InterviewSolution
Saved Bookmarks
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.
| 1651. |
Which arrangement of electron decides ferrimagnetism?A. `uarr uarr uarr uarr uarr `B. `uarr darr uarr darr`C. `uarr uarr uarr darr darr`D. None of these |
|
Answer» Correct Answer - c Ferrimagnation involves magnatic dipoles crientted in parallel and iniprallel direction it unequal number to give some net dipole moment |
|
| 1652. |
What type of stoichiometric defect is shown by: (a) `ZnS` (b) `AgBr` |
|
Answer» (i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size. AgBr crystals may show both Frenkel and Schottky defects. |
|
| 1653. |
What type of stoichiometric defect is shown by ZnS |
| Answer» ZnS shows Frenkel defect because its ions have large difference in size | |
| 1654. |
In closest packing of A type of atoms (radius `r_(A)`) the radius of atom B that can be fitted into octabedral voids isA. `1.155 r_(a)`B. `0.255 r_(a)`C. `0.414 r_(a)`D. `0.732 r_(a)` |
|
Answer» Correct Answer - C |
|
| 1655. |
Define the terms (i) F-centres (ii) n-type semi-conductors (iii) Ferrimagnetism. |
|
Answer» (i) F-centres (ii) n-type semi-conductors (iii) Ferrimagnetism |
|
| 1656. |
In the closest packing of atomsA. The size of `TV` is greater than that of `OV`.B. The size of `TV` is smaller than that of `OV`.C. The size of `TV` is equal to that of `OV`.D. The size of `TV` may be greater or smaller or equal to that of `OV` depending upon the size of atoms. |
|
Answer» Correct Answer - B For a given radius of anion `(r_(ɵ))` Radius ratio for `OV` and `TV` is as follows: `((r_(o+))/(r_(ɵ)))_(OV) gt ((r_(o+))/(r_(ɵ)))_(TV)` `[{:("For" OV,(r_(o+))/(r_(ɵ))=0.414-0.732),("For" TV,(r_(o+))/(r_(ɵ))=0.225-0.414):}]` Hence, size of `OV` is larger than that of `TV`. |
|
| 1657. |
In closest packing of A type of atoms (radius `R_(A)`), the radius of atoms B that can be fitted octahedral void isA. `0.155r_(A)`B. `0.125R_(A)`C. `0.414r`D. `0.732r_(A)` |
|
Answer» For octahderal void `(r_(B))/(r_(A))=0.414` or, `r_(B)=0.414r_(A)` |
|
| 1658. |
What is the co-ordination number of sodium in `Na_(2)O` ?A. 6B. 4C. 8D. 2 |
|
Answer» Correct Answer - B `Na_(2)O` has antifluorite structure `O^(2-)` ions constitute ccp lattice while `Na^(+)` ions occupy the tetrahedral voids. Each `O^(2-)` ion is in contact with 8 `Na^(+)` ions and each `Na^(+)` ion is in contact with `4O^(2-)` ions. The C.N. of sodium in `Na_(2)O=4` |
|
| 1659. |
For crystal sodium chloride, state (i) the type of lattice in which it crystallises. (ii) co-ordination number of each sodium ion and chloride ion in the crystal lattice. (iii) number of sodium and chloride ions present in a unit cell of sodium chloride. (iv) the structural arrangement of sodium chloride crystals. |
|
Answer» (i) Sodium chloride crystallises in expanded face centred cubic arrangement also known as cubic close packed arrangement. (ii) Co-ordination number of `Na^(+)` and `Cl^(-)` ions are in the ratio `6:6`. (iii) Unit cell of NaCl has four `Na^(+)` and four `Cl^(-)` ions. (iv) The structural arrangement of sodium chloride crystals is octahedral. |
|
| 1660. |
Co-ordination number for sodium metal isA. 11B. 12C. 8D. 10 |
| Answer» Correct Answer - C | |
| 1661. |
If the ratio of coordination no. of A to that iof B is x:y, then the ratio of no. of atoms of A to that no, of atoms of B in the unit cell isA. x:yB. y:xC. `x^(2):y`D. `y:x^(2)` |
|
Answer» Correct Answer - B |
|
| 1662. |
Statement `:` Due to Frenkel defect the density of the crystalline solid remains same. Explanation `:` In Frenkel defect, no cations or anions leave the lattice. |
|
Answer» Correct Answer - d |
|
| 1663. |
Assertion (A) : `CsCl` crystal, the coordination number of `Cs^(o+)` ion is `8`. Reason (R ) : `Cl^(ɵ)` ion in `CsCl` adopt `bcc` type of packing,A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
|
Answer» Correct Answer - C Correct Reason `Cs^(+)` ions in CsCl occupy bcc type close packing |
|
| 1664. |
Statement `:` Due to Frenkel defect the density of the crystalline solid remains same. Explanation `:` In Frenkel defect, no cations or anions leave the lattice.A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT -1B. If both the statement are TRUE and STATEMENT -2 is NOT the correct explanation of STATEMENT -1C. If STATEMENT -1 is the correcct and TRUE and STATEMENT -2 is FALSED. If STATEMENT -1 is the correcct and FALSE and STATEMENT -2 is TRUE |
|
Answer» Correct Answer - A |
|
| 1665. |
Statement `:` Due to Frenkel defect the density of the crystalline solid remains same. Explanation `:` In Frenkel defect, no cations or anions leave the lattice.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
| Answer» Correct Answer - A | |
| 1666. |
Which of the following is not true about the voids formed in 3 dimensional hexagonal close packed structure?A. A tetrahedral voids is formed when a sphere of the second layer is present above triangular void in the first layerB. All the triangular voids are not covered by the spheres of the second layerC. Tetrahedral voids are formed when the triangular voids in the second layer lie above the triangular voids in the first layer and the triangular shapes of these voids do not overlapD. Octahedral voids are formed when the triangular voids in the second layer exactly overlap with similar voids in the first layer |
| Answer» Correct Answer - C::D | |
| 1667. |
An element has atoic mass ` 93 " g mol" ^(-1)` and density ` 11.5 " g cm" ^(-3)` . If the edge length of its unit cell is 300 pm, identify the type of unit cell. |
|
Answer» Correct Answer - ` 38 . 4 g cm ^(-3) ` `p = (Z xx M)/(a^(3)xxN_(0)) , Z = ( p xx a^(3) xx N_(0))/M = (( 11.5 " g cm"^(-3)) (300xx 10^(-10)"cm")^(3)(6.023 xx10^(23) "mol"^(-1)))/(93" mol"^(-1)) = 2.01 =2 ` Hence, the type of unit cell in BCC. |
|
| 1668. |
Calculate the following: a. Number of `Zns` units in a unit cell of zine blende. b. Number of `CaF_(2)` unit cell of `CaF_(2)`. |
|
Answer» Correct Answer - 4 |
|
| 1669. |
What is the co-ordination number of sodium in `Na_(2)O` ? |
|
Answer» Correct Answer - 4 |
|
| 1670. |
In the closest packing of atomsA. Coordination number of particles placed in tetrahedral voids is smaller than octahedral voids.B. Size of tetrahedral void is larger than that of octahedral voidC. Size of voids depend upon size of atoms of atoms and tetrahedral void is smaller than octahedral voidD. Radius ratio for tetrahedral voids is smaller than octahedral void |
| Answer» Correct Answer - A::C::D | |
| 1671. |
Statement `:` Schottky defect is generally shown by the compounds with high co`-` ordination no. Explanation `:` Equal no. of cations and anions are missing from the lattice sites in Schottky defect.A. If both assertion and reason are correct and reason is correct explanation for assertionB. If both assertion and reason are correct but reason is not correct explanation for assertionC. If assertion is correct but reason is incorrectD. If both assertion and reason are incorrect |
|
Answer» Correct Answer - B Correct Explanation In compounds showing schottky defect, the cations are generally large in size |
|
| 1672. |
The cordination number of a metal crystallising in a hexagonal close-packed structure is:A. 12B. 4C. 8D. 6 |
|
Answer» Correct Answer - A In hcp, co-ordination number is 12. |
|
| 1673. |
In NaCl is doped with ` 10^(-4)` mol % of ` ScCl_(2)` , the concentration of cation vacancies will be ` (N_(A) = 6.02 xx 10^(23) mol^(-1))`A. ` 6.02 xx 10^(14) mol^(-1)`B. `6.02 xx 10^(15) mol^(-1)`C. ` 6.02 xx 10^(16) mol^(-1)`D. ` 6.02 xx 10^(17) mol^(-1) ` |
|
Answer» Correct Answer - b For each ` Sr^(2+)` ion introduced, one catio vacancy is created because ` 2 Na^(+)` ions are removed and one vacant site is occupied by ` Sr^(2+)` . Doping with ` 10^(-4) "mol"% " of " SrCl_(2)` means 100 moles of NaCl are doped with ` 10^(-4)` mole of ` SrCl_(2)` `SrCl_(2)` doped per mole of NaCl = `10^(-4) //100` ` 10^(-6)` mole =`10^(-6) xx ( 6.02 xx 10^(23)) Sr^(2+)` ions ` = 6.02 xx 10^(17) Sr^(2+)` ions Hence, concetration of cation vacancies ` 6.02 xx 10^(17) "mol"^(-1)` |
|
| 1674. |
KF and NaCl struture. If the distance between ` K ^(+) and F^(-)` is 269 pm, find the denisty of KF ( ` N_(A) = 6.02 xx 10^(23) "mol"^(-1)` a atomic mass of copper = 63.5 |
|
Answer» Correct Answer - `8.9 " g cm"^(-3)` |
|
| 1675. |
Nickle crystallise in a fcc unit cell with a cell edge length of 0.3524 nm. Calculate the radius of the nicke atomA. 0.1624 nmB. 0.1246 nmC. 0.2164 nmD. 0.1426nm |
|
Answer» Correct Answer - B For an fcc lattice , `r=(a)/(2sqrt2)=(0.3524)/(2sqrt2)` = 0.1246nm` |
|
| 1676. |
An element having an atomic radius of 0.14 nm crystallises in a fcc unit cell. What is the length of side of the cell ?A. 0.56 nmB. 0.24 nmC. 0.96 nmD. 0.4 nm |
|
Answer» Correct Answer - D For fcc unit cell , `r=(sqrt2)/(4)a` `rArr " " 0.14=(sqrt2)/(4)a` `a=(4xx0.14)/(sqrt2)=0.396nm=0.4nm` |
|
| 1677. |
Iron crystallises in a bcc system with a lattice parameter of 2.861 Å. Calculate the densityof iron in the bccc system (atomic weight of Fe = 56, `N_A=6.02xx10^23 mol^(-1)`A. `7.94 gcm^(-3)`B. `8.96 g cm^(-3)`C. `2.78 gcm^(-3)`D. `6.72 gcm^(-3)` |
|
Answer» Correct Answer - A `d=(ZM)/(N_Aa^3)` (for bcc, Z=2) `d_(Fe) = ((2)xx56.0gmol^(-1))/((6.02xx10^23mol^(-1)(2.861xx10^(-8))^3cm^3)` |
|
| 1678. |
Select the correct statement (s).A. Co-ordination no. of an atom at a lattice point in sample cubic arrangement is 6B. Co- ordination no. of an atom at octahedral site 8.C. Co-ordination no. of an atom at lattice point in hcp arrangment is 6D. Co-ordination no.of an atmo at octahedral site is 6 |
|
Answer» Correct Answer - A::B |
|
| 1679. |
In diamond structure ,carbon atoms form fcc lattic and `50%` tetrahedral voids occupied by acrbon atoms . Evergy carbon atoms is surrounded tetrachedral by four carbon atom with bond length 154 pm . Germanium , silicon and grey tin also crystallise in same way as diamond `(N_(A)=6xx 10^(23))` The side length of diamond unit cell is `("in pm")`:A. 154B. 1422.63C. 711.32D. 355.66 |
|
Answer» Correct Answer - D |
|
| 1680. |
In diamond structure ,carbon atoms form fcc lattic and `50%` tetrahedral voids occupied by acrbon atoms . Evergy carbon atoms is surrounded tetrachedral by four carbon atom with bond length 154 pm . Germanium , silicon and grey tin also crystallise in same way as diamond `(N_(A)=6xx 10^(23))` The density of daimond unit cell is `("in " gm//cm^(3))`A. 28.48B. 0.0556C. 0.445D. 3.56 |
|
Answer» Correct Answer - D |
|
| 1681. |
In diamond structure ,carbon atoms form fcc lattic and `50%` tetrahedral voids occupied by acrbon atoms . Evergy carbon atoms is surrounded tetrachedral by four carbon atom with bond length 154 pm . Germanium , silicon and grey tin also crystallise in same way as diamond `(N_(A)=6xx 10^(23))` The mass of diamond unit cell is:A. ` 96 amu`B. `96 gm`C. `144 amu`D. `144gm` |
|
Answer» Correct Answer - A |
|