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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.
| 951. |
` Br^(-)` ions form a close packed struture. If the radius of ` Br^(-) ` ions is 195 pm, calculate the radius of the cation theat just fits into the tetrahedral hole. Can a cation ` A^(+)` having a radius of 82 pm be slipped into the octahedral hole of the crystal ` A^(+)Br^(-)` ? |
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Answer» (i) Radius of the cation just fitting into the tetrahedral hole= Radius of the tetrahedral hole `= 0.225 xx r_(Br^(-)) = 0.225 xx 195 = 43.875 ` pm (ii) For the cation ` A^(+)` with radius = 82 pm. Radius ratio `= (r_(+))/(r_(-)) = (82"pm")/(195" pm") = 0.4205` As it lies in the range 0.414 - 0.732 , hence tha catio ` A^(+)` can be slipped into octahedral hole of the crystal ` A^(+)Br^(-)` |
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| 952. |
A comound is formed by two elements M and N. The element N froms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ? |
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Answer» Suppose the atoms N in the ccp = n No. of tetrahedral voids. = 2n . As ` 1/3` rd of the tetrahderal voids are occupied by atom M, therefore, no. of atoms M ` = (2n)/3` Ratio of M : N `= (2n)/3 : n = 2:3` Hence , the formula is ` M_(2)N_(3)` |
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| 953. |
Lead (II) sulphide crystal has NaCl structure. What is the distance betweeen `Pd^(2+) and S^(2-)` in PhS if its density is ` 12.7 " g cm^(-3)` ? (At .mass of Pb = 207) |
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Answer» Correct Answer - 330 pm Distance between ` Pb^(2+) and S^(2-) " ions" a/2` (as it has NaCl structure. ) |
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| 954. |
An element (atomic mass `= 100 g//mol`) having bcc structure has unit cell edge 400 pm .Them density of the element isA. `10.376 g//cm^(3)`B. `2.144 g//cm^(3)`C. `7.289 g//cm^(3)`D. `5.188 g//cm^(3)` |
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Answer» Correct Answer - d `rho = (n xx M)/(a^(3) xx N_(0) xx 10^(-30)) = (2 xx 100)/((400)^(3) xx (6.02 xx 10^(23)) xx 10^(-30))` `= 5.188 g//cm^(2)` |
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| 955. |
An element (atomic mass `= 100 g//mol`) having bcc structure has unit cell edge 400 pm .Them density of the element isA. `10.376g//cm^(3)`B. `5.188g//cm^(3)`C. `7.289g//cm^(3)`D. `2.144g//cm^(3)` |
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Answer» Correct Answer - B `rho=(nxxM)/(a^(3)xxN_(0)xx10^(-30))` `=(2xx100)/((400)^(3)xx(6.02xx10^(23))xx10^(-30))=5.188g//cm^(3)` |
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| 956. |
Silver crystallizes in fcc lattic. If the edge length of the cell is `4.07 xx 10^(-8) cm` and density is `10.5 g cm^(-3)`. Calculate the atomic mass of silver. |
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Answer» Edge length `(a)=4.077xx10^(-8)cm` `:. ` Volume of unit cell `(a^(3))=(4.077xx10^(-8))^(3)` `=67.77xx10^(-24)cm^(3)` In a `f.c.c.` uinit , there are four atoms per unit cell `(i.e., n=4)` Atomic mass of `Ag` `=("Density" xxAv. no .xx"Volume of unit cell")/(z)` `=(10.5xx6.023xx10^(23)xx67.77xx10^(-24))/(4)` `=107.15` |
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| 957. |
What fraction of calcium atoms lies on the surface of a cubie crystal that is 1.00 cm in length ? Calcium has fcc lattice with edge length 0.556 nm. |
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Answer» The number of unit cell present in the crystal =`(V_(("crystal")))/(V_(("unit cell")))=((1.0cm)^(3))/((0.556xx10^(-7)cm)^(3))=5.82xx10^(21)` Number of atoms present (Z = 4) = `4xx5.82xx10^(21)=23.28xx10^(21)` Number of unit cells on the surface `=("Area of six faces of crystal")/("Area of unit cell")` `(6xx(1cm)^(2))/((0.556xx10^(-7)cm)^(2))=1.94xx10^(15)` The number of atoms present on the surface of the unit cell = face centred atom + 1/4th of each of the four corner atoms = 2 Thus, the number of atoms present on the surface `=1.94xx10^(15)xx2=3.88xx10^(15)` Fraction of atoms on the surface `=(N_(("surface")))/(N_(("crystal")))=(3.88xx10^(15))/(23.28xx10^(21))=1.67xx10^(-7).` |
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| 958. |
If the radius of `Br^(-)` ion is 0.182 nn, how large can a cation be fit in its tetrahedral holes ? |
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Answer» Correct Answer - 0.0753 pm For tetrahedral void, `r^(+)//r^(-)=0.225-0.414` `r^(+)=(0.414)xxr^(-)` (For maximum radius of cation) ltBrgt `r^(+)=(0.414xx0.182"pm")=0.0753"pm"`. ltbr. Thus, cation of radius 0.0753 pm (maximum) can fit into each tetrahedral void. |
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| 959. |
By X -ray diffraction methods, the unit length of NaCl is observed to be 0.5627 nm. The density of NaCl is found to be `2.164 cm^(-3)`. What type of defect exists in the crystal ? Calculate the percentage of ` Na^(+) and Cl^(-)` ions missing . |
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Answer» Calculated density , ` p = (Z xxM)/(a^(3) xx N_(0)) = ( 4 xx 58.5 " g mol"^(-1))/((0.5627 xx 10^(-7) cm)^(3) xx ( 6.022 xx 10^(23) mol^(-1))) = 2.1809 " g cm"^(-3)` Observed density = ` 2.164 " g cm"^(-3)` As observed density is less than theoretically calculated valuem this means that some ` Na^(+) and Cl^(-1)` ions are missing from their lattice site , i.e, there is Schottky defect. Actual formula units of NaCl per unit cell be calculated as follows : ` Z = ( a^(3) xx p xx N_(0))/M = (( 0.5627 xx 10^(-7) cm)^(3) xx ( 2.164 cm^(-3)) xx ( 6. 022 xx 10^(23) mol^(-1)))/( 58.5 " g mol"^(-1)) = 3.968` Formula unit missing per unit cell = 4 - 3.968 = 0.032 % missing = ` ( 0.032)/4 xx 100 = 0.8 % |
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| 960. |
X-ray diffraction studies show that edge length of a unit cell of NaCl is 0.56 nm. Density of NaCl was found to be `2.16g//"cc".` What type of defect is found in the solid? Calculate the percentage of `Na^(+)` and `Cl^(-)` ions that are missing. |
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Answer» Correct Answer - Schottky defect; `2.375%` each Density of NaCl `(rho)=(ZxxM)/(a^(3)xxN_(0))=(4xx(58.5"g mol"^(-1)))/((0.56xx10^(-7)cm)^(3)xx(6.022xx10^(23)mol^(-1)))=2.212g cm^(-3)` The observed density `(2.16"g cm"^(-3))` is less than theoretical density. Therefore, the Solid has schottky defect. `Z=(a^(3)xxrhoxxNo)/(M)=((0.56xx10^(-7)cm)^(3)xx(2.16"g cm"^(-3))xx(6.023xx10^(23)"g mol"^(-1)))/((58.5"g mol"^(-1)))=3.905` Number of missing formula units `=4-3.905=0.095` Percentage of missing formula units `=(0.095)/(4)xx100=2.375%` Percentage of missing `Na^(+)` ions `=2.375%` Percentage of missing `K^(+)` ions `=2.375%` |
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| 961. |
Total no. of voids in `0.5` mole of a compound forming hexagonal closed packed structure are :A. `6.022xx10^(23)`B. `3.011xx10^(23)`C. `9.033xx10^(23)`D. `4.516xx10^(23)` |
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Answer» Correct Answer - C No. of voids per molecule `=3` No. of voids in `0.5` mole molecules `=3xxN_(0)xx0.5` `=3xx6.022xx10^(23)xx0.5` `=9.033xx10^(23)` |
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| 962. |
Total no. of voids in `0.5` mole of a compound forming hexagonal closed packed structure are :A. `6.022xx10^23`B. `3.011xx10^23`C. `9.033xx10^23`D. `4.516xx10^23` |
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Answer» Correct Answer - C In hexagonal close packed structure , there are 6 atoms per unit cell total number of voids = 6+12=18 `therefore` total number of voids per atom = `18/6=3` `therefore` In 1 mole compound the total number of voids `=3xx6.022xx10^23` `therefore` In 0.5 mole compound , the total number of voids `=3xx0.5xx6.023xx10^23` `=9.034xx10^23` |
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| 963. |
Point out the correct statement for the set of characteristics of ZnS crystal.A. Coordination number (4:4) : ccp , `Zn^(2+)` ions in the alternate tetrahedral voidsB. Coordination number (6:6) , hcp, `Zn^(2+)` ions in all tetrahedral voidsC. Coordination number (6:4) , hcp, `Zn^(2+)` ions in all octahedral voidsD. Coordination number (4:4) , ccp, `Zn^(2+)` ions in all tetrahedral voids |
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Answer» Correct Answer - D ZnS has coordination number 4 : 4 having ccp structure where `Zn^(2+)` ions occupy half of the tetrahedral voids . |
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| 964. |
CHARACTERISTICS, AMORPHOUS AND CRYSTALLINE SOLIDA. long rang solidsB. short rang solidsC. Disordered arrangementD. None of these |
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Answer» Correct Answer - a Crystalline solids have regular arrangement of constituent particle , sharp melting point and are anisotropic |
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| 965. |
Assertion: Anion vacancies in alkali halides are produced by heating the alkali halide crystals with alkali metal vapour. Reason: Electrons trapped in anion vacancies are referred to as F -centres.A. If both assertion and reason are true and the reason is the correct explanantion of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B On heating , the metal atoms deposit on the surface and finally they deffuse into the crystal and after ionisation the alkali metal ion occupies cationic vacancy where as electron occupies anionic vacancy. |
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| 966. |
Assertion: The presence of a large number of Schottky defects in NaCl lowers its density. Reason: In NaCl, there are approximately `10^(6)` Schottky pairs per `cm^(3)` at room temperature.A. If both assertion and reason are true and the reason is the correct explanantion of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - B When an atom or an ion is missing from its normal lattice site. Defect, due to missing density of crystal will be lowered. |
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| 967. |
In zinc blends structure the coordination number of `Zn^(2+)` ion isA. 2B. 4C. 6D. 8 |
| Answer» Correct Answer - B | |
| 968. |
Strontium chloride has a flurite structure, which of the statement is true for the structure of strontium chloride ?A. the strontium ions are in a body-centered cubic arrangementB. the strontium ions are in a face-centered cubic arrangementC. each chloride ion is at the centre of a cube of 8 strontium ionsD. each strontium ion is at the center of a tetrahedron of 4 chloride ions |
| Answer» Correct Answer - B | |
| 969. |
why are solids incomressible ? |
| Answer» the distance between the constituent partticles is very less in solids On briging them still closer repulsion will start between electron clouds of these particles hence , they cannot be brought further close together and are incomepressible . | |
| 970. |
Crystalline solid areA. SugarB. RubberC. PlasticD. Glass |
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Answer» Correct Answer - a Sugar is a crystalline solid , while glass rubber and plastic are amorphous solids |
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| 971. |
Amorphous substances show (A) short and long range order (B) short range order (C ) long range order(D ) have no sharp `M.P`A. A and C are correctB. B and C are correctC. C and D are correctD. B and D are correct. |
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Answer» Correct Answer - D Amorphous solids have short range order but no sharp m. pt. |
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| 972. |
Among solid the highest melting point is estrablished byA. Covalent solidsB. lonic solidsC. pseudo solidsD. Molecular solids |
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Answer» Correct Answer - b lonic solids have highest melting point due to string electrstic force of attraction |
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| 973. |
Amorphous substances show (A) short and long range order (B) short range order (C ) long range order(D ) have no sharp `M.P`A. B and D are correctB. B and C are correctC. C and D are correctD. A and C are correct |
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Answer» Correct Answer - a Amorphous solids have short rangs order but no sharp in melting point. |
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| 974. |
In which pair most efficient packing is present?A. hcp and bccB. hcp and ccpC. bcc and ccpD. bcc and simple cubic cell. |
| Answer» Correct Answer - B | |
| 975. |
In which pair most efficient packing is present?A. Hcp and bccB. hcp and ccpC. bcc and ccpD. bcc and simple cubic cell |
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Answer» Correct Answer - B packing efficiency is the percentage to total filled spaces by partices and it can be calculated as packing efficeency `("Volume occupied by spheres in the unit cell ")/("total volume of unit cell ")xx100` since packing efficiency for hcp or ccp is calculated to be 74 % which is maximum among all type of crystals. |
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| 976. |
The density of solid argon is `1.65g//mL ` at `-233^(@)C` . If the argon atom is assumed to be sphere of radius `1.54xx10^(-8)cm`, what percentage of solid argon is apparentaly empty space ? `(At. Wt. of Ar=40)`A. 26B. 32C. 68D. 52 |
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Answer» Correct Answer - C Volume of one atom of `Ar=(4)/(3)pir^(3)` Also, No. of atom in 1.65kg `=(1.65)/(40)xx6.023xx10^(25)` `therefore` Total volume of all atoms of Ar in solid state `=(4)/(3)pir^(3)xx(1.65)/(40)xx6.023xx10^(23)` `=(4)/(3)xx(22)/(7)xx(1.54xx10^(-4))^(3)xx(1.65)/(40)xx6.023=0.380cm^(3)` Volume of solid argon=`1cm^(3)` `therefore %` empty space `=([1-0.380])/(1)xx100=62%` |
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| 977. |
`K_(2)Cr_(2)O_(7)` is an example ofA. hexagonalB. triclinicC. cubicD. Orthorhombic. |
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Answer» Correct Answer - B For `K_(2)Cr_(2)O_(7)` that is, triclinic system, `a ne b ne c, alpha ne beta ne gamma = 90^(@)` |
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| 978. |
Copper belongs otA. cubic systemB. Tetragonal systemC. monoclinic systemD. Triclinic system |
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Answer» Correct Answer - A Copper belongs to cubic system. |
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| 979. |
Explain how much portin of an atom located at (a) corner and (b) body centre of a cubic unit cell is part of its neighouring unit cell.A. `1,(1)/(2)`B. `(1)/(2),1`C. `(1)/(8),1`D. `(1)/(8),(1)/(2)` |
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Answer» Correct Answer - C (i) A point laying at the corner of a unit cell is shared equally by eight unit cessl and therefore, only one-eight `((1)/(8))` of each such point belongs to the given unit cell. (ii) A body centred point belongs entirely to one unit cell since, it is not shared by any other unit cell . |
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| 980. |
Explain how much portin of an atom located at (a) corner and (b) body centre of a cubic unit cell is part of its neighouring unit cell. |
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Answer» (i) An atom located at the corner is shared by eight unit cells. Therefore, its contribution to a particular unit cell is `1//8`. (ii) An atom located at the body of the unit cell is not shared by any unit cell. It belongs to one particular unit cell only. |
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| 981. |
A crystalline solidA. AnisotropicB. IsotropicC. HardD. Dense. |
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Answer» Correct Answer - B Not isotropic actuallyrefers to anisotropic here. |
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| 982. |
A crystalline solidA. Changes abruptly from solid to liquid when heatedB. Has no definite melting pointC. Undergoes deformation of tis geometry easilyD. Has an irregular 3-dimensional arrangements |
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Answer» Correct Answer - A In crystalline solid there is perfect arrangement of the constituent particles only at 0 K. as the temperature increases the chance that a lattice site may be unoccupied by an ion increases. As the number of defects increases with temperature solid change in liquid. |
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| 983. |
In a hexagoanl crystal:A. `alpha = beta = gamma ne 90^(@), a=b=c`B. `alpha = beta = gamma = 90^(@), a=bne c`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 - D |
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| 984. |
Orthorhombic crystal has the following unit cell parameters:A. `a= b= c, alpha= beta= gamma=90^(@)`B. `a= bne c, alpha= beta= gamma=90^(@)`C. `ane bne c, alpha= beta=gamma=90^(@),`D. `a=bnec, alpha= beta=90^(@),gamma=120^(@)` |
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Answer» Correct Answer - C |
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| 985. |
Malleability and ductility of metals can be acounted due toA. the capacity of layers of metal ions to slide over the otherB. the interaction of electrons with metal ions in the otherC. the presence of electrostatic forcesD. the crystalline structure of metal |
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Answer» Correct Answer - A Malleability and ductility is tendency of metal ion layer Slide over the other layer. |
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| 986. |
Which of the following is not correct for ionic crystalsA. They posses high melting point and boiling pointB. All are electrolyte.C. Exhibit the property of isomorphism.D. Exhibit directional properties of the bond. |
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Answer» Correct Answer - D Ionic crystals exhibit non-directional properties of the bond. |
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| 987. |
Which are not true about the amorphous solids ?A. They melt over a range of temperatureB. They are anisotropicC. There is no orderly arrangement of particles.D. They are rigid and incompressible |
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Answer» Correct Answer - B Amorphous solids are isotropic, because these substances show same properties in all directions. |
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| 988. |
Malleability and ductility of metals can be accounted due toA. The capacity of layers of metal ions to slide over the otherB. The interaction of electrons with with metalions in the otherC. The presence of electrostatic forceD. The crystalline structure in metal. |
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Answer» Correct Answer - A When a metal is beaten, it does not break but is converted into sheet. It is said to possess the property of mallcability. Due to ductile nature, metals can be drawn into wires. These two properties of metals can be accounted due to the capacity of layers of metal ions to slide over the other. |
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| 989. |
The mineral haematite, ` Fe_(2)O_(3)` consists of a cubic close packed array of oxide ions with `Fe^(3+)` ions occupying intersitial positions. Predict whether the iron ions are in the octahdral or tetrahedral holes. Radius of ` Fe^(3+)` = 0.65Å . |
| Answer» ` r_(+)//r_(-) = 0.65 Å / 1.45 Å `which lies in the range 0.414 - 0.732 Hence, ` Fe^(3+)` ions will be in the octahedral holes. | |
| 990. |
Why do solids have a definite volume ? |
| Answer» The constituent particles of a solid have fixed positions and are not free to move about , i.e., they posses rigidity. That is why they have a definite volume. | |
| 991. |
Which defect cause decrease in the density of crystal?A. FrenkelB. SchottkyC. InterstitialD. F-centre |
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Answer» Correct Answer - B More is the Schottky defect in crystal, more is the decrease in density . |
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| 992. |
Assertion. CsCl has body-centred cubic arrangement. Reason . CsCl has one ` Cs^(+)` ion and `8 Cl^(-)` ions in its unit cell. |
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Answer» Correct Answer - d Correct R. CsCl has one ` Cs^(+)` ion and one `Cl^(-)` ion in its unit cell. |
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| 993. |
In CsCl structure, the coordination number of `Cs^(+)` isA. Equal to that of `Cl^(-)`, that is 6B. Equal to that of `Cl^(-)` that is 8C. Not equal to that of `Cl^(-)`, that is 6D. Not equal to that of `Cl^(-)`, that is 8 |
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Answer» Correct Answer - B `Cl^(-)` ions in CsCl adopt bcc type of packing. |
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| 994. |
Close packing is maximum in the crystal which isA. Simple cubicB. Face centredC. Body centredD. None |
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Answer» Correct Answer - B fcc = ccp and space occupied is 74%, i. e., maximum. |
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| 995. |
Close packing is maximum in the crystal which in the crystal which isA. Simple cubeB. bccC. fccD. none |
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Answer» Correct Answer - c The close packing in the crystal is `0.52,0.68 and 0.74` for simple cubic , body ceneters cubic , and face - centres cubic respectively i.e. the close packing is maximum in fcc |
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| 996. |
Given an alloy of Cu, Ag and in which Cu atoms consists the CCP arrangement. If the hyphothetical formula of the alloy is `Cu_(4)Ag_(3)Au`. What are the probable locations of Ag and Au atoms?A. Ag - All tetrahedral voids, Au -all octahedral voidsB. Ag - `3//8^(th)` tetrahedral voids, Au - `1//4^(th)` octahedral voidsC. Ag-1/2 octahedral voids, Au -1/2 tetrahedral voidsD. Ag -all octahedral voids, Au -all tetrahedral voids |
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Answer» Correct Answer - B `{:(Cu_(4),"Ag"_(3),Au),(darr,darr,darr),("From c.c.p.,",(3)/(8)th " of tetrhedral voids",(1)/(4) " of octahedral voids "[ :." No. of O-voids = 4"] ),(z = 4,[:. " No. of T -voids" = 8],):}` |
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| 997. |
The fraction of volume occupied by atoms in a body centered cubic unit cell is:A. 0.32B. 0.48C. 0.68D. 0.74 |
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Answer» Correct Answer - C |
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| 998. |
Square packed sheets are arranged on the top fo the other such that a sphere in the next layer rests on the center of a square in the previous layer. Identify the type of arrangement and find the coordination number:A. Simple Cubic, 6B. Face centered cubic, 8C. Face centered cubic 12D. Body centered cubic, 8 |
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Answer» Correct Answer - C unit cell is face centered cubic so coordinatin number is 12 |
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| 999. |
The fraction of total volume occupied by the atom present in a simple cubic isA. 0.48B. 0.52C. 0.55D. 0.68 |
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Answer» Correct Answer - B |
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| 1000. |
The fraction of total volume occupied by the atom present in a simple cubic isA. `(pi)/(4sqrt(2))`B. `(pi)/(4)`C. `(pi)/(6)`D. `(pi)/(3sqrt(2))` |
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Answer» Correct Answer - 3 For simple cubic unit cell we have `r=(a)/(2)` `Z=1` Therfore Packing fraction `=("Volume of one sphere")/("Volume of cubic unit cell")` `=((4)/(3)pir^(3))/(a^(3))=((4)/(3)pi((a)/(2))^(3))/(a^(3))` `(pi)/(6)` |
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