-0.414D. 0.414-0.732 |
Answer» Correct Answer - 3 | |
179. |
The maximum number of Bravais lattices is shown by tetragonal-type crystals. |
Answer» Correct Answer - F Maximum number of Bravais lattice is shown by orthorhombic-type crystal (four Bravais lattice, namely, primite, fc, bc, and end centred). |
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180. |
Which of the crystal systems has maximum number of Bravais lacttices?A. CubicB. MonoclinieC. TetragonalD. Orthorhombic |
Answer» Correct Answer - 4 It has four Bravais lattices: Primitive Body centered face centred and End centred. Cubic has three Bravais lattices , Primitive Body centred and face centred. Tetragonal has two Bravais lattices: Primitive and Body centred. |
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181. |
How many types of Bravais lattices can occur in crystalline solids?A. 7B. 14C. 32D. 230 |
Answer» Correct Answer - B There are 14 Bravais lattices (space lattice). |
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182. |
The number of space lattices possible for the crystallographic dimentions `alphacancel=betacancel=gamma`A. 1B. 2C. 3D. 4 |
Answer» Correct Answer - 1 | |
183. |
In case of a cubic system, the number of types of space lattices areA. 3B. 7C. 14D. 12 |
Answer» Correct Answer - 1 | |
184. |
The fourteen types of space lattices are collectively called ……. |
Answer» Correct Answer - 4,6 |
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185. |
Which of the crystal systems has more than one type of bravais lattices?A. HexagonalB. RhombohedralC. MonoclinicD. Triclinic |
Answer» Correct Answer - 3 Monoclinic has two Bravais lattices: Primitive and End centred. |
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186. |
Select the incorrect statement `:`A. Stoichiometry of crystal remains uneffected due to Schottky defect.B. Frenkel defect usually shown by ionic compounds havin low coordination number.C. `F-` centres generation is responsible factor for imparting the colour to the crystalD. Density of crystal always increases due to substitutional impurity defect. |
Answer» Correct Answer - 4 Density of crystal always increases due to substitutional impurity defect. |
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187. |
A compound formed by element `A` and `B` crystallizes in the cubic structure where `B` atoms are atg the face-centres. The formula of the compound isA. `AB_(3)`B. `AB`C. `A_(3)B`D. `A_(2)B_(2)` |
Answer» Correct Answer - A `{:("Number of A atoms" = 8xx1/8=1),("Number of B atoms" = 6xx1/2=3):}]"Formula"=AB_(3)`. |
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188. |
How many kinds of space lattices are possible in a crystal?A. `23`B. `7`C. `230`D. `14` |
Answer» Correct Answer - D | |
189. |
The coordination number of each atom in body centered cubic unit cell isA. `12`B. `6`C. `8`D. `5` |
Answer» Correct Answer - 3 Each sphere is touched by eight neighbors ,four in the layer below and four in the layer above.Alternatively a body centered atoms is in direct touch with `8` corner atoms and each corner atom is shared by `8` unit cells where it is touched by the body centered atoms. |
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190. |
In the spinel structure, oxides ions are cubical`-` closest packed whereas `1//8 th` of tetrahedral voids are occupied by `A^(2+)` cation and `1//2` of octahedral voids are occupied by `B^(2+)` cations. The general formula of the compound having spinel structure is `:`A. `A_(2)B_(2)O_(4)`B. `AB_(2)O_(4)`C. `A_(2)B_(4)O_(2)`D. `A_(4)B_(2)O_(2)` |
Answer» Correct Answer - B Effective no. of `O^(2-)` in a unit cell `=8xx(1)/(8)+6xx(1)/(2)=4` Effective no of `A^(2+)` in a unit cell `=8xx(1)/(8)=1` Effective no. of `B^(3+)` in a unit cell `=4xx=2" ":." "` general formula is `AB_(2)O_(4)` |
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191. |
Potassium crystallizes with aA. Face-centred cubic latticeB. Body-centred cubic latticeC. Simple cubic latticeD. Orthorhombic lattice |
Answer» Correct Answer - B | |
192. |
Potassium crystallizes with aA. face centered cubic lattic structureB. hexagonal closest packed structureC. primitive cubic lattice structureD. body centered cubic lattice structure |
Answer» Correct Answer - 4 At room temperature all the alkali metals posses body centred cubic lattice. |
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193. |
The atomic radius of strontium `(Sr)` is `215 p m` and it crystallizes with a cubic. Closest packing . Edge length of the cube is `:`A. 4.30pmB. 608.2pmC. 496.53pmD. none of these |
Answer» Correct Answer - B |
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194. |
The atomic radius of strontium `(Sr)` is `215 p m` and it crystallizes with a cubic. Closest packing . Edge length of the cube is `:`A. `430 p m`B. `608.2p m`C. `496.53p m`D. none of these |
Answer» Correct Answer - 2 The ccp of structure generates fcc unit cell So, `sqrt(2a)=4r` `a=608.2p m` |
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195. |
Calculate the packing fraction and density of diamond if a = 3.57Å . Diamond crystallizes in fcc lattice with some more carbon atoms in alternate tetrahedral voids. |
Answer» C-atoms per cell of fcc lattice =4 No .of tetrahedal voids = 4 As C -atoms are present in alternate tetrahedral voids. no. of C-atoms in tetrahedrl voids = 4 Total no. of C -atoms in the unit cell of diamed = 4+ 4 =8 ,i.e., Z=8 packing fraction =` ("Volume occupied by spheres")/("Volume of the unit cell") = (8 xx 4/3 pi r^(3))/a^(3) = 8 xx 4/3 xx (pir^(3))/a^(3)` As alternate tetrahedral voids are also occupied by C-atoms , it can by seen that ` sqrt3 a = 8r or a = (8r)/sqrt3` packing effeiciency = `(8 xx 4/3 pi r^(3))/ ((8r)/sqrt3)^(3) = 32/3 xx 22/7 xx (3sqrt3)/( 8 xx8xx8) = 0.43` density ` p = (ZM)/(a^(3)N_(0)) = ( 8 xx 12)/((3.57xx10^(-8))^(3) (6.023 xx 10^(23)) = 3.5"g/cm"^(3)` |
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196. |
Potassium crystallizes with aA. Face - centred cubic latticeB. Body - centered cubic latticC. Simple cubic latticD. Orlhorhombic lattice |
Answer» Correct Answer - b Potassium `(K)` has bcc lattice. |
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197. |
The packing fraction of the element that crystallizes in simple cubic arrangement isA. `(pi)/(4)`B. `(pi)/(6)`C. `(pi)/(3)`D. `(pi)/(2)` |
Answer» Correct Answer - b `P.E. = ((4)/(3) pi r^(3)xx1 )/(a^(3)) = ((4)/(3) pi r^(3))/((2r)^(3)) = (pi)/(6)` |
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198. |
Potassium crystallizes is a bcc lattice the coordination number of potassiium in potassium metal isA. `0`B. `4`C. `6`D. `8` |
Answer» Correct Answer - d For bcc lattice, co- crdination number is `8`. |
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199. |
In a compound atoms of element `Y` form cup lattice and those of element `X` occupy `2//3^(rd)` of lettabedral voids .The foemula of the compound will beA. `X_(3)Y_(4)`B. `X_(4)Y_(3)`C. `X_(2)Y_(3)`D. `X_(2)Y` |
Answer» Correct Answer - b In a cup array number of tetrahefral voids is swich the number of atom `y:x: :1 [(2)/(3) xx 2] = 3:4` `rArr X_(3)Y_(4)` |
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200. |
In corrundum, oxid ions are arranged in `h.c.p.` array and the aluminum ions occupy two`-`thirds of octahedral voids. What is the formula of currundum ? |
Answer» Correct Answer - `Al_(2)O_(3);` |
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