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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.
| 1151. |
Silicon is aA. SemiconductorB. InsulatorC. ConductorD. None of these |
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Answer» Correct Answer - A Si and Ge are semiconductors which are bad conductor of electricity at room temperature but a condcutor at high temperature . Hence , Si and Ge are semiconductors. |
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| 1152. |
Which of the following unit cell having maximum number of atoms ?A. bccB. hcpC. fccD. cubic |
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Answer» Correct Answer - C In hcp - atoms in crystals are packed in ABAB pattern while in bcc - Number of atoms present `=8xx1/8+1=1+1=2` Number of atoms present in fcc `=8xx1/8+1/2xx6=1+3=4` therefore , fcc has highest number of atoms per unit cell |
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| 1153. |
A metallic element has a cubic lattice. Each edge of the unit cell is 2Å. The density of the metal is 2.5 g `cm^(-3)`.The unit cells in 200g of the metal areA. `1xx10^24`B. `1xx10^20`C. `1xx10^22`D. `1xx10^25` |
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Answer» Correct Answer - D Number of unit cell `=("mass of metal")/("mass of one unit cell")` Given, edge length of unit cell = 2Å `= 2xx10^(-8)` cm Mass of metal = 200 g Density of metal = `2.5gcm^(-3)` Volume of unit cell = `("edge length")^3=(2xx10^(-8))^3` `=8xx10^(-24)` Mass of one unit cell = volume x density `=8xx10^(-24) xx 2.5=20xx10^(-24)` `therefore` number of unit cell in 200 g metal `=("mass of metal")/("mass of one unit cell")=(200)/(20xx10^(-24))` `=10xx10^24=1.0xx10^25` |
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| 1154. |
The number of atoms contained in a fcc units cell of a monatomic substance isA. `1`B. `2`C. `4`D. `6` |
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Answer» Correct Answer - 3 A face-centered cubic unit cell is a cubic unit celll in which there are lattice points at the centre of each face of the unit cell is addition to those at the corners.Exclusive contribution of lattice points all the corners per unit cell =`8xx1//8=1` Exclusively contribution of lattice points at the face centres per unit cell`=6xx1//2=3` Thus total number of atoms per fcc unit cell `=1+4=4` |
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| 1155. |
Titanium (Ti) crystallizes in fcc lattice. It reacts with C or H intersititally and these elements occupy the voids of host lattice (i.e., Ti) radius that can be occupies in TVs without Casing strain in the host lattice. Calculate the fromula fo titanium hydride and tianium carbide.A. 0.225B. 0.414C. 0.732D. 0.9 |
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Answer» Correct Answer - B For Ovs, `(r_(o+))/(r_(Ti))=0.414`. |
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| 1156. |
A metallic element has a cubic lattice. Each edge of the unif cell is `2A^(@)`. The density of the metal is `2.5gcm^(-3)`. The unif cells in 200g of metal areA. `1xx10^(24)`B. `1xx10^(20)`C. `1xx10^(22)`D. `1xx10^(25)` |
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Answer» Correct Answer - 4 `v=a^(3)=8xx10^(-24)c c` mass of 1 unit cell =`vxxd=8xx10^(-24)xx2.5` no. of unit cells in 200 grams of metal `=(200)/(8xx10^(-24)xx2.5)=1xx10^(25)` |
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| 1157. |
Titanium (Ti) crystallizes in fcc lattice. It reacts with C or H intersititally and these elements occupy the voids of host lattice (i.e., Ti) radius that can be occupies in TVs without Casing strain in the host lattice. What is the maximum ratio of "foreign" atom radius (i.e., H atom) to host atom (i.e ., Ti) radius that can be occupies in TVs without causing a strain in the host lattice.A. 0.225B. 0.414C. 0. 732D. 0.9 |
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Answer» Correct Answer - A For `TVs, (r_(o+))/(r_(-))=(r_(H))/(r_(Ti))=0.225` |
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| 1158. |
If the radius of `K^(+)` and `F^(-)` are 133pm and 136pm respectively, the distance between `K^(+)` and `F^(-)` in KF isA. 269pmB. 134.5pmC. 136pmD. 3pm |
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Answer» Correct Answer - 1 `r_(k^(+))r_(F^(-))=269 "pm"` |
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| 1159. |
MgO crystallizes in a cubic type crystal system. The ionic radii for `Mg^(2+) and O^(2-)` are 0.066 abd 0.140nm respectively One can conclude that the `Mg^(2+)` ions occypy:A. a cubic hole in a simple structureB. every tetrahedral hole in a close packed structureC. an octahedral hole in a cubic packed structureD. every other tetrahedral hole in a close paked structure |
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Answer» Correct Answer - C |
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| 1160. |
`MgO` exists in a rock- salt type unit cell .Each `Mg^(+2)` ion will be is contact with be in contact withA. `4 O^(-2)` ionsB. `6 O^(-2)` ionsC. `8 O^(-2)` ionsD. `2 O^(-2)` ions |
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Answer» Correct Answer - b `C.N` of cation and anion is six in rock salt structure . |
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| 1161. |
An example of a substance possessiong gaint covalent structure is :A. Iodine crystalB. SilicaC. Solid carbon dioxideD. White phosphorus |
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Answer» Correct Answer - B Silica has giant covalent structure. |
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| 1162. |
How many lattice points are there in one unit cell of body centered cubic lattice ? |
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Answer» In body - centered cubic unit cell The number of corner atoms per unit cell = 8 corners `xx 1/8` per corner atom `= 8 xx (1)/(8) = 1 ` atom Number of atoms at body center = `1 xx 1 = 1` atom `therefore` Total no. of lattice points = 1 + 1 = 2 |
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| 1163. |
Which is not related to amorphous solids.A. do not have sharp melting pointsB. rigid and hardC. Not bound by plane surfacesD. give diffraction bands. |
| Answer» Correct Answer - 4 | |
| 1164. |
Which one of the following is the most correct statement?A. Brass is an interstitial alloy, while steel is a subsitutional alloyB. Brass is a substitutional alloy while steel is an interstitial alloyC. Brass and steel are both substitutional alloysD. Brass and steel are both interstitial alloys |
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Answer» Correct Answer - C Brass, Cu = 80% , Zn = 20% substitutional alloy. Steel is an interstitial alloy because it is an alloy of Fe with C, C atoms occupy the interstitial voids of Fe crystal. |
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| 1165. |
In which of the following `8 :8` coordination is found?A. CsClB. MgOC. `Al_(2)O_(3)`D. All the these |
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Answer» Correct Answer - A Each `Cs^(+)`n is surrounded by eight `Cl^(-)` ions is CsCl crystal lattice because its co-ordination number is 8 : 8. |
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| 1166. |
How are interstitial solids formed ? |
| Answer» Interstitial solids are formed when the vacant spaces also called interstities in the lattice of a crystal are filled by small atoms of non-metals like hydrogen (H), boron (B), carbon (C), nitrogen (N) etc. | |
| 1167. |
Which of the following is a molecular crystal?A. SiCB. NaClC. GraphiteD. Ice |
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Answer» Correct Answer - D Ice is a molecular crystal. |
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| 1168. |
A metal crystallizes into two cubic phases, face-centred cubic and body-centred cubic, which have unit cell lengths `3.5` and `3.0 A`, respectively. Calculate the ration of densities of fcc and bcc. |
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Answer» `f.c.c.` unit cell length `=3.5`Å `b.c.c.` unit cell length `=3.0`Å Density in `f.c.c. (z_(1)xxat.wt.)/(V_(1)xxAv. no .)` Density in `b.c.c. (z_(2)xxat.wt.)/(V_(2)xxAv. no .)` `rho_(f.c.c.)/(rho_(b.c.c.))=(z_(1))/(z_(2))xx(V_(2))/(V_(1))` Now, `z_(1)` for `f.c.c.=4,` Also, `V_(1)=a^(3)=(3.5xx10^(-8))^(3)` `z_(2)` for `b.c.c.=2,` Also, `V_(2)=a^(2)=(3.0xx10^(-8))^(3)` `(rho_(f.c.c.))/(rho_(b.c.c.))=(4xx(3.0xx10^(-8))^(3))/(2xx(3.5xx10^(-8)))=1.259` |
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| 1169. |
The `"OLIVINE"` series of minerals consists of crystal in which `Fe^(2+)` and `Mg^(2+)` ions may substitute for each other causing susbstitutional impurity defects without changing the volume of unit cell. In `"OLIVINE"` series of minerals, `O^(2-)` ions exist as fcc with `Si^(4+)` occupying one-fourth of `OVs` and divalent metal ions occupying one-fourth of `OVs` and divalent metal ions occupying one-fourth of `TVs`. The density of "forsterite" (magnesium silicate) is `3.21 g cm^(-3)` land that of "fayalite" (ferrous silicate) is `4.34 g cm^(-3)`. If in "forsterite mineral" bivalent `Mg^(2+)` ions are to be replaced by unipositive `Na^(o+)` ions, and if `Na^(o+)` ions are occupying half of `TV_(s)` in`fcc` lattice, the arrangement of rest of the constituents is kept same, then the formula of the new solid is:A. `Na_(2)SiO_(4)`B. `Na_(2)SiO_(3)`C. `Na_(4)SiO_(4)`D. `Na_(2)Si_(2)O_(6)` |
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Answer» Correct Answer - C Number of `Na^(o+)` ions in fcc `= 1/2 xx TV = 1/2 xx 8` `= 4//"unit cell"` So, formula of solid `=overset(+1xx4)(Na_(4))Si^(4+)O_(4^(2-)) rArr Na_(4)SiO_(4)` |
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| 1170. |
Percentage of void space in AB solid having rock salt structure if `(r_(+))/(r_(-)) = (1)/(2)` having cation anion contact. Given `pi = 3.15` |
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Answer» Correct Answer - 0.3 `r_(+) + r_(-) = (a)/(2) rArr (3r)/(2) = (a)/(2) , r - =(a)/(3) and r_(+) = (a)/(6)` Packing fraction `= (4 XX (4)/(3) PI (r_(+)^(3) + r_(-)^(3)))/(a^(3)) = (4 xx (4)/(3) pi [((a)/(6))^(3) + ((a)/(3))^(3)])/(a^(3)) = 0.7` Percentage void = 30% |
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| 1171. |
`AX,AY,BX`, and `BY` have rock salt type structure with following internuclear distances: `|{:("Salt","Anion-anion","Cation-anion"),(,"distance in"Å,"distance in"Å),(AX,2.40,1.70),(AY,1.63,1.15),(BX,2.66,1.88),(By,2.09,1.48):}|` Ionic radii of `A^(o+)` and `B^(o+)`, respectively, areA. `0.35` and `0.68 Å`B. `0.68` and `0.35 Å`C. `1.20` and `0.80 Å`D. `0.80` and `1.20 Å` |
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Answer» Correct Answer - A `Y^(ɵ)` is small than `X^(ɵ)`. The difference in internucler distanc between `AX` and `BX` indicates that the cation must not be touching apl the anions in `AX` (smaller cation, larger anion) [see `Q.10`, Fig.(a)]. There must ve anion-anion contact is this compound. `:.` Anion-anion ditance = `2r_(ɵ)` The remaining radius ratio is represented in (see `Q.9` Figure]. b. Radius of `X^(ɵ)(r_(ɵ)) = (2.40)/(2) = 1.20Å` Radius of `B^(o+)(r_(o+)) = (1.88 - 1.20)Å = 0.68Å` c. Similarly, radius of `Y^(ɵ)(r_(ɵ)) = (1.48 - 0.68) = 0.80 Å` Radius of `A^(o+)(r_(o+)) = (1.15 - 0.80)Å = 0.35 Å` Radius of `A^(o+)` and `B^(o+)`, respectively, are `= 0.35` and `0.68Å` |
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| 1172. |
The radius of `Na^(+)` is 95pm and that of `Cl^(-)` is 181 pm. The edge length of unit cell in NaCl would be (pm).A. 181B. 95C. 276D. 552 |
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Answer» Correct Answer - 4 Edge length =`2(r_(c)+r_(a))` |
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| 1173. |
The crystal `AB`(rock salt structure) has molecular weight `6.023Y u`, where `Y` is an arbitrary number in `u`. If the minimum distance between cation and anion is `Y^(1)/(3) nm` and the observed density is `20 kg m^(-3)`. Find a. The density in `kg m^(-3)`. and b. The type of defect |
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Answer» a. Density, `rho = (Z_(eff) xx Mw)/(a^(3) xx N_(A))` Edge length `a = 2Y^(1)//(3) xx 10^(-9) m` For rock salt structure, `Z_(eff) = 4` `M = 6.023 Y u = (6.023)/(1000) Y kg` Density, `rho` `(4 xx (6.023)/(1000) xx10^(-3) xx Y)/(6.023 xx 10^(23) xx (2Y^(1//3) xx 10^(-9))^(3))` `= (40)/(8) = 5 kg m^(-3)` b. Observed density `= 20 kg m^(-3)` Calculated density `= 5 kg m^(-3)` The observed density is higher than the calculated density, hence, the solid must have non-stoichiometric defect. |
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| 1174. |
Consider the following incomplete sentence (i) When most liquids are cooled, they eventually freeze and form …. (ii) Ice is a….. These are complated byA. column = I supercooled liquid ,column = II crystalline solidB. column = I crystalline solid ,column = II crystalline solidC. column = I supercooled liquid ,column = II supercooled liquidD. column = Icrystalline solid ,column = II supercooled liquid |
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Answer» Correct Answer - b I. Liquid `overset("Freezing")overset("Point")hArr` solid formed is crystaline in nature. II. ice is also a crystalline solid. |
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| 1175. |
Calculate the approximate number of unit cells present in 1 g of gold. Given that gold cyrstallises in a face centred cubic lathce (Given atomic mass of gold = 197 u).A. `6.02xx10^(23)`B. `7.64xx10^(20)`C. `3.82xx10^(20)`D. `15.28xx10^(20)` |
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Answer» Correct Answer - B 1 Mole of gold = 197 g `mol^(-1)=6.02xx10^(23)` atoms `:.` 1g gold `=(6.02xx10^(23))/(197)` atoms As fcc contains 4 atoms per unit cell, therefore, number of unit cells `=(6.02xx10^(23))/(197xx4)=7.64xx10^(20)` |
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| 1176. |
At room temperature, sodium crystallized in a body`-`centred cubic lattrice with `a=4.24Å`. Calculate theoretical density of sodium `(` at wt. of`Na =23)`. |
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Answer» Correct Answer - `1.002g cm^(-3);` |
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| 1177. |
Thallium chloridde, `TlCl` crystallized in either a simple cubic lattice or a face `-` centred cubic lattice of `Cl^(c-)` ions with `Tl^(+)` ions in the holes. If the density of the solid is with `Tl^(+)` ions in the holes. If the density of the solide is `9.00g cm^(-3)` and edge of the unit cell is `3.85xx10^(-8) cm`, what is the unit cell geometry ? |
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Answer» Density `=("Formula mass" xxz)/(Av. no . Xx"Volume")` `:. 9=(240xxz)/(6.023xx10^(23)xx(3.85xx10^(-8))^(3))` `:. Z=1.288` Thus, `TlCl` is a cubic cell. |
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| 1178. |
Following three planes `(P_(1),P_(2),P_(3))` in an FCC unit cell are shown `:` Consider the following statement and choose the correct option that follow`:` `(i) ` `P_(1)` contains no voids of three dimensions. `(ii)` `P_(2)` contains only Octahedral voids. `(iii)` `P_(3)` contains both Octahedral and Tetrahedral voids.A. All are trueB. Only (i) & (ii) are trueC. (i) & (iii) are trueD. only (iii) is true |
| Answer» Correct Answer - A | |
| 1179. |
Which of the following statements is not correct?A. The co`-`ordination number of each type of ion in `CsCl` crystal is 8B. A metal that crystallises in `b.c.c.` structure has a co`-` ordination no. 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 p m` `(r_(Na^(+))=95p m, r_(Cl^(-))=181p m)` |
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Answer» Correct Answer - b In `b.c.c.` structure , co`-`ordination no. is eight. |
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| 1180. |
The radius of the `Na^(2+)` is `95p m` and that of `Cl^(-)` ion is `181 p m`. Predict the co`-` ordination number of `Na^(+) :`A. 4B. 6C. 8D. unpredictable |
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Answer» Correct Answer - B `(r_(Na^(+)))/(r_(Cl^(-)))=(95)/(181)=0.524, i.e., ` between `0.414` to `0.732` and thus co`-` ordination no. `=6` |
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| 1181. |
Following three planes `(P_(1), P_(2), P_(3))` in an fcc unit cell are shown in the figure below. Consider the following statements and choose the correct option/options that follow: A. `P_(1)` contains no three dimensional voids .B. `P_(2)` contains only octahedral voids.C. `P_(3)` contains both octahedral and tetrahedral voids .D. `P_(3)` contains tetrahedral voids . |
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Answer» Correct Answer - A::B::C::D A `P_(1)` represents one of the clo-packed layer having triangular voids only. B. `P_(2)` contains location of `Ovs` (edge centers of unit cell). C. `P_(3)` contains 3OVs locations (one at body center and two at edge center. Also, plane `P_(3)` contains the body diagonals, hence it contains TVs location (TVs lie at body diagonal). |
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| 1182. |
Silicon doped with arsenic is an example of `:`A. `p-` type conductorB. `n-` type conductorC. `n-p-`type conductorD. none of these |
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Answer» Correct Answer - b Follow answer |
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| 1183. |
The mass of a unit cell of an element is the product of the atomic mass of the element and ……… divided further by…… |
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Answer» Correct Answer - 6.8 |
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| 1184. |
An ionic solid `A^(o+)B^(Θ)` crystallizes as an bcc structure. The distance between cation and anion in the lattice is `338 pm`. The edge length of cell isA. `338` pmB. `390.3` pmC. `292.7` pmD. `507` pm |
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Answer» Correct Answer - B For bcc, `(r_(o+) + r_(ɵ)) = (sqrt(3))/(2)a` `338 "pm" = (sqrt(3))/(2)a rArr a = 390.3 "pm"` |
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| 1185. |
An ionic solid `A^(o+)B^(Θ)` crystallizes as an bcc structure. The distance between cation and anion in the lattice is `338 pm`. The edge length of cell isA. `390.3` pmB. `100` pmC. `190.3` pmD. `195.15` pm |
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Answer» Correct Answer - A In a body centred cubic unit cell, the distance between the cation and anion `(r_(+)+r_(-))` is related to the edge length (a) by `r^(+)+r^(-)=sqrt(3)xx(a)/(2)therefore a=(2)/sqrt(3)(r^(+)+r^(-))7` It is given that `(r^(+)+r^(-))` =338 pm `therefore a=(2)/sqrt(3)xx338`pm `thereforea=(2)/sqrt(3)xx338` pm `=(676)/(1.732)=390.3`pm |
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| 1186. |
A metal crystallises in `b.c.c.` lattice. The per cent fraction of edge length not covered by atom is `:`A. `11.4%`B. `12.4%`C. `13.4%`D. `14.4%` |
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Answer» Correct Answer - c In `b.c.c.,` radius of atom `r=(sqrt(3)a)/(4` Also, edge length of unit cell `=a` `:. ` Edge length not covered by atom `=a-2r` `=a-(sqrt(3))/(2).a` `=a((2-sqrt(3))/(2))` `:. %` fraction not covered `(a((2-sqrt(3))/(2)))/(a)xx100=13.4%` |
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| 1187. |
A metal crystallises in bcc. Find % fraction of edge length not covered and also % fraction of edge length covered by atom isA. `10.4%`B. `13.4%`C. `86.6%`D. `11.4%` |
| Answer» Correct Answer - B::C | |
| 1188. |
A metal crystallizes in `bcc` lattice. The percent fraction of edge length not covered by atom isA. `10.4%`B. `13.4%`C. `12.4%`D. `11.4%` |
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Answer» Correct Answer - B For bcc, `r = (sqrt(3))/(4)a ` Edge length not covered by atom `= a - 2r` `= a-2xx (sqrt(3))//(4)a` `= a[(2-sqrt(3))/(2)]` `:.%` of fraction not covered `= (a[(2-sqrt(3))/(2)])/(a) xx 100` `= 0.314 xx 100 = 13.4%` |
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| 1189. |
In an atomic bcc lattie what fraction of edge is not covered by atoms?A. 0.32B. 0.16C. 0.134D. 0.268 |
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Answer» Correct Answer - C |
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| 1190. |
Aluminium metal forms a cubic face centred closed packed crystal structure. Its atomic radius is `125xx10^(-12)`m. (a) Calculate the length of the side of the unit cell. (b) How many unit cells are there in `1.0m^(3)` of aluminium? |
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Answer» (a) For a face centred cubic lattice (fcc). Radius (r) `=(a)/(2sqrt2)` `a=rxx2sqrt2=125xx10^(-12)xx2sqrt2m` `=125xx2xx1.414xx10^(-12)=354xx10^(-12)m` (b) Volume of unit cell `(a)^(3)=(354xx10^(-12))^(3)m^(3)=4.436xx10^(-29)m^(3)` No. of unit cells in `1.0m^(3)` of Al `=((1.0m^(3)))/((4.436xx10^(-29)m^(3)))=2.25xx10^(28)` |
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| 1191. |
Iron changes its crystal structure from body-centred to cubic close- packed structure when heated to ` 916^(@)C`. Calculate the ratio of the density of the bcc crystal to that of ccp crystal, assuming that the metallic radius of the atom does not change. |
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Answer» In the bcc packing, the space occupied is 68% of the total volume available while in ccp, the space occupied is 74% . This means that for the same volume, masses of bcc and ccp are in the ratio of 68 : 74 . As the volume is same , ratio of density is also same, viz , 68 : 74 . i.e,x ` (d(ccp))/(d(ccp)) = 68/74 = 0.919` Alternatively Density ` (p) = (Z xx M)/( a^(3) xx N_(0))` For bcc, ` Z = 2, r= ( sqrt3 a)/4 or a_("bcc") = (4r)/sqrt3` For fcc, ` Z = 4 , r = a/(2sqrt2) or a_("fcc") = 2 sqrt2 r ` ` p_("bcc")= (2xxM)/((a_("bcc"))^(3) xx N_(0)) and p_("fcc") = ( 4 xx M)/((a_("fcc"))^(3)xx N_(0))` `p_("bcc")/(P_("fcc"))= 2/((a_("bcc"))^(3))xx ((a_("fcc"))^(3))/4 = 2/((4r//sqrt3)^(3))xx ((2sqrt2r)^(3))/4 = (2xx3sqrt3)/(64 r^(3)) xx (16sqrt2 r^(3))/4 = 3/8sqrt6 = 0.919` |
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| 1192. |
When heated above `916^(@)C`, iron changes its bcc crystalline from to fcc without the change in the radius of atom . The ratio of density of the crystal before heating and after heating is :A. 1.069B. 0.918C. 0.725D. 1.231 |
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Answer» Correct Answer - B |
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| 1193. |
An element occurs in two crystalline form `alpha` and `beta`. The `alpha`-from has an `fcc` with `alpha = 3.68 Å` and `beta`-from has a `bcc` with `a = 2.92 Å`. Calculate the ratio of their densities.A. 1B. 2C. 3D. 4 |
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Answer» `Z_(eff)` for fcc=4,`Z_(eff)` for bcc=2 Atomic volume of `alpha-`form `=((3.68xx10^(-8))^(3)xxN_(A))/(4)` Atomic volume of `beta-`form `((2.92xx10^(-8))^(3)xxN_(A))/(2)` (As Aw is same, element is same), so the density ratio is `rho_(alpha):rho_(beta)=v_(beta),v_(alpha)=((2.92)^(3))/(2),((3.68)^(3))/(4)=12.448:12.459=1:1` |
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| 1194. |
`{:("Compound", " Magnetic property") , ("(A) NaCl" , "(p) Ferrimagnetic"),("(B) MnO" , "(q) Paramagnetic"),("(C) CrCl"_(3) , "(r) Ferromagnetic"),("(D) CrO"_(2) , "(s) Diamagnetic"),("(E) MgFe"_(2)O_(4) , "(i) Antiferromagnetic"):}`A. A-p,B-r,C-q,D-t,E-sB. A-t,B-q,C-r,D-p,E-sC. A-r,B-t, C-q,D-p,E-sD. A-s,B-t,C-q,D-r,E-p |
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Answer» Correct Answer - (A-r ; B-p,r,s; C-r, D-q ) NaCl - Diamagnetic , ` CrCl_(3)` - Paramagnetic , ` CrO_(2)` - Ferromagnetic MnO - Antiferromagnetic , ` MgFe_(2)O_(4)` - Ferrimagnetic |
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| 1195. |
How many chloride ions are there around sodium ion in sodium chloride crystal?A. `6`B. `8`C. `4`D. `3` |
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Answer» Correct Answer - a In `NaCI`crystal each `Na^(+)` is surrounded by `6 CT` ions and each chiede ion surrounded by `6 Na^(+) `ions |
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| 1196. |
A binary solid `(A^(+)B^(+))` has a rock sell structure .If the edge length is `400 pm` and radius of cation is 75 pm the radius of amion attion isA. 100 pmB. 125 pmC. 250 pmD. 325 pm |
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Answer» Correct Answer - b Rock salt - simple cubic `r_(A)^(+) + r_(B)^(-) = (a)/(2)` `75 + r_(B)^(-) = (400)/(2)` `r_(B)^(-) = 200 - 75 = 125` pm |
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| 1197. |
Which of the following is not a cystalline solids?A. PolyurethaneB. CopperC. Potassium nitrateD. Benzoic acid |
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Answer» Correct Answer - 1 Crystalline solids are those whose atoms,molecules or ions have an ordered arrangement extending over a long range ,This order on the atomic level is also seen on the visible level because crystalline solids usually have flat faces and distant angles. Glass ,rubber polymers and plastics are typical examples of amorphous solids .Polyurethane being a polymer is a non=crystalline solids. Crystals range in size from those large enough for their faces or cleavage planes to be evident at a glance to those so small that they seem to be powerful as an electron microscope or `X` ray diffraction techiniques,may be used ot detect their crystalline structure. The way in which a crystal grows affects its size .Size is inversely proportional to the rate of growth.i.e, if a crystal grows slowly from the liquids state or form solutionsthere is greater opportunity for large crystals to form,Crystals exist in a variety of shapes.The way in which a crystal of a given substance is formed may affect its shape .For example a cystal of `NaCl` grown by subspension in the centrer of a solution of `NaCl` will have a symmterical cubic shape because `NaCl` deposits at an equal rate on all sides of the crystal.i.e, all faces of the crystal grown equally.But if the crystal is grown on the button of the container ,then the crystal growns faster horizonatlly then vertically because material is not deposited on the face that touches the bottom. The shape of a crystal of `NaCl` can be changed even further by ther variations in the conditions of growth. |
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| 1198. |
Assertion: Crystalline solids are anisotropic. Reason: Cystalline solids are not as closely packed as ionic solids.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 |
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Answer» Correct Answer - C Correct Reason Anisotropy is due to different arrangement of constituent particles in different directions |
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| 1199. |
A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 ` g cm^(-3)` , then the number of atoms present in 256 g of the crystal is ` N xx 10^(24)` . The value of N is |
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Answer» Correct Answer - c `p ( Z xx M)/( a^(3) xx N_(0) xx 10^(-30)` Where a is in pm and M in g ` Mol^(-1)` ` 8 = ( 4 xx M)/((400)^(3) xx 6 xx 10^(23) xx 10^(-30))) or M = 76.8` Moles in 256 g = ` 256/76.8 = 3.33` No, of atoms = ` 3.33 xx 6 xx 10^(23)` `= 2xx 10^(24)` |
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| 1200. |
Xenon crystallizes in the face-centred cubic lattice and the edge of the unit cell is `620` pm. What is the nearest neighbour distance and what is the redius of xenon atom?A. 438.5 pmB. 219.25 pmC. 420 pmD. 261.5 pm |
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Answer» Correct Answer - A As xenon crystallises in fcc lattice , nearest neighbour distance (d) `=a/sqrt2=(620)/(sqrt2)=620/1414 = 438.5` |
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