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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.
| 751. |
The number of tetrahedral voids in the unit cell of a face-centred cubic lattice of similar atoms isA. 4B. 6C. 8D. 10 |
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Answer» Correct Answer - C In F.C.C. structure tetrahedral void = No. of corner = 8 |
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| 752. |
In the body centered unit cell, the lattice point are present at the:A. corners of the unit cell onlyB. corners and centre of the unit cellC. corners and centre of each face of the unit cellD. corners and at one set of faces of unit cell |
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Answer» Correct Answer - B |
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| 753. |
Tetragonal crystal system has the following unit cell dimensions `:`A. cubicB. tertragonalC. monoclinicD. rhombohedral |
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Answer» Correct Answer - B |
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| 754. |
In AgBr, there can occurA. only schottky defectB. only Frenkel defectC. both (1) and (2)D. None of these |
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Answer» Correct Answer - C AgBr show schottky and frenkel defect |
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| 755. |
Graphite cannot be classified as ………A. conducting solidB. network solidC. covalent solidD. ionic solid |
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Answer» Correct Answer - A |
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| 756. |
Graphite cannot be classified as :A. conducting defectB. network solidC. covalent solidD. ionic solid. |
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Answer» Correct Answer - D Graphite is not an ionic solid. |
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| 757. |
Identfity molecular solide, covalent solid, ionic solid: `P_(4)(s), S_(8)(s), SiC (s), Al_(2)O_(3)(s), He(s), Al_(2)Cl_(6)(s)`. |
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Answer» Molecular solide `rarr P_(4)(s), S_(8)(s), He(s), Al_(2)Cl_(6)(s)` Covalent solid `rarr SiC` Ionic solid `rarr Al_(2)O_(3)(s)` |
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| 758. |
The lattice parameters of a given crystal are `a = 5.62 Å , b = 7.41 Å and c= 9.48Å`. The three cordinate axes are mutually perpendicular to each other. The crystal is:A. tetragonalB. orthorhombicC. monoclinicD. trigonal |
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Answer» Correct Answer - B `a!= b!= c" & "alpha = beta = gamma = 90^(@)` the crystal system is orthorhombic |
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| 759. |
The lattice site in a pure crystal cannot be occupied by :A. moleculeB. ionC. electronD. atom |
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Answer» Correct Answer - C Each point in a lattice is known as lattice point which are either atom or molecule or ion which are joined together by a straight line to bring out geometry of lattice in pure crystal constituents are arranged in fixed stoichiometric ratio. Hence, existence of free electrons are not possible. |
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| 760. |
When carbon are trapped into the crystal of iron, the defect is known as :A. Schottky defectB. Frenkel defectC. Stoichiometric defectsD. Interstitial defect |
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Answer» Correct Answer - D It is a fact. |
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| 761. |
Ice crystallizes in a hexagonal lattice. At ascertain low temperature, the lattice constants are a `= 4.53 Å and c = 7.41 Å`. The number of `H_(2)O` molecules contained in a unit cell (`d ~~ 0.92 g cm^(-3)` at the given temperature) isA. 4B. 8C. 12D. 24 |
| Answer» Correct Answer - A | |
| 762. |
Brass is an example of a the defectA. Schottky defectB. Frenkel defectC. Interstitial defectD. Substitution impurity defect |
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Answer» Correct Answer - D It is an alloy of copper and zinc metal |
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| 763. |
In `AgBr` crystal , the ion size lies in the order `Ag^(+) lt lt Br^(-)` The `AgHt` crystal should have the following characheristicsA. defect less (perfect) crystalB. Scgittky defectC. Frenkel defectD. Both Schottky and Frenkel defect |
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Answer» Correct Answer - D It is a fact. |
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| 764. |
Why does `ZnO` show increased electrical conductivity and turns yellow on heating?A. Frenkel defectB. Metal excess defectC. Metal deficiency defectD. Schottky defect |
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Answer» Correct Answer - B Zinc oxide is white in colour at room temperture. On heating it loses oxygenand turns yellow `ZnOoverset("heating")(to)Zn^(2+)+(1)/(2)O_(2)+2e^(-)` Now there is excess of zinc in the crystal and its formula becomes `Zn_(1+x)O`. The excess `Zn^(2+)` ions move to intrestitial sites and the electrons to neighbouring interstital sites . |
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| 765. |
In a ccp structure, the `:`A. first and third layers are repeatedB. first and fourth layers are repeatedC. second and fourth layers are repeatedD. first, third and sixth layers repeated |
| Answer» Correct Answer - B | |
| 766. |
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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| 767. |
Due to Frenkel defect, the density of the ionic solidsA. IncreasesB. DecreasesC. Does not changeD. Changes. |
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Answer» Correct Answer - C Since no ions are missing from the crystal as a whole, there is no effect of density. |
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| 768. |
In a covalent solid the lattice points are occupied byA. atomsB. ionsC. moleculesD. electrons |
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Answer» Correct Answer - A Atoms occupy the lattice points in the unit cell of a covalent solid. |
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| 769. |
Which of the following statements are correctA. The coordination number of each type of ion in CsCl crystal is 8B. A metal that crystallises in bcc structure has a coordination number of 12C. A unit cell of ionic crystal shares some of its ion with other unit cellsD. The length of the unit cell in NaCl is 552 pm. `(gamma_(Na) = 95 pm, gamma_(Cl^(-) = 181 pm)` |
| Answer» Correct Answer - A::C::D | |
| 770. |
Which of the following is/are correct ?A. Schottky defect lowers the densityB. Frenkel defect increases the dielectric constant of the crystalsC. Stoichiometric defects make the crystals electrical conductorsD. In the Schoottky defect, equal number of extra cations and electrons are present in the interstitials sites |
| Answer» Correct Answer - A::B::C | |
| 771. |
Which of the following statements are true about semiconductors ?A. Silicon doped with electron rich impurity is a p-type semiconductorB. Silicon doped with an electron rich impurity is an n-type semiconductorC. Delocalised electrons increase the conductivity of doped siliconD. An electron vacancy increases the conductivity of n-type semiconductor |
| Answer» Correct Answer - B::C | |
| 772. |
Consider the figure given for solid XY. Answer the following questions : The number of XY units per unit cell is:A. `4`B. `3`C. `3`D. `8` |
| Answer» Correct Answer - A | |
| 773. |
Consider the following partially labelled figure (graph is upto scale) for an ideal binary solution of benzene and toluence and answer the following questions. A. tetrahedral voidB. Octahedral voidC. triangle voidD. cubical void |
| Answer» Correct Answer - B | |
| 774. |
In the calcum fluaride structure, the coordination bumber of the cations and anions are respectively ,A. 6 and 6B. 8 and 4C. 4 and 4D. 4 and 8 |
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Answer» Correct Answer - B `Ca^(2+)` ion is contact with eight `F^(-)` ions and `F^(-)` ion and is in contact with four `Ca^(2+)` ions |
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| 775. |
Co-ordination number for copper (Ca) isA. 1B. 6C. 8D. 12 |
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Answer» Correct Answer - D Copper has cubic close packed crystal structure. In this structure one Cu atom is attached to another 12 spheres (Cu-atom) in metallic copper. |
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| 776. |
The number of atoms present in unit cell of a monoatomic substance of simple cubic lattic isA. 6B. 3C. 2D. 1 |
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Answer» Correct Answer - D In a simple cubic structure. `z=(1)/(8)xx8` (atoms one at a corners) z=1 |
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| 777. |
In the calcum fluaride structure, the coordination bumber of the cations and anions are respectively ,A. 6,6B. 8,4C. 4,4D. 4,8 |
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Answer» Correct Answer - B The `Ca^(2+)` ions are arranged in (ccp) arrangement i.e. `Ca^(2+)` ions are present at all corners and tat the centre of each face of the cube, the fluoride ions occupy all the tetrahedral sites. This is 8:4 arrangement i.e., each `Ca^(2+)` ion is surrounded by `8F^(-)` ions and each `F^(-)` ion by four `Ca^(2+)` ions. |
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| 778. |
When heated above `916^(@)C`, iron changes, its crystal structure from body centred cubic to cubic closed packed structure. Assuming that the metallic radius of an atom does not change, calculate the ratio of the density of the bcc crystal to that of ccp crystal. |
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Answer» Let the density of iron in bcp structure `=rho_(1)` Let the density of iron in ccp structure =`rho_(2)` The efficiency of packing in bcp `=68%` The efficiency fo the packing in ccp `=74%` The density of the crystal is related to the efficiency of the packing as `(rho_(1))/(rho_(2))=((68%))/((74%))=0.919`. |
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| 779. |
Silver crystallises in a face centred cubic unit cell. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face centred atom is touching the four corner atoms.) |
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Answer» For face centred cubic unit cell (fcc) Radius (r) =`(a)/(2sqrt2)=(("409 pm"))/(2sqrt2)` `=(("409 pm"))/(2xx1.414)=144.6"pm"` |
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| 780. |
In the cubic lattice given below, the three distances between the atoms `A-B, A-C`, and `A-G` are, respectively, A. `a, sqrt2a, sqrt3a`B. `a, sqrt3a, sqrt2a`C. `(a)/(2), (a)/(sqrt2), (sqrt3a)/(2)`D. `a, (sqrt3a)/(2), sqrt2a` |
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Answer» Correct Answer - A i. Edge length `= AB = AD = BC = CD = a` ii. AC `= sqrt((AB^(2)) + (BC)^(2)) = sqrt((a^(2)) + a^(2)) = sqrt(2)a` iii. AG `= sqrt((AC^(2)) + (CG)^(2)) = sqrt(2a^(2) + a^(2)) = sqrt(3)a` |
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| 781. |
Following three planes `(P_(1), P_(2), P_(3))` in an fcc unit cell are shown in the figure below. Consider the following statements and choose the correct option/options that follow: A. `P_(1)` contains no three-dimensional voids.B. `P_(2)` contains only octahedral voids.C. `P_(3)` contains both octahedral and tetrahedral voidsD. All of these |
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Answer» Correct Answer - A::B::C::D a. `P_1` represents one of the close-packed layer having triangular voids only. b. `P_2` contains of `Ovs` (edge centres of unit cell) c. `P_3` contains `3 Ovs` locations (one at body centre and two at edfe centre). Also, plane `P_3` contains the body diagonals, hence it contains `TVs` location (`TVs`lie at body diagonal). |
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| 782. |
A solide `A^(+)B^(-)` has the `B^(-)` ions arranged as below. If the `A^(+)` ions occupy half of the tetrahedral sites in the structure. The formula of solid is `:` A. `AB`B. `AB_(2)`C. `A_(2)B`D. `A_(3)B_(4)` |
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Answer» Correct Answer - a In a closed packed structure, the number of tetrahedral voids per atom of the crystal is two. Since half of the tetrahedral voids are occupied by `A^(+)` , the number of `A^(+)` is same as that of `B^(-)` in the crystal . Thus formula is `AB`. |
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| 783. |
In a face-centred cubic crystal, the neighbour distance is …… times the edge of the crystal. |
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Answer» Correct Answer - 0.433 |
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| 784. |
An element crystallizes as body `-` centred cubic lattic. Its density is `7.12g cm^(-3` and the length of the side of the unit cell is `2.88Å`. Calculate the number of atoms present is `288g` of the element.A. `1.693 X 10^(24)`B. `1.693 X 10^(23)`C. `3.386 X 10^(23)`D. `3.386 X 10^(24)` |
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Answer» Correct Answer - D Volume of unit cell `=a^(3)=(2.88overset(@)A)=(2.88xx10^(-6)cm)^(3)` `=23.887xx10^(-24)cm^(3)` Volume of 288g of the element `=("Mass")/("Density")=(288g)/(7.12g cm^(-3))=40.449cm^(3)` `=("volume of the element")/("volume of the unit cell")=(40.449)/(23.887xx10^(-24))=1.693xx10^(24)` `because` Each unit cell of b.c.c. lattice contains 2 atoms `therefore` number of atoms in 288g of element =No. of unit cells xx no. of atoms per unit cell `=1.693xx10^(24)xx2=3.386xx10^(24)` |
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| 785. |
In a `f.c.c.` arrangement of `A` and `B` atoms, where `A` atoms are at the corners of the unit cell and `B` atoms at the face `-` centres, one of the `A` atom is missing from one corner in each unit cell. The formula of compound is `:`A. `A_(24)B_(7)`B. `A_(7)B_(24)`C. `A_(7)B_(28)`D. `A_(28)B_(7)` |
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Answer» Correct Answer - b No. of atom `A` from the corner of unit cell `=(7)/(8)` No. of atom `B` from the face of unit cell `=3` Thus `A:B::(7)/(8):(3)/(1)` or `A:B::7:24` `:.` Formula is `A_(7)B_(24)` |
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| 786. |
Crystalline solids are anisotropc in nature. What does this statement mean ? |
| Answer» The statement means that crystalline solids have different properties such as electrical conductivity, thermal conductivity, mechanical strength, refractive index etc. in different directions. | |
| 787. |
What type of semiconductor results when highly purified silicon is doped with arsenic? |
| Answer» Silicon has 4 valence electrons while arsenic has 5 valence electrons. Doping silicon with arsenic leads to extra electrons which are free to conduct electric current. This result in n-type semi-conductors. | |
| 788. |
Following three planes `(P_(1),P_(2),P_(3))` in an FCC unit cell are shown `:` Consider the following statement and choose the correct option that follow`:` `(i) ` `P_(1)` contains no voids of three dimensions. `(ii)` `P_(2)` contains only Octahedral voids. `(iii)` `P_(3)` contains both Octahedral and Tetrahedral voids.A. All are trueB. Only `(i) & (ii)` are trueC. `(i) & (iii)` are trueD. Only `(iii)` is true. |
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Answer» Correct Answer - 1 Refer theory octahedral & tetrahedral voids about positions of `P_(1),P_(2)` and `P_(3)` |
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| 789. |
The arranegement `ABC, ABC, ABC …….` is referred asA. Octahedral close packingB. hexagonal close packingC. tetrahedral close packingD. cubic close packing |
| Answer» Correct Answer - 4 | |
| 790. |
In solid ammonia, each `NH_(3)` molecule has six other `NH_(3)` molecules as nearest neighbours. `DeltaH` sublimation of `NH_(3)` at the melting point is `30.8 k J mol^(-1)`, and the estimated `DeltaH` sublimation in the absence of hydrogen bonding is `14.4 kJ mol^(-1)`. the strength of a hydrogen bond is `NH_(3)` isA. `5.47 kJ " mol"^(-1)`B. `10.93 kJ " mol"^(-1)`C. `16.40 kJ " mol"^(-1)`D. `-16.4 kJ " mol"^(-1)` |
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Answer» Correct Answer - A Total strength of all H-bonds `=30.8-14.4=16.4` kJ `mol^(-1)` There are six nearest neighbours but each H-bond involves two molecules. Hence the effective neighbours `=3` `:.` Strength of H-bond `=(16.4)/3` `=5.47 kJ "mol"^(-1)` |
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| 791. |
The radius of a divaent cation `A^(2+)` is 94pm and of divalent anion `B^(2-)` is 146pm. The compound AB has:A. NaCl structureB. Linear structureC. CsCl structureD. ZnS structure |
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Answer» Given , radius of cation of 94 pm, radius of anion = 146pm For `M^(2+)X^(2-)`, radius ratio , `(r_+)/(r_-)=94/146=0.6438` Since , the radius ratio is in between 0.41 to 0.732 , it has NaCl structure. |
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| 792. |
At what angle for the first - order diffraction, spacing between two planes respectively is `lambda and (lambda)/(2)`?A. `0^(@), 90^(@)`B. `90^(@), 0^(@)`C. `30^(@), 90^(@)`D. `90^(@), 30^(@)` |
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Answer» Correct Answer - C It first case `2d sin theta=nlambda` `2lambda sin theta = lambda (n=1)` ` sin theta =1//2, theta=30^(@)` In second case `2d sin theta=nlambda` `2lambda//2 sin theta=nlambda` `sin theta =1, theta=90^(@)` |
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| 793. |
At what angle for the first - order diffraction, spacing between two planes respectively is `lambda and (lambda)/(2)`?A. `0^(@),90^(@)`B. `90^(@),0^(@)`C. `30^(@),90^(@)`D. `90^(@),30^(@)` |
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Answer» Correct Answer - 3 `nlambda-2d sin theta` |
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| 794. |
Which of the following statement is not true about amorphous solids?A. On heating they may become crystalling at certain temperatureB. They may become crystalling on keeping for long timeC. Amorphous solids can be moulded by heatingD. They are anisotropic in nature |
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Answer» Correct Answer - D Amorphous solids are isotropic in nature. These can be moulded by heating. Moreover , they become cyrstalling on standing for a long time or on heating. |
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| 795. |
Solid `CO_(2)` isA. Ionic crystalB. Covalent crystalC. Metallic crystalD. Molecular crystal |
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Answer» Correct Answer - D Solid `CO_(2)` is molecular crystal. |
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| 796. |
Formula mass of `NaCl` is `58.45g mol^(-1)` and density of its pure form is `2.167g cm^(-3)`. The average distance between adjacent sodium and chloride ions in the crystal is `2.814xx10^(-8)cm`. Calculate Avogadro constant. |
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Answer» Density `=("Formula mass"xxz)/(Av. no . xxVolume )` For `f.c.c.` structure of `NaCl` , `z=4(` for `f.c.c.), d=2.167g//cm^(3), ` formula mass `=58.45g//mol,` edge length `=2(r^(+)+r^(-))` `=2xx2.814xx10^(-8)=5.628xx10^(-8)` `:.` Volume `=a^(3)=(5.628xx10^(-8))^(3)` `=1.78xx10^(-22)cm^(3)` `:. 2.167=(58.45xx4)/(Av. no . xx1.78xx10^(-22))` `:. Av. no . =6.06xx10^(23)` |
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| 797. |
Sodium metal crystallises in body centred cubic lattic with cell edge `4.29 Å` .What is the radius of sodium atom ? |
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Answer» Radius of `Na=(sqrt(3)a)/(4)(` if `b.c.c` lattice `)` `=(sqrt(3)xx4.29)/(4)=1.8574A` |
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| 798. |
Assertion :- (a) the total number of atoms present in a simple cubic unit cell is one . Reasn :-(R ) simple cubic cell has atoms at its corners , each of which is shered between eight adjecent adjeacent unit cells.A. Assertion and reason both are correct statements and Reason is correct explantion for Assertion .B. Asserton and Reason both are correct statement but Reason is not correct explanation for assertion .C. Assertion is correct statement but reason is worng statement .D. Assertion is Wrong statement but Reason is correct statement . |
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Answer» Correct Answer - A in simple cubic unit cell atom is present at corners having contribution 1/8 Hence total number of atoms present per units cell in scc is `(1)/(8)xx8=1` |
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| 799. |
Which one of the following is not a close packing ?A. hcpB. ccpC. bccD. fcc |
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Answer» Correct Answer - C bcc is not as close packing arrangment as hcp and ccp. |
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| 800. |
Assertion :- (a) the total number of atoms present in a simple cubic unit cell is one . Reasn :-(R ) simple cubic cell has atoms at its corners , each of which is shered between eight adjecent adjeacent unit cells.A. Assertion and reason both are corrent statements and reason is correct explanation for assertion.B. Assertion and reason both are corrent statements but reason is not correct explanation for assertion.C. Assertion is corrent statement but reason reason is wrong statement.D. Assertion is wrong statement but reason is correct statement. |
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Answer» Correct Answer - A Reason is the corrent explanation for assertion. |
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