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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.
| 851. |
Why is phosphours doped silicon a semi-conductor? |
| Answer» Phosphorus is pentavalent and silicon is tetravalent. Silicon doped with phosphours results in the foumation of p-type semi-conductor. They leads to increase in its electrical conductivity. | |
| 852. |
The `Ew` of an element is `13`. It forms an acidic oxide which `KOH` forms a salt isomorphous with `K_(2)SO_(4)`. The `Aw` of element is a. `13`, b. `26`, c. `52`, d. `78` |
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Answer» `MSO_(4)` and `K_(2)SO_(4)` are isomorphous. Thus, valency of `S` and `M` should be same `= 6`. `:. Aw = Ew xx "Valency"` `= 13 xx 6 = 78`. |
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| 853. |
Which of the following statements are not true ?A. Vacany defect results in a decrease in the density of the substanceB. Interstitial defects results in a n increase in the density of the substaceC. Impurity defect has no effect on the density of the substanceD. Frenkel defect results in an increase in the density of the substance |
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Answer» Correct Answer - C,D Statements(C) and (D) can be correctly written as (C) Impurity defect cahgnes the density of substance as impurity has different than the ion present on prefect crystal e.g., when `SrCl_(2)` is added to the NaCl crystal, it causes impurity defect, (d) Frenkel defect results neither decrease nor increase in density of substance. |
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| 854. |
In an ionic compound `A^(+)X^(-)` , the radii of ` A^(+) and X^(-)` ions are 1.0 pm and 2.0 pm rspecitvely . The volume of the unit cell of the crystal AX will beA. `27 "pm"^(3)`B. ` 64 "pm"^(3)`C. ` 125 "pm"^(3)`D. ` 216 "pm"^(3)` |
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Answer» Correct Answer - C ` (r_(A^+))/(r_(X^(-))) = 1/2 = 0.5 `,As it lies in the range that of NaCl. 0.414 - 0.732 , AX has octahedral structure like that of NaCl. Hence, edge length `= (a) =2 (r_(+) + r_(-))` = 2 (1 + 2) pm = 6 pm Volume of the unit cell = ` a^(3) = ( 6 "pm")^(3) = 216 "pm"^(3)` |
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| 855. |
Which one of the following statements are true about metals?A. Valence band overlap with conduction bandB. The gap between valence band and conduction bank is negligibleC. The gap between valence badn and conduction band cannot be determinedD. Valence band and conduction badn cannot be determined |
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Answer» Correct Answer - A,B,D In metal, valence badn overlap with conduction band. The gap between valence badn and conduction band is negligible and valence band may remain partially filled. |
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| 856. |
A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion `(Y^(-))` will beA. 275.1B. 322.5 pmC. 241. 5 pmD. 165. 7 pm |
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Answer» Correct Answer - B NaCl and face-centred cubic arrangement of ` Cl^(-)` ions and `Na^(+)` ions are present in the octahedral voids. Hence, for such a solid , radius , of cation =` 0. 414 xx " radius of the anion " ( or (r_(+))/(r_(-)) = 0.414)` i.e, ` r_(+) = 0.414 xx r_(-)` `or = r_(-) = ( r_(+))/(0.414) = 100/(0. 414) = 241. 5 "pm" ` |
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| 857. |
A compound is formed by elements `A` and `B`. This crystallises in the cubic structure when atoms `A` are at the corners of the cube and atoms `B` are at the centre of the body.The simplest formula of the compound isA. ABB. `AB_(2)`C. `A_(2)B`D. `AB_(4)` |
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Answer» Correct Answer - A A atoms are at eight corners of the cube. Therefore, the number of A atoms in the unit cell `=(8)/(8)=1`, atoms B per Unit cell =1. Hence the ddformula is AB. |
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| 858. |
Body centred cubic lattice has a coordination number ofA. 4B. 8C. 12D. 6 |
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Answer» Correct Answer - B Body centred cubic lattice has a co-ordination number 8. |
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| 859. |
In a multilayered close - packed stractureA. there are twice as many tetrahedral holes as there are closed packed atomsB. there are as many tetrahedral holes as there are closed packed atomsC. there are twice as many octahedral holes as there are closed packed atomsD. there are as many tetrahedral holes as there are octahdralpacked atoms |
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Answer» Correct Answer - a No. of ocubachdral holes = Na of close packet atoms No. of ocubachdral holes` = 2 xx` No. of close packet atoms. |
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| 860. |
The number of octahedral and terhedral sides in a cubical closed packed array of `N` spheres respectively isA. `N and 2N`B. `N//2 and N`C. `2N and N`D. `4N and 2N` |
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Answer» Correct Answer - a Number of octabedral sites = numer of sphares in packing Number of octabedral volts `2= xx` numer of sphares in packing Number of octabedral volts `2= xx` oculadral volts so, the answer is `N and 2N`. |
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| 861. |
All nobe gases crystallise in the ccp kstructure exceptA. HeliumB. NeonC. ArgonD. Krypton |
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Answer» Correct Answer - A All noble gas crytallizes in the ccp structure except He which has hcp structure. |
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| 862. |
The following is not a function an impurity present in a crystalA. Establishing thermal equilibriumB. Having tendency to diffuseC. Contritbuting to scatteringD. Introducing new electronic energy levels. |
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Answer» Correct Answer - A Inpurity present in a crystal does not establish, thermal equilibrium. |
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| 863. |
Which one of the following crystal does not exhibit Frenkel defect?A. AgBrB. AgClC. KBrD. ZnS. |
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Answer» Correct Answer - C KBr exhibits Schottky defect and not Frenkel defect. |
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| 864. |
Which of the following crystal lattice has the miniumuum empty space?A. simple cubicB. Body centred cubicC. Face centred cubicD. Simple tertragonal |
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Answer» Correct Answer - C |
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| 865. |
Which of the following statements is not correct.A. The units of surface tension are dynes `cm^(-1)`B. The units of viscosity coefficient of a liquid are "Poise"C. CsCl crystallizes in body central cubic type of lattice.D. The coordination number of `S^(2-)` in ZnS is 6 |
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Answer» Correct Answer - D In zinc blende (ZnS) half number of tetrahedral holes are filled by zinc atoms. |
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| 866. |
The empty space availablein scc crystal lattice isA. `5.87r^(3)`B. `3.81r^(3)`C. `4.37r^(3)`D. `3.94r^(3)` |
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Answer» Correct Answer - B Empty space = Total volume - volume occupied by 1 atom `=((4)/(sqrt(3))r)^(3)-(2xx(4)/(3)pir^(3))=3.81r^(3)`. |
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| 867. |
The space in which atoms are not present in unit cell isA. In sc `48%`B. In fcc `26%`C. In bcc `32%`D. In hexagonal `26%` |
| Answer» Correct Answer - A::B::C::D | |
| 868. |
Which of the following having having their radius ratio between`0.414` and `0.732`, i.e., for `NaCl` structure, have their radius ratio not in this range but posses `NaCl`-type structure?A. LiBrB. KClC. RbClD. BaO |
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Answer» Correct Answer - A::B::C::D For `NaCl, r_(o+)//r_(ɵ) = 0.414 - 0.732` But radius ratio for `LiBr, KCl, RbCl`, nd `BaO` are `0.34, 0.38, 0.77` and `0.83`, respectively, which are not in the range of `0.414 - 0.732`. |
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| 869. |
In the fluorite structure if the radius ratio is `(sqrt(3)/(2)-1)` how many ions does each cation touch?A. `4` anionsB. `12` cationsC. `8` anionsD. No cations |
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Answer» Correct Answer - B::C `(r_(o+))/(r_(ɵ)) = sqrt(3/2) - 1 = 0.225` Hence, it is the limiting case where cation in the void of fcc structure is not distorted. So, number of cations surrounding the particular cation `= 12`. But at the same time `8` anions (present in `TVs`) touch the particul,ar cation. |
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| 870. |
The volume of atom present in a face-centred cubic unit cell of a metal (`r` is atomic radius ) isA. `(16)/(3)pir^(3)`B. `(20)/(3)pir^(3)`C. `(24)/(3)pi^(3)`D. `(12)/(3)pir^(3)` |
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Answer» In fcc, no of atoms=4 volume occupied atoms `=4[(4)/(3)pir^(3)]=(16)/(3)pir^(3)` |
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| 871. |
The number of octahedral sites per sphere in fcc structure isA. 8B. 4C. 2D. 1 |
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Answer» Correct Answer - C |
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| 872. |
If the ratio is in the range of `0.414- 0.732` , then the coordination number will beA. `2`B. `6`C. `4`D. `8` |
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Answer» Correct Answer - b The radius ratio for coordation number `4,6` and `8` lies in between the rabgs `0.225- 0.414, 0.141- 0.732 and 0.732 - 1` respectively. |
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| 873. |
If the ratio is in the range of `0.414- 0.732` , then the coordination number will beA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - C The radius ratio for coordunation number 4, 6 and 8 lies in between the ranges `0.225 - 0.414` `0.414 - 0.732` and `0.732 - 1.000` respetively . |
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| 874. |
The packing fraction for a body-centred cube isA. `0.42`B. `0.53`C. `0.68`D. `0.82` |
| Answer» Correct Answer - C | |
| 875. |
The number of unit cells in the `Ca` atom lies on the surface of a cubic crystal that is `1.0 cm` in length isA. `9.17 xx 10^(23)`B. `9.17 xx 10^(22)`C. `2 xx 9.17 xx 10^(23)`D. `2 xx 9.17 xx 10^(22)` |
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Answer» Correct Answer - B Volume of crystal `= (1.0 cm)^(3) = (10^(-2)m)^(3) = 10^(-6) m^(3)` Number of unit cells `= (10^(-6)m^(3))/(10.9 xx 10^(-3) m^(3)) = 9.17 xx 10^(22)` |
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| 876. |
Cesium chloride forms a body-centred cubic lattice. Cesiumm and chloride ions are in contact along the body diagoanl of a cell. The length of the side of the unit cell is `412 pm` and `Cl^(Θ)` ion has a radius of `181` pm. Calculate the radius of `Cs^(o+)` ion. |
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Answer» Body diagonal `= sqrt((sqrt2a)^(2) + a^(2)) = sqrt3a` `= 2(r_(Cs^(o+)) + r_(Cl^(Θ)))` `= 1.732 xx 412` pm `= 713.58` pm `:. r_(Cs^(o+)) = (713.58)/(2)-r_(Cl^(Θ))` `= (356.79 - 181)` pm `= 175.8` pm |
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| 877. |
For a cubiccrystal, the face diagonal is `3.5 Å`. Calculate the face length. |
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Answer» Face diagonal `= sqrt(a^(2) + a^(2)) = sqrt2a` Face length `(a) = ("Face diagonal")/(sqrt(2)) = (3.50 Å)/(sqrt(2)) = (3.50 Å)/(1.414)` `= 2.47 Å` |
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| 878. |
If the length of the body for `CaCl` which crystallizes into a cubic strructure with `Cl^(Θ)` ions at the corners and `Cs^(o+)` ions at the centre of the unit cell is `7 Å` and the radius of he `Cs^(o+)` ions is `1.69Å`, what is the radius of `Cl^(ɵ)` ions? |
| Answer» Length of the body diagonal `= 2(r_(Cl^(Θ)) + r_(Cs^(o+)))` | |
| 879. |
A compound is formed by elements `A` and `B`. This crystallises in the cubic structure when atoms `A` are at the corners of the cube and atoms `B` are at the centre of the body.The simplest formula of the compound isA. `AB`B. `AB_(2)`C. `A_(2)B`D. `AB_(4)` |
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Answer» Correct Answer - 1 Exclusive contribution atoms `A` per unit cell `=8xx1//8=1` Exclusive contribution of atoms `B` per unit cell `=1xx1=1` Thus the empiritical fourmula is `AB`. |
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| 880. |
On doping `Ge` with a little of `In or Ga` one getsA. `p`-type semiconductorB. `n`-type semiconductorC. insulatorD. rectifier |
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Answer» Correct Answer - 1 The electrical conductivity of a semiconductor can be greatly increased by doping with impurities. Doping the Ge crystal with`In` or `Ga` (Group `13` element) produces is `p` type semiconductor. Each In or `Ga` atom contributes only three valence electrons to bonding orbitals in the valence band, and therefore a hole localized near each `In` or `Ga` atom. Thermal energy is enough to separate the negatively charge `In` or `Ga` atom form the hole, delocalizing the latter in this case the charge carries are the holes , which are positive and the crystal is doped p-type. |
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| 881. |
On doping `Ge` with a little of `In or Ga` one getsA. n-type semi - conductorB. n-type semi - conductorC. insulatorD. rectifer |
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Answer» Correct Answer - B Ge is a group 14 element, while in is a group 13 element. On doping certain holes will be created to form p-type semi-conductors |
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| 882. |
Consider a corner atom of Ist layer of an HCP unit cell showing alternate . A A layers . Arrange the distances in ascending order . A. `d_(a) lt d_(b) = d_(f) = d_(g) lt d_(d)`B. `d_(a) gt d_(b) = d_(f) = d_(g) gt d_(D)`C. `d_(a) lt d_(b) lt d_(f) lt d_(8) lt d_(g)`D. `d_(a) gt d_(b) gt d_(f) gt d_(g) gt d_(D)` |
| Answer» Correct Answer - A | |
| 883. |
Extremely pure samples of `Ge` and `Si` are non`-`conductors, but their conductivity increases suddenly on introducting `…………………….` in their crystal latticeA. arsenicB. boronC. both `(a)` and `(b)`D. none of these |
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Answer» Correct Answer - c Doping of elements of group `14(Ge ` and `Si)` with group `15 (As)` elements produces excess of electrons and show `n-` type conduction, the symbol `n` indicating flow of negative charge in them. Doping of elements of group `14(Ge` and `Si)` with group `13(B)` elements produces hole `(` electron deficiency `)` in the crystal and shows `p-`type conduction, the symbol `p` indicating flow of positive charge. |
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| 884. |
The shortest distance between `I^(st) and V^(th)` layer of HCP arrangement is:A. `8 sqrt((2)/(3)) r`B. `4sqrt((3)/(2))r`C. `16(sqrt12)/(3)r`D. `8sqrt((3)/(2))r` |
| Answer» Correct Answer - A | |
| 885. |
The crystals are bonded by plane faces `(f)` straight edges `(e)` and interfacial angle `(c)`. The relationship between these is `:`A. `f+c=e+2`B. `f+e=c+2`C. `c+e=f+2`D. none of these |
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Answer» Correct Answer - a `f+c=e+2,` where `f` is plane face, `c` is interfacial angle and `e` is straight edges. |
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| 886. |
Select the correct statement(s).A. Solids with `F`-centres are paramagnetic.B. Ferrimagnetic character of `Fe_(3)O_(4)` at room temperature changes to paramagnetic character at `850 K`.C. Anti-ferrimagetic `V_(2)O_(3)` changes to paramagnetic at `150 K`.D. Non-stoichiometric `Cu_(2)O` is a `p`-type semiconductor. |
| Answer» Correct Answer - A::B::C::D | |
| 887. |
Volume of HCP unit cell is:A. `24 sqrt2 r^(3)`B. `8sqrt2 r^(3)`C. `16 sqrt2 r^(3)`D. `24 sqrt3 r^(3)` |
| Answer» Correct Answer - A | |
| 888. |
Select the correct statement(s) about three-dimensional `hcp` system.A. The number of atoms in `hcp` unit cell is six.B. The volume of `hcp` unit cell is `24sqrt2r^(3)`.C. The empty space in `hcp` unit cell is `26%`.D. The base area of `hcp` unit is `6sqrt3r^(2)`. |
| Answer» Correct Answer - A::B::C::D | |
| 889. |
In which of the following systeams interfacial angles `alpha = gamma = 90^(@)` but `beta != 90^(@)`?A. MonoclinicB. RhombohedralC. TriclinicD. Hexagonal |
| Answer» Correct Answer - A::B | |
| 890. |
In which of the following systeam primitives `a != b != c`?A. OrthorhombicB. MonoclinicC. TriclinicD. Hexagonal |
| Answer» Correct Answer - A::B::C | |
| 891. |
Why FeO in non-stochiometric with the formula `Fe_(0.95)O` ? |
| Answer» Some `Fe^(2+)` ions are rpelaced by ` Fe^(3+) ` ions :` 3 Fe^(2+) -= 2 Fe^(3+)` to maintain electrical neutrality | |
| 892. |
Ferromagnetic Substances |
| Answer» Because of the re-orientation of the electron spins. | |
| 893. |
If `r_(o+)//r_(ɵ)` for a crystal is `0.50`, it has ……….. structure. |
| Answer» Correct Answer - Octahedral | |
| 894. |
The compounds having the general molecular formula `A^(2+)Fe_(2.04)` are called ……….. . |
| Answer» Correct Answer - Ferrites | |
| 895. |
Unlike paramagenetic substance, ferromagnetic substances show ………. even if the external magnetic field is removed. |
| Answer» Correct Answer - Permanent magnetism | |
| 896. |
The number of atoms touching a particular atom in a crystal is called its ………. |
| Answer» Correct Answer - Coordination number | |
| 897. |
The `ABC ABC` type packing is called ……….. . |
| Answer» Correct Answer - Cubic close packing `(ccp)` | |
| 898. |
Moldydenum `("At. mass=96g//mol"^(-1))` crystallizes as bcc crystal. If density of crystal is `10.3g//cm^(3)`, then radius of Mo atoms `(use N_(A)= 6 xx 10^(23))`:A. 111PMB. 314PMC. 135.96PMD. none of these |
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Answer» Correct Answer - C |
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| 899. |
The number of atoms present per unit cell in simple, fcc and bcc are ………, ……….., and …………, respectively. |
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Answer» Correct Answer - 2 |
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| 900. |
An element `X (At,wt = 80 g//mol)` having fcc structure, calculate the number of unit cells in `8g of X`A. `0.4xxN_(A)`B. `0.1 xxN_(A)`C. `4 xxN_(A)`D. none of these |
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Answer» Correct Answer - D |
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