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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.
| 1551. |
Packing refers to the arrangement of constituent units in such a way that the forces of attraction among the constituent particles is the maximum and the contituents occupy the maximum available space. In two dimensions, there are hexagonal close packing and cubic close packing. In three dimentions, there are hexagonal, cubic as well as body centred close packings. A certain oxide of a metal M crystallises in such a way that `O^(2-)` ions occupy hcp arrangement following ABAB... pattern. The metal ions however occupy `2//3` rd of the octahedral voids. The formula of the compound is :A. `M_(2)O_(3)`B. `M_(3)O`C. `M_(8//3)O_(3)`D. `MO_(2)` |
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Answer» Correct Answer - A Let the no. of metal atoms (M) in hcp lattice `=n` The no. of octahedral voids is the same as the no. of atoms in close packed arrangement `:.` No of octahedral voids =n Since only `2//3` of these voids are occupied by metal atoms M `:.` No of metal atoms in hcp lattice `=(2n)/3` Ratio of `M:=2 n//3 n=2//3: 1=2:3` `:.` Formula of metal oxide `=M_(2)O_(3)` |
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| 1552. |
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 empty space in this hcp unit cell isA. `74%`B. `97.6 %`C. `32%`D. `26%` |
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Answer» Correct Answer - D Packing fraction `=("Volume of atom in one unit cell")/("Volume of unit cell")` `=(6xx4/3pir^(3))/(24sqrt(2)r^(3))=pi/(3sqrt(2))=0.74` Packing fraction `=74%` Empty space `=26%` |
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| 1553. |
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 `=sqrt(2/3)4r` Base area `=(6xxsqrt(3))/4(2r)^(2)` Volume of unit cell `=(6xxsqrt(3))/4(2r^(2))sqrt(2/3)4r=24sqrt(2)r^(3)` |
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| 1554. |
A mineral having the formula `XY_(2)` crystaillises in the cubic closed packed lattic with the A- atoms occupaying the lattice point .Fraction of the betradral sites occupired by B- atoms isA. `1.00`B. `0.52`C. `0.68`D. `0.74` |
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Answer» Correct Answer - a `AB_(2)` cryswtanllises as cubic packed (ccp) This `A^(2+)` is surrounded by edge `B^(-)` and `B` is surrounding by four `A^(2 +)` Thus coordination number of `A^(2+) = 8` Coordination number of `B^(-) = 4` Number of atoms as trtahoudral in small of unit of atoms `B= 4` Thus fraction occepaid `= (4)/(4) = 1 (100%)` |
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| 1555. |
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 empty space in this hcp unit cell isA. `74%`B. `48.6%`C. `32%`D. `26%` |
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Answer» Correct Answer - D Refer Section 1.22. |
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| 1556. |
`CsBr` has bcc stucture with edge length `4.3` pm .The shortest interionic distance in between Cs and Br isA. `3.72 Å`B. `1.86 Å`C. `7.44 Å`D. `4.3 Å` |
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Answer» Correct Answer - a Distance between `Cs^(+)` and `Br^(-) = (a sqrt(3))/(2) = (4.3 xx 1.732)/(2)` `= 3.723 Å` |
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| 1557. |
`CsBr` has bcc stucture with edge length `4.3` pm .The shortest interionic distance in between Cs and Br isA. 3.72B. 1.86C. 7.44D. 4.3 |
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Answer» Correct Answer - A `R_(+) + r_(-) = (sqrt3a)/(2) = (sqrt3 xx 4.3)/(2) = 3.72 Å ` |
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| 1558. |
In a solid `AB` having the `NaCl` structure, A atom occupies the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid isA. `AB_(2)`B. `A_(2)b`C. `A_(4)B_(3)`D. `A_(3)B_(4)` |
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Answer» Correct Answer - D `AB` has `NaCl`- type structure (fcc systems). `:.` Number of `A` atoms `= 8 ("corner") xx (1)/(8) ("per corner share") + 6 ("faces")` `xx (1)/(2) ("per face centre share")` `= 1 + 3 = 4//"unit cell"` Number of `B` atoms `= 12 ("edge") xx 1/4 ("per edge centre share")` `+1 ("body centre")` `= 1+3=4//"unit cell"` Number of A atoms removed (face centred tom of one axis) `= 2("faces") xx 1/2 ("per edge centre share")` `= 1//"unit cell"` Number of A atoms left `= 4 -1 = 3//"unit cell"`. Formula `= A_(3)B_(4)` |
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| 1559. |
`CsBr` has `bcc` structure with edge length of `43 pm`. The shortest interionic distance between cation and anion isA. `37.2`B. `18.6`C. `74.4`D. `43` |
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Answer» Correct Answer - A Body diagonal `= sqrt(3)a` Distance between `Cs^(o+)` and `Br^(ɵ)` is `("Body diagonal")/(2) = (sqrt(3)a)/(2) = (sqrt(3) xx 43)/(2) = 3.724` |
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| 1560. |
The coordination number of a metal crystallizing in a hexagonal close-packed structure isA. `12B. `4`C. `8`D. `6` |
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Answer» Correct Answer - A In hcp, a particel as shown here is surrounded by `12` particles, six in its own plane and three each above and below the plane. |
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| 1561. |
The coordination number of a metal crystallizing in a hexagonal close-packed structure isA. `4`B. `6`C. `8`D. `12` |
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Answer» Correct Answer - d In `hcp` coordination number is `12`. |
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| 1562. |
The coordination number of a metal crystallizing in a hexagonal close-packed structure isA. `6`B. `8`C. `4`D. `12` |
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Answer» Correct Answer - d Coordination number in hcp struture is `12`. |
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| 1563. |
The coordination number of a metal crystallizing in a hexagonal close-packed structure isA. 12B. 4C. 8D. 6 |
| Answer» Correct Answer - A | |
| 1564. |
Statement 1: In any ionic solid [MX] withschotty defects, the number of positive and negative ions are same Statement 2: Equals number of cation and anion vacancies are present .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 assertion is false but reason is true. |
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Answer» Correct Answer - A Schottky defect is defined as a vacancy developed for anion and cation site, so cation and anion vacancy will be same in number. Therefore an ionic solid MX with Schottky defects will still have the same number of anions and cations. |
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| 1565. |
Identify the correct statement(s)A. CsCl changes to NaCl structure on heatingB. NaCl changes to CsCl structure on applying pressureC. Coordination number increases on applying pressureD. Coordination number increase on heating. |
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Answer» Correct Answer - A::B::D (a) is correct when heating is done, the coordination number decreases and CsCl changes to NaCl structure because on applying pressure coordination number increases. (c) Is wrong because coordination number increases on applying pressure . (d) is wrong becasue coordination number decreases on heating. |
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| 1566. |
A metallic crystal crystallizes into lattice containing a sequence of layers AB,AB,AB. . .. . . . .. ... The percentage of free space in this lattice isA. 0.74B. 0.26C. 0.32D. 0.48 |
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Answer» Correct Answer - B It is FCC packing Volume occupied `=(4(4)/(3)m^(3))/(a^(3)) [4r=sqrt(2)arArra=2sqrt(2)r]` `=(4xx(4)/(3)m^(3))/((2sqrt(2)r)^(3))=(16)/(3)xx(1)/(16sqrt(2))pi=(pi)/(3sqrt(2))=(3.142)/(3xx1.414)=0.74` Volume occupied is 74%, free space =26% |
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| 1567. |
The coordination number of a metal crystallizing in a hexagonal close packed structure isA. 4B. 12C. 8D. 6 |
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Answer» Correct Answer - B In hcp, co-ordination no. is 12. |
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| 1568. |
Which of the following statement is correct for `CsBr_(3)`A. It is a covalent compoundB. It contains `Cs^(3+)` and `Br^(-)` ionsC. It contains `Cs^(+)` and `Br_(3)^(-)` ionsD. It contain `Cs^(+),Br^(-)` and lattice, `Br_(2)` molecule. |
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Answer» Correct Answer - C `CsBr_(3)` consist of `Cs^(+)` and `Br_(3)^(-)` ions. |
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| 1569. |
Which of the following are not correct statement for `CsBr_(3)`.A. It is a covalent compoundB. It contain `Cs^(3+)` and `Br^(-)` ionsC. It contains `Cs^(+)` and `Br_(3)^(-)` ionsD. It contains `Cs^(+),I^(-)` and lattice `I_(2)` molecule. |
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Answer» Correct Answer - A::B::D (abd)(a),(b) and (d) are correct answer because they are not correct statement (c) is correct statement because it contains `Cs^(+)` and `Br^(3-)` ions. |
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| 1570. |
The vacant space in bcc lattice unit cell is :A. 0.68B. `52.4%`C. `60.4%`D. 0.32 |
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Answer» Correct Answer - A The space filled by atomsin bcc is 68% . |
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| 1571. |
The vacant space in bcc lattice unit cell is :A. 0.23B. 0.26C. 0.32D. none of these |
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Answer» Correct Answer - C In bcc structure 68% of the avaliable volume is occupied by spheres. Thus, vacant space is 32% |
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| 1572. |
The vacant space in bcc lattice unit cell isA. 0.23B. 0.26C. 0.32D. 0.74 |
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Answer» Correct Answer - B It is a fact. |
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| 1573. |
Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?A. 108 pmB. 128 pmC. 157 pmD. 181 pm |
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Answer» Correct Answer - B Since copper crystallises in a face centred cubic lattice, atomic radius `=a/(2sqrt(2))` `361/(2xx1.414)=127.6~=128` |
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| 1574. |
Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?A. 128B. 157C. 181D. 108 |
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Answer» Correct Answer - A For a fcc lattice , the radius of atom is `r=(a)/(2sqrt2)= (361)/(2xxsqrt2)=127.6=128` pm |
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| 1575. |
If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:A. `4/sqrt(3) a`B. `sqrt(3)/4 a`C. `sqrt(3)/2 a`D. `2/sqrt(3) a` |
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Answer» Correct Answer - C It is half of the body diagonal and is given as `=sqrt(3)/2 a` |
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| 1576. |
If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:A. `sqrt(3)/(4)a`B. `sqrt(3)/(2)a`C. `(2)/sqrt(3)a`D. `(4)/sqrt(3)a` |
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Answer» Correct Answer - B Half of body diagonal, `sqrt(3a)/(2)`. |
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| 1577. |
`Fe^(III) (Fe^(II) Fe^(III))O_(4)` represent an inverse `2 : 3` spinel structre. |
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Answer» Correct Answer - T The inverse spinel is represented by the formula `AB_(2)O_(4)` containing `A^(2+)` and `B^(3+)` ions. |
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| 1578. |
Copper crystallises in fcc lattice with a unit cell edge of 361 pm. The radius of copper atom isA. 108 pmB. 128 pmC. 157 pmD. 181 pm |
| Answer» Correct Answer - B | |
| 1579. |
Copper crystallises in fee with a cell length of 361 pm .What is the radius of copper atoms?A. `127.6` pmB. 157 pmC. 181 pmD. 109 pm |
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Answer» Correct Answer - a For fcc`n = 4 and r = (a)/(2sqrt(2)) = (360)/(2sqrt(2)) = 127.6` pm |
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| 1580. |
Copper crystallises in fcc with a cell length of 361 pm .What is the radius of copper atoms?A. `108 p m`B. `127 p m`C. `157 p m`D. `181 p m` |
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Answer» Correct Answer - B For `f.c.c.` structure, Radius of atom `=(a)/(2sqrt(2))=(361)/(2sqrt(2))=127.56p m` |
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| 1581. |
In diamond, `CN` of carbon is `4` and its unit cell has `8` carbon atoms. |
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Answer» Correct Answer - T See text. |
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| 1582. |
the number of tetrahedral voids per unit cell in NaCl crystal is ………… .A. 4B. 8C. twice the number of octahedral voidsD. four times the number of octahedral voids |
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Answer» Correct Answer - B,C NaCl has fcc arrangment of `Cl^(-)` ions. Thus, Number of `Cl^(-)` ions in pakcing per unit cell=4 Number of tetrahedral voids `=2xx` No. of particles present in close packing `=2xx4=8` |
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| 1583. |
Amorphous solids can also be callled ………….. .A. pseudo solidsB. true solidsC. super cooled liquidsD. super cooled solids |
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Answer» Correct Answer - A,C Amorphous solid has short range order which has a tendency to flow very slowly. Hence, it is also known as pseudo solids or super cooled liquids. Glass panes fiexed to windows or doors of old bulidings are invariably observed to be thicker at bottom than at the top. These are examples of amorphous solids. |
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| 1584. |
In a face centred cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centres. Once of the A atom is missing from one corner in unit cell. The simplest formula of compound isA. `A_7B_3`B. `AB_3`C. `A_7B_24`D. `A_(7//8)B_3` |
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Answer» Correct Answer - C Number of atoms of A from corners of unit cell `=1-1/8=7/8` Number of atoms of B from faces of unit cell = 3 Thus , A : B :: 7/8 : 3 or 7 : 24 . Thus , formula is `A_7B_24` . |
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| 1585. |
The correct statement regarding defects in crystalling solids.A. Frenkel defect is a dislocation defectB. Frenkel defect is found in halides of alkaline metalsC. Schottky defects have no effect on the density of crystalline solidsD. Frenkel defects decrease the density of crystalline solids. |
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Answer» Correct Answer - 1 Frankel defect is shown by ionic solid in which there is a large different in the size of ions (and hence the coordination number is low).for example `ZnS,AgCl,AgBr` and `Agl` due to small size of `Zn^(2+)` and `Ag^(+)` ions. In defect the smaller ion (usually cation) is dislocated from its normal site to an interstitial site.This creates a vacancy defect at its origin site and are interstitial defect at its new location.Therfore Frenkel defect is also called dislocation defect. it does not change the density of the solid. |
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| 1586. |
If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:A. `(2)/(sqrt(3))a`B. `(4)/(sqrt(3))r`C. `(sqrt(3))/(4)a`D. `(sqrt(3))/(2)a` |
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Answer» Correct Answer - 4 The distance between the body centered atom and one corner atom in the cube is actually the nearest neighbour distance the distance between centres of touching spheres. In body centered cubic unit cell ,the atom at the centre is in touch with the other two atoms diagonally arranged at the corners,if a is the lenght of the side of a cube , the length of a body diagonal is equal to `sqrt(3)a` .it is also equal to `4r`,where `r` is the radius of the sphere (atom) as all the three sphere (atoms) along the diagonal touch each other Therefore `sqrt(3)a=4r` `r=(sqrt(3))/(4)a` The nearest neighbour distance `(d)` is `2r`. Therefore `d=2r=2((sqrt(3))/(4)a)=sqrt(3)/(2)a` Note that nearest neighbouring distance `d` is half of the body diagonal. |
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| 1587. |
`ZnS` exists in two different form: zinc blende and wurtzite. Both occur in `4 : 4` coordination compounds. Zinc blende has an fcc structure and wurtizite has an hcp structure. |
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Answer» Correct Answer - T Factual statement. |
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| 1588. |
Wurtzite has …… formula units per unit celll wherease zinc blende has ……. Formula units per unit celll. |
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Answer» Correct Answer - remains unchanged, decreases |
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| 1589. |
A given metal crystalline out with a cubic structure having edge length of `361` pm .if there are four metal atoms in one unit cell, what is the radius of metal atom?A. 127 pmB. 80 pmC. 108 pmD. 40 pm |
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Answer» Correct Answer - A In fcc unit cel (Z=4) `4r=sqrt(2) a` `r=(sqrt(2)a)/(4) = (1.414xx361)/(4)` 127.25 pm `~~` 127 pm |
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| 1590. |
A given metal crystallises out with a cubic structure having edge length of `361 "pm"` .If there are metal atoms in one cell, what is the radius of one atoms?A. `80` pmB. `108` pmC. `40` pmD. `127` pm |
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Answer» Correct Answer - d `Z = 4,i.e.` structure is fcc Hence `r = (a)/(2sqrt(2)) = (361)/(2sqrt(2))` ` = 127.65 "pm" = 127` pm |
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| 1591. |
A given metal crystalline out with a cubic structure having edge length of `361` pm .if there are four metal atoms in one unit cell, what is the radius of metal atom?A. `108p m`B. `40p m`C. `127p m`D. `80p m` |
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Answer» Correct Answer - 3 Since there are four metal atoms in one unit cell the metal crystalline in `FC C` lattice.For `FC C` unit cell `4r=sqrt(2)a` `r=(sqrt(2))/(4)a=(sqrt(2))/(2.2)a=(sqrt(2)a)/(2sqrt(2)sqrt(2))` `=(a)/(2sqrt(2))` `=(361p m)/(2sqrt(2))` `=127p m =(sqrt(2)=1.414)` |
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| 1592. |
A given metal crystllizes out with a cubic structure having edge length of 361 pm. If there are four metal stoms in one unit cell. What is the radius of one atom ?A. 80 pmB. 108 pmC. 40 pmD. 127 pm |
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Answer» Correct Answer - a Z = 4 means structure is fcc, for fcc. ` r = a/(2sqrt2) = ( 361)/( 2 xx 1.414) = 121 . 65 "pm" = 127 "pm" ` |
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| 1593. |
Assertion (A) : Zinc blende and wurtzite both have `fcc` arrangement of `S^(2-)` ions. Reason (R ) : A unit cell of both has four formula units of `ZnS`.A. if both (A) and ® are correct , and ® is correct explanartion of (AB. if both (A) and ® are correct , but ® is nmot the correct explanation ofC. if (A) is correct , but ® is incorrect.D. if both (A) and ® are icorrect . |
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Answer» Correct Answer - D Correct(A): Zinc blende has fcc arrangement of `S^(-2)` ions while wurtzite has hcp arrangement of `S^(2-)` ions. Correct(R): A unit cell of Zinc blende has 4 formula units while that of wrtzite has 6 formula units of ZnS. |
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| 1594. |
`AB` crystallizes in a body centred cubic lattice with edge length `a` equal to `387p m` .The distance between two oppositely charged ions in the lattice is :A. 300pmB. 335pmC. 250pmD. 200pm |
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Answer» Correct Answer - B For BC C `r^(+)+r^(-)=sqrt(3a)/(2)therefore r^(+)r^(-)=(sqrt(3)xx387)/(2)` pm `=335.14` pm=335pm |
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| 1595. |
A given metal crystalline out with a cubic structure having edge length of `361` pm .if there are four metal atoms in one unit cell, what is the radius of metal atom?A. 127 pmB. 80 pmC. 108pmD. 40pm |
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Answer» Correct Answer - A In FC C unit cell (Z=4) `4r=sqrt(2)a` `r=(sqrt(2)a)/(4)=(1.414xx361)/(4)=127.25` Pm |
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| 1596. |
a metal crystallizes with a face-centered cubic lattice.The edge of the unit cell is `408` pm. The diameter of the metal atom is :A. 288 pmB. 408pmC. 144pmD. 204pm |
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Answer» Correct Answer - 1 In FCC, diameter `=(a)/sqrt(2)` |
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| 1597. |
a metal crystallizes with a face-centered cubic lattice.The edge of the unit cell is `408` pm. The diameter of the metal atom is :A. 288pmB. 408pmC. 144pmD. 204pm |
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Answer» Correct Answer - A For C CP `sqrt(2)a=4R` `(sqrt(2)xx408)/(2)=2R` (2R=Diameter) Diameter=288.5 |
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| 1598. |
a metal crystallizes with a face-centered cubic lattice.The edge of the unit cell is `408` pm. The diameter of the metal atom is :A. `288p m`B. `408 p m`C. `144p m`D. `204p m` |
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Answer» Correct Answer - 1 In a face -centered cubic lattice. the atom of the face centre is in touch with the other two atoms diagonally arranged at the corners. The length of the face diagonal is `sqrt(2)a` As all the three spheres along the face diagonal touch each other, the length of the body diagonal is also equal to `4r` where `r` is the radius of the sphere (atom). Therefore `4r=sqrt(2)a` or `r=(sqrt(2))/(4)a=(a)/(2sqrt(2))` the diameter `(d)` of the metal atom is `2r`.Therefore `d=2r=(a)/(sqrt(2))=(408p m)/(1.414)` `=288p m` |
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| 1599. |
Assertion : The number of tetrahedral voids is double the number of octahedral voids Reason : The size of the tetrhedral voids is half of that of the ochedral voidA. 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 |
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Answer» Correct Answer - C Correct Reason The size of the tetrahedral void is smaller but not half that of the octahedral void |
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| 1600. |
Statement -1 : The octahedral voids have double the size of the tetrahdedral voids of the tetrahedral voids in a crystal. Statement -2 : The number of tetrahedral voids is bouble the number of octahedral voids in a crystal.A. Statement 1 is True , Statementy -2 True , Statement -2 is a correct explanation for statement -1 .B. Statement 1 is True , Statement -2 is True , Statement -2 in NOT a correct explanation of statement -1.C. Statement -1 is True , Statement -2 is False.D. Statement -1 is False , Statement -2 is True. |
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Answer» Correct Answer - a Correct Statement -1 . Octahedral voids are larger in size than tetrahedral voids but not double in size. Statement-2 is correct. |
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