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.
| 1501. |
What is the C.N of `Cl^(-)` ions in NaCl crystals ? |
| Answer» Correct Answer - Six | |
| 1502. |
The total number of identical spheres required in cubic close packing arrangement of a unit cell isA. 6B. 8C. 12D. 14 |
|
Answer» Correct Answer - C It is a fact. |
|
| 1503. |
In the crystals structures of sodium chloride, the arrangement of `Cl^(-)` ions isA. fccB. bccC. Both fcc and bccD. none of these |
|
Answer» Correct Answer - A In NaCl, the arrangement of `Cl^(-1)` ions is fcc. |
|
| 1504. |
If the corrdination number of an element in its crystal lattice is 8, then packing is :A. fccB. hcpC. bccD. none of the above |
|
Answer» Correct Answer - C bcc arrangment hasa C. N. of 8. |
|
| 1505. |
Body -centred cubic lattice has a corrdination number ofA. 8B. 10C. 6D. 4 |
| Answer» Correct Answer - A | |
| 1506. |
If `a != b != c and alpha = beta = gamma = 90^(º)`, the crystal system isA. MonoclinicB. TriclinicC. HexagonalD. othorhombic |
| Answer» Correct Answer - D | |
| 1507. |
Which of the following crystal will exhibit the relation a=b=c and `alpha=gamma=90^@` ?A. DiamondB. CinnabarC. CalciteD. Ice |
|
Answer» Correct Answer - A Diamond is a cubic crystal system that has a=b=c intercept and `alpha= beta = gamma = 90^@` crystal angles |
|
| 1508. |
The crystal system of a compound with unit cell dimensions `a=0.387,b=0.387` and `c=0.504` and `alpha=beta=90^(@)` and `gamma=120^(@)`isA. cubicB. hexagonalC. OrthorhombicD. rhombohedral |
|
Answer» Correct Answer - 2 In hexagonal crystal axis distance or edges lengths are releated as `a=bnec` and axist angles are related as `alpha=beta=90^(@),gamma=120^(@)` In both cubic and rhombohedral `a=b=c` and in orthorhombic `ane bne c` |
|
| 1509. |
The crystal system of a compound with unit cell dimensions a=0.387,b=0.387 and c=0.504 nm and `alpha=beta=90^(@)and gamma=120^(@)` isA. CubicB. HexagonalC. OrthorhobicD. Rhombohedral |
|
Answer» Correct Answer - B For the given crystal, `a=bnec` `alpha = beta = 90^@` `gamma = 120^@` these are the characteristic of a hexagonal system |
|
| 1510. |
The crystal system of a compound with unit cell dimensions `a=0.387,b=0.387` and `c=0.504` and `alpha=beta=90^(@)` and `gamma=120^(@)`isA. CubicB. HexagonalC. OrthorhombicD. Rombohedral |
|
Answer» Correct Answer - B For crystal, if `a=b ne c, alpha = beta=90^(@), gamma =120^(@)` it belongs to hexagonal system |
|
| 1511. |
Assertion. No compound has both Schottky and Frenkel defects. Reason. Both defects change the density of the soild . |
|
Answer» Correct Answer - d Correct A. Some compounds may have both Schottky and Frenkel defect ( e.g., AgBr). Correct R. Schottky defect lowers the density while Frenkel defect does not. |
|
| 1512. |
Assertion. No compound has both Schottky and Frenkel defects. Reason. Both defects change the density of the soild .A. AgBrB. ZnOC. NaClD. KCl |
| Answer» Correct Answer - 1 | |
| 1513. |
In a face centred cubic lattice, atom `A` occupies the corner positions and atom `B` occupies the face centred positions. If one atom of `B` is missin from one of the face centred points,, the formula of the compound is `:`A. `AB_(2)`B. `A_(2)B_(3)`C. `A_(2)B_(5)`D. `A_(2)B` |
|
Answer» Correct Answer - 3 Effective no. of A atoms `=(1)/(8)xx8=1` Effective no. of B atoms `=(1)/(2)xx5` (One is missing) `=(5)/(2)` Therefore formula is `(A_(1)B_((5)/(2))=A_(2)B_(5))` |
|
| 1514. |
Lithium forms body centred cube structure .The length of the side of its unirt cell is 351 pm Atomic radius of the lithium will beA. 75pmB. 300pmC. 240pmD. 152pm |
|
Answer» Correct Answer - 4 For BCC, `sqrt(3)a=4r, r=(sqrt(3)xx351)/(4)=152` pm |
|
| 1515. |
Lithium crystallizes in a body - centred cubic crystal .If the length of the unit cell if lithium is `315`pm , the atomic radius of the lithium will beA. `300.5` pmB. `240.8` pmC. `151.8` pmD. `75.5` pm |
|
Answer» Correct Answer - c For bcc `r =(sqrt(3)a)/(4) = (sqrt(3))/(4) xx 351 = 151.8` pm |
|
| 1516. |
Lithium forms body centred cube structure .The length of the side of its unirt cell is 351 pm Atomic radius of the lithium will beA. `151.8 "pm"`B. `75.6 "pm"`C. `300.5 "pm"`D. `240.8 "pm"` |
|
Answer» Correct Answer - A Since lithium crystallises in a body centred cubic crystal, Atomic radius `(r)=(sqrt(3)a)/4=a/(2sqrt(2))` `=sqrt(3)/4xx351"pm"=151.8 "pm"` |
|
| 1517. |
Which of the following statements are not true ?A. Vacancy defect results in a decrease in the density of the substanceB. Interstitial defect results in an increase in the density of the substanceC. Impurity defect has no effect on the density of the substanceD. Frankel defect results in an increase in the density of the substance. |
| Answer» Correct Answer - C::D | |
| 1518. |
An excess of potassium ions makes KCL crystals appear violet or lilac in colour since …………. .A. some of the anionic sites are occupied by an unpaired electronB. some of the anionic sites are occupied by a pair of electronsC. there are vacancies at some anionic sitesD. F-centres are created which impart colour to the crystals. |
| Answer» Correct Answer - A::D | |
| 1519. |
An excess of potassium ions makes KCL crystals appear violet or lilac in colour since …………. .A. some of the anionic sites are occupied by an unpaired electronB. some of the anionic sites are occupied by an pair of electronsC. there are vacancies at some anionic sitesD. F - centres are created which impart colour to the crystals |
|
Answer» Correct Answer - A,D When KCl crystals are heated it leads the diposition of potassium ion on surface fo KCl the `Cl^(-)` ions diffuse to the surface of crystal and loss electron by potassium atom to from `K^(+)` ion relared electron occupies anionic site which is known as F - centre and imparta colour to the crystal . |
|
| 1520. |
Find the simplest formula of a solid containing `A` and `B` atoms ia cubic arrangement In which `A` occuples corner and `B` the centre of the faces of unit cell. If the side length is `5 Å`, estimate the density of the solid assuming atomic weights of `A` and `B` as `60` and `90`, respectively. |
|
Answer» Simplest formula `AB_(3), Mw = 60 + 3 xx90 = 330` Volume of unit cell `= (5 xx 10^(-8))^(3) = 125 xx 10^(-24) mL` One cell has one `AB_(3)`, molecule So volume of one mole `= 125 xx 10^(-24) xx 6.023 xx 10^(23)` `= 75.287 mL` Density `= (330)/(75.287) = 4.38g cm^(-3)` |
|
| 1521. |
The vacant space in body `-` centred cubic lattice `b.c.c.` unit cell is about`,`A. `32%`B. `10%`C. `23%`D. 46%` |
|
Answer» Correct Answer - a In `b.c.c.` structure , `68%` of the available volume is occupied by spheres. Thus vacant space is `32%`. |
|
| 1522. |
A spinal is an important class of oxides consisting of two types of metal ions with oxides ions arranged in CCP layer. The normal spinal has one-eighth of the tetrahedral holes occupied by one type of metal ion and one-half of the octahedral holes occupied by another type of metal ion. such a spinal is formed by `Zn^(2+), Al^(3+) and O^(2-)` in the tetrahedral holes. if formula of the compound is `Zn_(x)Al_(y) O_(z)`, then find the value of `(x + y + z)` ? The type of packing generated by type (I) is:A. hexagonal close packingB. squre close packingC. cubic close packingD. body centered packing |
|
Answer» Correct Answer - `7 = (1 + 2 + 4)` `O_(4) Zn_((1)/(8) xx 8) Al_((1)/(2) xx 4)` So, formula is `ZnAl_(2)O_(4)` |
|
| 1523. |
The correct statement for the molecule `CsI_(3)` isA. It is a covalent moleculeB. It contains `Cs^(+)` and `I_(3)^(-)`.C. It contains `Cs^(3+)` and `I^(-)` ionsD. It contains `Cs^(+),I^(-)` and lattice `I_(2)` molecule. |
| Answer» Correct Answer - B | |
| 1524. |
A solid has a structure in which `X` atoms are located at cubic corners of unit cell, `O` atom are at the edge centres and `Y` atoms at cube centre. Then the formula of compound is `X_(a)Y_(b)O_(c)` If two atoms of `O` missing from any of two edge centres per unit cell, then the molecular formula is `X_(a)Y_(b)O_(z)`. Then, find the value of `(x + y + z) - (a + b+ c)`. |
|
Answer» Correct Answer - 4 a. Number of X atoms `=8xx(1)/(8)=1`/unit cell. Number of Y atoms=1/unit cell. Number of O atoms `=12xx(1)/(4)=3` /unit cell Formula is : `XYO_(3)rArrX_(a)Y_(b)O_(c)` b. Number of O atoms missing from two edge centers per unit cell `=2xx(1)/(4)=(1)/(2)` /unit cell Number of O atoms left `=3-(1)/(2)=2.5`/unit cell Formula is `XYO_(2.5)rArrX_(2)Y_(2)O_(5)rArrX_(x)Y_(y)O_(z)` `therefore` Total value of `(x+y+z)-(a+b+c)=(2+2+5)-(1+1+3)=4` |
|
| 1525. |
In an ionic solid `r_((+)) = 1.6 Å and r_((-)) = 1.864 Å`. Use the radius ratio rule to determine the edge length of the cubic unit cell in `Å` |
|
Answer» Correct Answer - 4 `(r_(+))/(r_(-)) = (1.6)/(1.864) = 0.858` So, it is CsCl type unit cell So `sqrt3 a = 2 (r_(+) + r_(-))` So `a = (2(1.864 + 1.6))/(sqrt3) Å = 2 xx 2 Å = 4Å` |
|
| 1526. |
In an ionic solid `r_((+))=1.6A` and `r_((-))=1.864A`. Use the radius ratio to determine the edge length of the cubic unit cell in `A`.A. 4B. `2sqrt(3)`C. `3sqrt(3)`D. `(4)/(sqrt(3))` |
|
Answer» Correct Answer - 1 `(r_(+))/(r_(_))=(1.6)/(1.864)=0.858` So, it is `CsCl` type of unit cell So `sqrt(3)a=2(r_(+)+r_(_))` So, `a=(2(1.864+1.6))/(sqrt(3))Å=2xx2Å=4Å` |
|
| 1527. |
While placing the second layer over that first layer , if the sphere of the second layer is above the void of the first layer, the void formed isA. TetrahedralB. OctahedralC. Both a and bD. Hexagonal |
|
Answer» Correct Answer - A If the sphere of the second layer is above the void of the first layer, the void formed is tetrahedral. |
|
| 1528. |
In an antiflourite structure, cations occupyA. Centre of cubeB. Tetrahedral voidsC. Corners of cubeD. Octahedral voids |
|
Answer» Correct Answer - B In an antifluorite structure, metal ion (cation) fill half of the tetrahedral voids . |
|
| 1529. |
Copper has a face`-` centred cubic structure with a unit`-` cell edge length of `3.61Å`. What is the size of the largest atom which could fit into the intersectices of the copper lattice without distorting if ? |
|
Answer» Correct Answer - `0.53Å ` `r_("octahedral") = 0.414 R` For FCC `4R = sqrt2a` `R = (sqrt2a)/(4)` `r = (0.414 sqrt2 a)/(4) = (0.414 sqrt2 xx 3.61)/(4) = 0.53Å ` |
|
| 1530. |
Silicon dopend with electron-rich impurity forms…….A. p-type semiconductorB. n-type semiconductorC. intrinsic semiconductorD. insulator |
|
Answer» Correct Answer - B |
|
| 1531. |
The numbers of tetrahedral and octahedral holes in a ccp array of 100 atoms are respectivelyA. 200 and 100B. 100 and 200C. 200 and 200D. 100 and 100 |
| Answer» Correct Answer - 1 | |
| 1532. |
In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids ?A. `0.543`B. `0.732`C. `0.414`D. `0.637` |
|
Answer» Correct Answer - 1 `(r_("tetrahedral"))/(r_("octahedral"))=(0.225R)/(0.414R)=0.543` |
|
| 1533. |
In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids ?A. 0.543B. 0.732C. 0.414D. 0.637 |
|
Answer» Correct Answer - A `(r_("tetrahedral"))/(r_("octahedral")) = (0.225 R)/(0.414R) = 0.543` |
|
| 1534. |
Which of the following exhibit Frenkel defect?A. AgBrB. AgClC. KBrD. ZnS. |
|
Answer» Correct Answer - A::B::D (a),(b),(d) show Frenkel defects because cation and anion differ in size appreciatly whereas (c) does not. |
|
| 1535. |
Consider the plane shown in the figure `:` A rock sale `(NaCl)` lattice is cut across this plane in which anions occupy fcc positions. It passes through centers of how many cations ?A. zeroB. 4C. 6D. 12 |
|
Answer» Correct Answer - 1 Cations are present at edge centers and body center through which this plane does not pass. |
|
| 1536. |
The `CoCl_(2)` lattice isA. Body centred cubic closed packB. Body centred tetragonalC. There are 16 formula units per unit cell.D. The number of molecules which the basis consist is 2 |
|
Answer» Correct Answer - A::B::C Choice (a) is correct because in `CaCl_(2)` there is a body centred cubic closed packing. Choice (b) is correct because there is tetragonal body centre observed. Choice (c) is also correct because there are 16 `CaCl_(2)` present in one `CaCl_(2)` unit lattice. |
|
| 1537. |
In a AB unit cell (Rock salt type) assuming `A^(+)` forming fcc :A. The nearest neighbout of `A^(+)` is `6B^(-)` ionB. The nearest neighbour of `B^(-)` is `6A^(+)` ionC. The nearest neighbour of `B^(-)` is `6A^(+)` ionD. The packing fraction of AB crystal is `(sqrt(3)pi)/(8)` |
|
Answer» Correct Answer - A::B::C In AB which in NaCl type crystal (a) Here `A^(+)` is surrounded by 6 `B^(-)` because `A^(+)` is in octahedral void of `B^(-)`. Is in octahedral void of `A^(+)`. (c) The second neighbouring of `A^(+)=12A^(+)` ions The packing fraction of `A^(+)=sqrt(3)/(8)pi` and not of the whole All crystal. Choice (a) is correct Choice (b) is correct Choice (c) is correct, Choice (d) is incorrect. |
|
| 1538. |
In the crystal lattice of `CsCl,CS^(+)` ions are present at theA. `Cs^(+)` forms a simple cubic lattice. `Cl^(-)` forms a simple cubic latticeB. `Cl^(-)` occupies body centre of `Cs^(+)`C. `Cs^(+)` occupies body centre of `Cl^(-)`D. It is impossible for `Cl^(-)` to occupy body centre of `Cs^(+)` because the body centre void of `Cs^(+)` is smaller than the `Cl^(-)` ion size. |
|
Answer» Correct Answer - A::B::C In CsCl, the coordination number is 8. Here `Cl^(-)` forms BCC packing with `Cs^(+)` occupying body centre voids or `Cs^(+)` forms SCC packing with `Cl^(-)` occupying body centred voids. |
|
| 1539. |
Which of the following crystals do not crystallise in HCP structure.A. NaB. BeC. CaD. Ba |
|
Answer» Correct Answer - A::C::D Na has BCC packing, Be has a HCP packing Ca has FCC packing Ba has a FCC packing. |
|
| 1540. |
Assertion: Every 4th layer is identical to first layer in ccp pattern of three dimension arrangement Reason: ABCABC…. Arrangement is present in ccpA. 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 | |
| 1541. |
It is stated that ZnS does not crystallise in the NaCl structure it is due to:A. The `r^(+)//r^(-)` ratio is `0.402` too low to avoid anion anion contact as in the NaCl stuctureB. ZnS is water insoluble, NaCl is water solubleC. ZnS is water soluble, NaCl is water insolubleD. Zn belongs to d-block, Na belongs to s-block |
| Answer» Correct Answer - A | |
| 1542. |
If three elements P, Q and R crystallise in a cubic unit cell with P atoms at the corners, Q atoms at the cubic centre and R atoms at the centre of each face of the cube, then write the formula of the compound. |
|
Answer» Contribution made by the atoms of P present in all eight conrners of the cube `=1//8xx8=1` Contribution made by the atoms of Q present in the centre of body = 1 Contribution made by the atoms of R present at all the six faces `=1//2xx6=3` `therefore` Formula of the compound = `PQR_(3)` |
|
| 1543. |
Atoms A, B, C and D are present at corners, face centres, body centres and edge centres and respectively in a cubic unit cell. Find the formula of compound. |
|
Answer» Contribution by atoms A = `8xx1//8=1` Contribution by atoms B = `6xx1//2=3` Contribution by atoms C = `1xx1=1` Contribution by atoms D = `12xx1//4=3` `A:B:C:D=1:3:1:3` Formula of compound = `AB_(3)CD_(3)`. |
|
| 1544. |
Schottky defect lowers the density of ionic crystals while Frenkel defect does not. Discuss.A. lowB. 1.3C. 1.5D. slightly less than unity |
|
Answer» Correct Answer - A Frenkel defect is observed in crystals in which the radius ratio is low . |
|
| 1545. |
Schottky defect lowers the density of ionic crystals while Frenkel defect does not. Discuss. |
| Answer» In Schottky defect, certain cations and anions are missing from the crystal lattice. Therefore, the density of the crystal decreases. However, in Frenkel defect, the ions do not leave the lattice but they simply change their positions from lattice points to the interstitial spaces. As a result, the density of the crystal remains unchanged. | |
| 1546. |
Assertion :No compound has both schottky and frenkel defect Reason : schottky defect change the density of the solidA. If both assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertionC. If assertion is true bur reason is falseD. If assertion is false bur reason is true |
|
Answer» Correct Answer - d Only schottky effect decreases the density of solids |
|
| 1547. |
Due to schottky defect the density of the crystal A… whereas due to frenkel defect it B…, here , A and B refers toA. Decreases-A, remains unchanged - BB. Decreases - A , increases - BC. Increases - A, remains unchanged - BD. Remains changed - A, decreases - B |
|
Answer» Correct Answer - A Due to schottky defect, the density of the crystal decreases whereas due to frenkel defect, it remains unchanged . |
|
| 1548. |
Assertion: In a close packing of spheres, a tetrahedral void is surrounded by four spheres whereas an octahedral void is surrounded by six spheres. Reason: A tetrahedral void has a tetrahedral shape while an octahedral void has an octahedral shapeA. 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 the assertion and reason both are false. |
|
Answer» Correct Answer - C Tetrahedral void is so called because it is surrounded by four spheres tetrahedrally while octahedral void is so called because it is surrounded by six spheres octahedrally. |
|
| 1549. |
Schottky defect in a crystal is observed whenA. Density of crystal is increasedB. Unequal number of cations and anions are missing from the latticeC. An ion leaves its normal site and occupies and interstitial siteD. Equal number of cations and anions are missing from the lattice. |
|
Answer» Correct Answer - D Schottky defect is due to missing of equal number of cations and anions. |
|
| 1550. |
Assertion: Cyclic silicates and chain silicates have the same general molecular formula. Reason: In cyclic silicates, three corners of each `SiC_(4)` tetradedron are shared while in chain silicates only two are shared with other tetrahedra.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 the assertion and reason both are false. |
|
Answer» Correct Answer - C Two corners per tetrahedron one shared in both the cases. |
|