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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.
| 1001. |
Which of the following statements is not correct?A. the number of Bravais lattices in which a crystal can ebh categorized in `44`.B. The fraction of the total volume occupaied by the atoms in a primitive cell is `0.48`.C. Molecular solids are generally volatile.D. The number of `C` atoms in an unit cell of diamond is `4`. |
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Answer» Correct Answer - 2,4 In a simple cubic lattice the edge length (or side of the cube) `a` and the radius of each particle `r` are related as `a=2r` The volume of the cubic unit cell =`a^(3)` `=(2r)^(3)` `=8r^(3)` Since a simple cubic unit cell contains only `1` atom the volume of the occupied space `=4//3pir^(3)` `:.` Packing efficiency `=("volume of one atom")/("Volume of the cubic unit cell")xx100%` ltb rgt `=((4)/(3)pir^(3))/(8r^(3))xx100%` `=(pi)/(6)xx100%` `= 52.4%` Here option `(2)` is not correct . In diamond carbon atom have fcc arrangement with one `C`-atom on each body diagonal (i.e, occupying alternate tetrahedral voids). Structure of diamond is similar to that of zince blende `(ZnS)` Net contribution of `C`-atoms on the corners. `=8xx(1)/(8)=1` Net contribution of `C`-atoms on face centres `=6xx(1)/(2)=3` Net contribution of `C`-atoms on `4` body diagonals `=4xx1=4` therefore total number `C`-atoms per unit cell of diamond is `1+3+4=8`. Hence option `(4)` is also not correct. |
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| 1002. |
With Which one of the following elements silicon should be doped so as to give p-type of semiconductor?A. GermaniumB. ArsenicC. SeleniumD. Boron |
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Answer» Correct Answer - 4 To get `p`-type of semiconductor, we should dope `Si` with electron deficient species such as boron that has fewer than four valance electron. Boron atoms replaces some silicon atoms. After a boron atom forms bonds to four silicon atoms, one of the boron silicon bonds has only one electron in it. We can think of this situation as a vacancy , or positively charge "hole" in the bonding orbital. An electron from a neighbouring atom can move to occupy this hole Now a hole exists on the neighbouring atom and an electron from another atom can move into it. and so forth.the result is the movement of positively charged holes in the semiconductor . Boron -doped sillicon is an example of a p-type semiconductor because the charge is carried by positive holes. Doping with electron rich impurities such as arsenic and selenium will result in an n-type semiconductor. |
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| 1003. |
Consider the following statements I. bcc structure has maximum packing efficiency II. Percentage of total space filled by the particles is called packing efficiency Which of the above statements is /are true ?A. Only IB. Only IIC. Both I and IID. Neither I nor II |
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Answer» Correct Answer - B bcc does not have maximum packing efficiency while percentage of total space filled by particles is packing efficiency |
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| 1004. |
Lead metal has a density of `11.34 g//cm^(3)` and crystallizes in a face-centered lattice. Choose the correct alternativesA. the volume of one unit cell is `1.214 xx 10^(-22) cm^(3)`B. the volume of one unit cell is `1.214 xx 10^(-19) cm^(3)`C. the atomic radius of lead is 175 pmD. the atomic radius of lead is 155.1 pm |
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Answer» Correct Answer - A::C Density `= (Z xx M)/(N_(A) xx "volume")` ltrbgt so, Volume `= (4 xx 207)/(6.02 xx 10^(23) xx 11.34) = 1.213 xx 10^(-22) cm^(3), 4r = asqrt2` `r = (4.95 xx 10^(-8) xx sqrt2)/(4) = 175 pm` Volume `= a^(3) = 1.213 xx 10^(-22)` ltrbgt So, `a = (1.213 xx 10^(-24))^(1//3) rArr a = 4.95 xx 10^(-8) cm` |
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| 1005. |
Which of the following statement (s) for crystal having schottky defect is/are correctA. Schottky defect arises due to absence of cation & anion from position which they are expected to occupyB. The density of crystal having shottky defect is smaller than that of perfect crystalC. Schottky defect are more common in co-valent compound with higher co-ordination numberD. The crystal having shottky defect is electrically neutral as a whole |
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Answer» Correct Answer - A::B::D Schottky defect is ony observed in ionic compound |
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| 1006. |
Given that interionic distance in `Na^(+), F^(-)` crystal is `2.31 Å and r_(F^(+)) = 1.36 Å`, which of the following predictions will be rightA. `r_(Na^(+))//r_(F^(-)) ~~ 0.7`B. coordination number of `Na^(+)` = coordinatin number of `F^(-) = 6`C. `Na^(+), F^(-)` will have rock salt type crystal structureD. effective nuclear charge for `Na^(+) and F^(-)` are equal |
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Answer» Correct Answer - A::B::C `Na^(+) " & " F^(-)` are isoelectronic hence they will have same screening const (s) but not the effective nuclear charge `:. R_(Na^(+))//r_(F^(-)) ~~ 0.7` (coordination = 6, rock structure) |
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| 1007. |
An ionic compound AB has fluorite type structres. If the radius `B^(-) is 200"pm, then the ideal radius of" `A^(+)` would be:A. 82.8pmB. 146.4pmC. 40pmD. 45pm |
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Answer» Correct Answer - D |
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| 1008. |
In a cubic,`A` atoms are present on alternative corners, `B` atoms are present on alternate faces, and `C` atoms are present on alternalte edges and body centred of the cube. The simplest formula of the compound isA. `A_(2)BC_(4)`B. `AB_(2)C_(4)`C. `ABC_(4)`D. `ABC_(2)` |
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Answer» Correct Answer - B Number of `A` atoms at alternate corners `= 4` corner `xx (1)/(8)` per corner share `= (1)/(2)` Number of `B` atoms at alternate face `= 2` faces `xx (1)/(2)` per face center share `= 1` Number of `C` atoms at alternate edges and at body centre `=` (`4` edges `xx (1)/(4)` per edge centre share) `+ 1` atom at body centre `= 1 + 1 = 2` Thus, formula `A_((1)/(2))B_(1)C_(2)` Simplifying, `AB_(2)C_(4)`. |
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| 1009. |
Name the parameters that characterise a unit cell . |
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Answer» A unit cell is characterised by i) its dimensions along the three edges , a, b, and c . ii) angles between the edges , `alpha` (between b and c) `beta` (between a and c ) and `gamma` (between a and b) . Thus , a unit cell is characterised by six parameters a , b , c, `alpha , beta and gamma` . Illustration of parameters of a unit cell : |
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| 1010. |
A crystalline solid `XY_(3)` has ccp arrangement for its element Y. The element X occupies :A. `66%` of tetrahedral holesB. `33%` of tetrahedral holesC. `66%` of octahedral holesD. `33%` of octahedral holes |
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Answer» Correct Answer - D The ratio of X and Y is `1:3` `:.` X will occupy `1//3` rd of octahedral voids `:.%` of octahedral voids occupied by X `=1/3xx100=33.3%` |
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| 1011. |
A perfect crystal is one in whichA. the free energy is minimumB. entropy is zero at room temperatureC. each atom is vibrating on its correct lattice position in the crystal structureD. each atom is in rest at its correct lattice position in the crystal structure |
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Answer» Correct Answer - 4 At zero Kelvin, most of the ionic crystals possess no defect as a perfect ordered arrangement of constituent particles in a crystal is possible only at `0 K` . In a crystalline solid, there is a regular arrangment of constitutent particles. However this arrangement is usually not found to be perfect, because a solid contains a large number of small crystals and some of these may not have a perfect regular arrangement of the constituent particles. Imperfection or defeat describes any departure from perfectly ordered arragnement of constituent particles in crystal. |
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| 1012. |
Which of the following defects is present in KCl crystals ?A. FrenkelB. SchottkyC. LinearD. Impurity |
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Answer» Correct Answer - B is the correct answer Both `K^(+)` and `Cl^(-)` ions have comparable ionic radii |
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| 1013. |
Crystals are imperfect because the presence of defects upto a certain concentrationA. increases `G` and decreases `S`B. decreases `G` and increase `S`C. increases both `G` and `S`D. decreases both `G`and `S` |
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Answer» Correct Answer - 2 On account of crystal defects ,`DeltaG` is `-ve` while `DeltaS` is `+Ve` . Defects may arise due to the heat absorbed by the crystals from the surrounding or due to the presence of impurities in the crystal. |
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| 1014. |
which of the following decribes atomic imperfections?A. Lattice imperfections extended along surfacesB. Lattice imperfections extended along linesC. Departures from the ordered and periodic arrangement in the vicnity of a particle or a group of particles.D. Deviation from periodicity extended over microscope regions of the crystals. |
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Answer» Correct Answer - 3 First one describes plane defects ,second one describe line defects (also called dislocations) while the fourth one describes lattice imperfections. |
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| 1015. |
following is the fact about the newly discover superconductor fo `C_(60)` (fullerence).the alkali mental fulleride superconductor `M_(3)C_(60)` has a cubic closesr - packed (face0centered cubic) arrangement of nearly spherical `C_(2-)^(60)` anions with `M^(+)` cations in the holes between the larger `C_(3-)^(60)` ions, the holes are of two types - octahedral hoels , which are surrounded octachedral by six `C_(3-)^(60)` ions. tetarahedral holes , which are surrounded tetrahedrally by for `C_(3-) ^(60)` ions . the ionic radii of `Na^(+), K^(+) and Rb^(+)` are 97 , 133 and 147 pm , respectivley . which of these ions will fit into the octanhedrral holes ? (radius of `C_(3-)^(60)` is about 350 pm )A. `Na^(+)`B. `K^(+)`C. `Rb^(+)`D. all of these |
| Answer» Correct Answer - C | |
| 1016. |
The structue of sodium chloride crystal is `:`A. body`-`centred cubic latticeB. face`-`centred cubic latticeC. octahedralD. square planar |
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Answer» Correct Answer - b `NaCl` has `f.c.c.` structure. Thus each unit cell as `4Na^(+)` and `4Cl^(-)` in it. It has `6-6` co`-`ordination no. |
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| 1017. |
A solid has a structure in which `W` atoms are located at the corrent of a lattice `O` atoms at the center of edges and `Na` atoms at the center of the cube.The formula for the compound isA. `NaWO_(2)`B. `NaWO_(3)`C. `Na_(2)WO_(3)`D. `NaWO_(4)` |
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Answer» Correct Answer - b in a unit cell, `W` atoms at the corner `= (1)/(8) xx 8 = 1` `O` atoms at the center of edge `= (1)/(4) xx 12 = 3` Na atoms at the center of the cube `= 1` `W : O : Na = 1 : 3: 1` , hence formula `= NaWO_(3)` |
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| 1018. |
The number of atom per unit in a simple cuhic, face - centered cubic and body - centered cubic are ….respectivelyA. `1,4,2`B. `4,1,2`C. `2,4,1`D. `4,8,2` |
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Answer» Correct Answer - a `sc= (8)/(8) = 1` `fc c = (8)/(8) + (6)/(2) = 4 bc c = (8)/(8) + (1)/(1) = 2` |
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| 1019. |
In a face contact cubic cell, combination of are atom at a face of the unit cell isA. `1//2`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - a Since half of the stone is inside the unit cell. |
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| 1020. |
In the above question, if an external magnetic field is applied, then the answer is `:`A. `(A)`B. `(B)`C. `(C)`D. `(D)` |
| Answer» Correct Answer - `-`do`-` | |
| 1021. |
The total number of lattic arrangements in different crystal system isA. `3`B. `8`C. `7`D. `14` |
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Answer» Correct Answer - c The seven basic crystal lattice arrangement are cubic tetragonal orthombic monoclinic hexagonal rhombohedral and triclinic. |
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| 1022. |
Which of the following gases are of paramagnetic nature ?A. OxygenB. `N_(2)O_(4)`C. `CO_(2)`D. `H_(2)` |
| Answer» `O_(2)` gas has two unparied electrons. | |
| 1023. |
The total number of lattic arrangements in different crystal system isA. 7B. 3C. 10D. 14 |
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Answer» Correct Answer - D The lattice arrangement (14) are also known as Bravais lattice. |
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| 1024. |
Which of the following materials is not ferromagnetic ?A. IronB. CobaltC. NickelD. Copper |
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Answer» Correct Answer - d Ferromagnetism is due to spontaneous alignment of unpaired electrons dipoles. `Cu` has only one unpaired electrons. |
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| 1025. |
For a solid with the structure shown in Fig, the coordination number of the points of the points A and , respectively are A. `6,8`B. `8,8`C. `6,6`D. `4,6` |
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Answer» Correct Answer - C It is fcc packing. The co-ordination numbers of the points A and B are 6,6 respectively |
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| 1026. |
When anion leaves the normal lattice site and electron occupies interstitial sites in its crystal lattice, It is called:A. Schottky defectB. Frenkel defectC. Metal excess defectD. Stoichiometric defect |
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Answer» Correct Answer - C |
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| 1027. |
For the structure of solid given below, if the lattice points represent `A^(+)` ions and the `B^(-)` ioins occupy the tetrahedral voids, then coordination number of `A` is `:` A. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - d It is fluorite type structure having co`-`ordination number 8 for `A`. |
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| 1028. |
For the structure given below, the site marked as `S` is a `:` A. tetrahedral voidB. cubic voidC. octahedral voidD. none of these |
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Answer» Correct Answer - c It shows octahedral voids `(` a type of cubic void in `f.c.c.` arrangement`)`. |
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| 1029. |
In a crystal, some ions are missing from normal sites. This is an example of `:`A. `F-` centresB. interstitial defectC. Frenkel defectD. Schottky defect |
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Answer» Correct Answer - d Schottky defects are arised when one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes. These are more common in ionic compounds with high co`-` ordination number and having almost similar size of cations and anions. |
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| 1030. |
For a solid with the following structure, the coordination number of the point `B` is `:` A. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - d `B` occupies octahedral voids and thus co`-`ordination number is 6 |
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| 1031. |
Which of the following statements is(are) correctA. In semi conductor , electrical resistivity decreases with a rise in temperatureB. In metallic conductor , electrical resistivity increases with a rise in temperatureC. In super conductor , electrical resistivity decreases gradually with the lowering temperatureD. A unit cell of an ionic crystal shares some of its ions with other unit cells. |
| Answer» Correct Answer - A::B::C::D | |
| 1032. |
Correct Statement is/areA. Individual crystals with their ordered atomic lattices exist but are so small as to be unrecognizable except under a microscope is said to be polycrystalline .B. A metallic sample may appear to be isotropic , even through single crystal is anisotropic and is said to be microcrystalline .C. The natural crystals of minerals formed by geological processes are often large when compared to crystals formed in the laboratory precipitation reactions (formed very rapidly from solution that are greately super saturated )D. As a result of continous redissolving and reprecipitation , a precipitate of virtually amorphos material can be converted to a polycrystalline substance. |
| Answer» Correct Answer - A::B::C::D | |
| 1033. |
Which of the following statement(s) is/are correct ?A. The coordination number of each type of ion in CsCl crystal is 8B. A metal that crystallizes in bcc structure has a coordination number of 12C. A unit cell of an ionic crystal shares some of its ions with other unit cellsD. The length of the unit cell in NaCl is 552 pm. `(r_(Na)=95 pm, r_(Cl^(-))=181 pm)` |
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Answer» Correct Answer - A::C::D The crystal to CsCl has bcc(body centred cubic) structure. In such an arrangement the coordination number of both is 8. in case of NaCl, two interpenetrating fcc crsytal lattices are present, out other of `Cl^(-)` ions only. Each `Na^(+)` ion is located half-way between two `Cl^(-)` ions and each `Cl^(-)` ion is located half way between two `Na^(+)` ions. in a unit cell of NaCl, `Cl^(-)` occupy corners as also the face centres and `Na^(+)` ions are located at octaheral voids. On each of a unit cell we have two `Cl^(-)` ions and one `N^(+)` ion. Hence, `a=2(r_(Na^(+))+r_(Cl^(-)))=2(95+181)`pm=552pm Hence (A,C,D) are correct options. |
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| 1034. |
`TiO_(2)` is well known example of `:`A. Triclinic systemB. Tetrahedral systemC. Monoclinic systemD. None of these |
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Answer» Correct Answer - B `TiO_2` has tetragonal system with five planes of symmetry and five axes of symmetry . |
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| 1035. |
Correct statement/s/is /areA. The crystal structure is observed by associating with each lattice point as identical structral moety.B. A unit cell can be thought of all the fundemental regions from which the entire crystal may be construted by purely translational displacements.C. An infininte number of different unit cells can described the same lattice . But we normally choose with sides have shortest lengths and that are mostlty perpendicular to one another .D. Unit cell are classified into seven crystal systems by repeating the rotational symmetry elements they posses . |
| Answer» Correct Answer - A::B::C::D | |
| 1036. |
Correct Statement is/areA. The high electrical conductivity of metals is readily explained if the valence electrons are free to move in an applied electric field .B. The high thermal conductivity of metals is consequence of free electrons , which can acquire large thermal kinetic energy move rapidly through the crystal and there by transport heat .C. In metals , because the valence electrons are not localised and the metallic bonding is not strongly directional , bonding forces need not completely disrupted when crystals is distorted .D. The electronic structure of metals differ from other substances in that the valence electrons of the metallic atoms are not localized at each atom , but belong to the crystal as a whole . |
| Answer» Correct Answer - A::B::C::D | |
| 1037. |
As regards magnetic behaviour , `TiO_(2)` is …….. |
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Answer» Correct Answer - ferromagnetic |
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| 1038. |
Cooling curves for amorophous and crystalline substances represented as follows Collect statement regards (A) and (B) graphsA. Graph (A) represents amorophous and Graph (B) represents crystal line substancesB. Graph(A) represents crystalline and graph (B) represents amorophous substancesC. In graph(B) , two points a , b represents beginning and end of the process of crystallation .D. In graph(B) , Temperature remains constant from `a rarr b` , because as the process of crystallisation is accompanied by some absorbation of energy , which campensates for the loss of heat and causes the temperature to remains constant . |
| Answer» Correct Answer - A::C::D | |
| 1039. |
Substances which show permanent magnetism even in the absence of magnetic field are called…. |
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Answer» Correct Answer - zero |
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| 1040. |
In a hexaonal system system of cycstals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are refular hexagons, and three atoms are sandwiched in between them. A space-cilling model of this structure, called hexagonal close-paked is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spherres are then placed overt the first layer so that they toych each other and represent the second layer so that they toych each other and present the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is convered with a third layer identical to the bottom layer in relative position. Assume the radius of every sphere to be `r`. The voume of this hcp unit cell isA. `24sqrt(2)r^(3)`B. `16sqrt(2)r^(3)`C. `12sqrt(2)r^(3)`D. `(64)/(3sqrt(3))r^(3)` |
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Answer» Correct Answer - A Height of unit cell `=4rsqrt((2)/(3))` Base area `=6xxsqrt((3)/(4)(4r)^(2)` volume=heightxx base area `4rsqrt((2)/(3))xx6xxsqrt((3)/(4))(2r)^(2)=24sqrt(2)r^(3)` Hence, (1) is correct option |
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| 1041. |
Magnetite is …….as regards magnetic behaviour . |
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Answer» Correct Answer - piezoelectricity |
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| 1042. |
Antiferromagnetic subtance have….. Magnetic moment. |
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Answer» Correct Answer - ferrimagnetic |
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| 1043. |
If the electrical resistance of a typical substance suddenly drops to zero then the subtance is calledA. ConductorB. SuperconductorC. InsulatorD. Semiconductor . |
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Answer» Correct Answer - B The substance which possesses zero resistance is called superconductor. |
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| 1044. |
Electrical resistance of mercury becomes almost zero atA. 20kB. 10kC. 25kD. 4k |
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Answer» Correct Answer - D Most of the metals have their transition temperature (i.e., the temperature at which a substance starts to behave as super conductor) in the range of 2-5k. |
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| 1045. |
When an electron in an excited Mo atom fall from L to k shell, an X -ray is emitted. These X -rays are diffracted at angle of `7.75^(@)` by planes with a separation of `2.64overset(@)A`. What is the difference in energy between K-shell and L-shell in Mo assuming a first-order diffraction `(sin 7.75^(@)=0.1349)`A. `36.88xx10^(-19)J`B. `27.88xx10^(-16)J`C. `63.88xx10^(-17)J`D. `64.88xx10^(-16)J`. |
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Answer» Correct Answer - B `2d sin theta=n lambdarArrlambda=2d sin theta=2xx2.64xx10^(-10)xxsin7.75^(@)` `=0.7123xx10^(-10)m` `E=(hc)/(lambda)=(6.62xx10^(-34)xx3xx10^(8))/(0.7123xx10^(-10))=27.88xx10^(-16)J` |
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| 1046. |
Calculate the distance between 111 planes in a crystal of Calculate the distance between 111 planes in a crystal of Ca. the answer isA. 1.61nmB. 0.610nmC. 0.321 nmD. None of these |
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Answer» Correct Answer - C We have, `d=(a)/(sqrt(h^(2)+k^(2)+I^(2))):d_(111)=(0.556)/(sqrt(I^(2)+I^(2)+I^(2)))` =0.321 nm. |
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| 1047. |
How do the spacings of the threee planes `100`, `110`, and `111` of cube lattice vary? |
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Answer» Applying the formula `d_(hkl) = (a)/(sqrt((h^(2) + k^(2) + l^(2)))` `d_(100) = (a)/(sqrt((1^(2) + 0^(2) + 0^(2))))= a` `d_(110) = (a)/(sqrt((1^(2) + 1^(2) + 0^(2))))= (a)/(sqrt2)` `d_(111) = (a)/(sqrt(1^(2) + 1^(2) + 1^(2)) )= (a)/(sqrt3)` Thus, `d_(100) : d_(110) : d_(111) : = 1 : (1)/(sqrt2) : (1)/(sqrt3) = 1 : 0.707 : 0.577` |
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| 1048. |
Calculate the miller indices of crystal. Planes which cut through the crystal axes atA. `(2a, 3b, c)`B. `(a, b, c)`C. `(6a, 3b, 3c)`D. `(2a, -3b, -3c)` |
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Answer» Following the procedure given above, we prepare the tables as follow: i. `{:(a,b,c),(2,3,1),(1//2,1//3,1),(3,2,6):}` Hence, the miller indices `(326)` ii. `{:(a,b,c),(1,1,1),(1,1,1),(1,1,1):}` Hence, the miller indices are `(111)`. iii. `{:(a,b,c),(6,3,3),(1//6,1//3,1//3),(1,2,2):}` Hence, the miler indices are `(122)`. iv. `{:(a,b,c),(2,-3,-3),(1//2,-1//3,-1//3),(3,-2,-2):}` Hence, the miller indices are `(3overline(22))`. |
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| 1049. |
The number of tetrahedral voids in the unit cell of a face- centred lattice of similar atoms isA. 4B. 6C. 8D. 12 |
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Answer» Correct Answer - B A face- centred cubic lattice has 4 atom per unit cell in the packing . No. of tetrahedral voids is double the number of atoms in the packing . |
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| 1050. |
silicon doped with electron rich impurity forms ………. .A. p-type semiconductorB. n-type semiconductorC. intrinsic semiconductorD. insulator. |
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Answer» Correct Answer - B n-type semi-conductors are formed because of delocalised electrons. |
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