94420 + Interview Questions in Chemistry in Class 12

1.	In a face centred lattice with all the positions occupied by A atoms the body centred octahedral hole in it is occupied by an atom B of an appropriate size for such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound.
Answer» ANSWER :`0.75`

Discussion

2.	In a face centred cubic unit cell of close packed atoms, the radius of atom ( r ) is related to the edge length (a) of the unit cell by the expression
Answer» `R = ( a)/( sqrt(2))` `r = ( a)/( 2)` `r = ( a)/( 2 sqrt(2))` `r = ( sqrt(3)a)/(4)` ANSWER :C

Discussion

3.	In a face centred cubic lattice the number of nearest neighbours for a given lattice point are:
Answer» 6 8 12 14 Answer :C

Discussion

4.	In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is :
Answer» `A_(2) B_(3)` `A_(2) B_(5)` `A_(2)B` `AB_(2)` Solution :No. of .A. atoms at corners `= 8 xx ( 1)/( 8 ) =1 ` No. of .B. atoms at FACE CENTRES `= 5 xx ( 1)/( 2) = ( 5)/( 2)` Formula `=A_(1) B_(5//2) ` or `A_(2) B_(5)`

Discussion

5.	In a face centred cubic lattice, a unit cell is shared equally by low many unit cells ?
Answer» 8 4 2 6 Solution :Infcc, a UNIT CELL is SHARED equally by six unit cells.

Discussion

6.	In a face centred cubic lattice, a unit cell is shared equally by how many unit cells ?
Answer» 2 4 6 8 Solution :Each UNIT CELL is SHARED by 6 unit CELLS

Discussion

7.	In a face-centred cubic cell, an atom at the face centre is shared by:
Answer» 6 UNIT CELLS 1 unit cells 4 unit cells 2 unit cells ANSWER :D

Discussion

8.	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: One of the A atom is missing from one corner in unit cell . The simplest formula of compund is :
Answer» `A_7B_3` `Ab_3` `A_7B_24` `A_7/8B_3` ANSWER :C

Discussion

9.	In a face centred cubic arrangement of A and B atoms in which ‘A’ atoms are at the corners of the unit cell and .B. atoms are at the face centers, one of the ‘A’ atoms is missing from one corner in the unit cell. The simplest formula of the compound is
Answer» `AB_(3)` `A_(7)B_(24)` `A_(7)B_(8)` `A_(7)B_(8)` SOLUTION :NUMBER of A atoms at the corners = `7 xx (1)/(8)= (7)/(8)` No. of B atoms at face centres = `6 xx (1)/(2) = 3` RATIO of `A: B = (7)/(8) : 3 = 7:24` `:.` Formula of compound is `A_(7)B_(24)`

Discussion

10.	In a face centred cubic arrangement of A and B atoms, A atoms are at the corners of the unit cell and B atoms at theface centres. One of the A atom is missing from one corner in the unit cell. The simplest formula of the compound is
Answer» `A_(7)B_(3)` `AB_(3)` `A_(7)B_(3)` `A_(7//8)B_(3)` Solution :If ONE atom (A) is missing from one corner, no of ATOMS A in unit CEL `=7xx1/8=7/8` no. of atoms B in unit cell `=6xx1/2=3` `A:B=7/8j:3`.so, simplest FORMULA is `A_(7)B_(24)`.

Discussion

11.	In a face centered unit cell (fcc) the number of atoms present
Answer» 4 2 3 5 Answer :A

Discussion

12.	In a Face centered Cubic lattice, the packing fraction is
Answer» 0.52 0.68 0.74 0.65 Answer :C

Discussion

13.	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 centred. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is :
Answer» `a_(7)b_(8)` `a_(7)b_(3)` `ab_(3)` `a_(7)b_(24)` Answer :D

Discussion

14.	In a dry cell , zinc container is protected from the atmosphere by covering it with
Answer» a METAL cap paste of `NH_(4)Cl` and `ZnCl_(2)` cardboard metal cap Answer :C

Discussion

15.	In a Duma's method 0.3g of an organic compound gave 50ml of nitrogen collected at 27^(@)C and 715mm pressure . The percentage of N is (aqueous tension at 27^(@)C=15mm) :
Answer» <P>`19.46%` `21.46%` `17.46%` `36.64%` SOLUTION :VAPOUR pressure of GAS `=715-15` `=700mm` To calculate the volume of `N_(2)` at S.T.P `{:(V_(1)=50mL,V_(2)=?),(p_(1)=700mm,p_(2)760MM),(T_(1)=300K,T_(2)=273K):}` Applying `(p_(1)V_(1))/(T_(1))=(p_(2)V_(2))/(T_(2))` or `V_(2)=(p_(1)V_(1)T_(2))/(p_(2)T_(1))` `:.V_(2)=(700xx50xx273)/(760xx300)` `=41.9mL` `22400mL` of nitrogen at S.T.P.weigh `=28g` `41.9mL` of nitrogen at S.T.P.weigh `=(28xx41.9)/(22400)` `=0.0524g` Percentage of nitrogen `=(0.0524)/(0.3)xx100=17.46%`

Discussion

16.	In a dry cell the depolarizer is :
Answer» `MnO_2` `ZN` CHARCOAL POWDER `NH_4Cl` ANSWER :A

Discussion

17.	In a determination of P an aqueous solution of NaH_(2)PO_(4) is treated with a mixture of ammonium and magnesium ammonium phosphate Mg(NH_(4))PO_(4).6H_(2)O. This is heated and decomposed to magnesium pyrophosphate, Mg_(2)P_(2)O_(7) which is weighed . A solution of NaH_(2)PO_(4) yeilded 1.11 g of Mg_(2)P_(2)O_(7). What weight of NaH_(2)P_(2)O_(7) was present originally ? (P = 31)
Answer» ANSWER :`1.20` GM

Discussion

18.	In a double sulphate salt K_(2)SO_(4)Al_(2)(SO_(4))_(3)4xx xH_(2)O. The value of x is
Answer» SOLUTION :(6) DOUBLE sulphate salts are `K_(2)SO_(4). Al_(2)(SO_(4))_(3).24H_(2)O`. `:. K_(2)SO_(4).Al_(2)(SO_(4))_(3).6xx4H_(2)O`.

Discussion

19.	In a dilute aqueous H_(2)SO_(4), the complex disaquadioxalato ferrate (II) is oxidised by MnO_(4)^(-). For this reaction, the ratio of the rate of change of [H^(+)]. For this reaction, the ratio of the rate of change of [H^(+)] to the rate of change on [MnO_(4)^(-)] is:
Answer» Solution :The BALANCED chemical equation is: `MnO_(4)^(-)+[Fe(H_(2)O)_(2)(C_(2)O_(4))_(2)]^(2-) + 8H^(+) to MN^(2+) + Fe^(3+) + 4CO_(2)+H_(2)O` `r_(H^(+))/r_(MnO_(4)^(-))=8/1`

Discussion

20.	In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present
Answer» 3 double bonds , 9 single bonds 6 double bonds , 6 single bonds 3 double bonds , 12 single bonds Zero double bonds , 12 single bonds Solution :REFER to the STRUCTURE of cyclotrimetaphoric acid. No. of P = O double bonds = 3 and No. of P-O single bonds = 9.

Discussion

21.	In a cyclotrimetaphosphoric acid molecule, how many P-Osingle and double bonds are present ?
Answer» 3 DOUBLE bonds, 9 single bonds 6 double bonds, 6 single bonds 3 double bonds , 12 single bonds ZERO double bonds, 12 single bonds Solution :The structure of cyclotrimetaphosphoric acid is :

Discussion

22.	In a cubic unit cell:
Answer» a = B = c, `alpha = BETA = GAMMA = 90^(@)` a = b = c, `alpha = beta = gamma = 120^(@)` a ne b ne c, `alpha ne beta ne gamma = 90^(@)` a = b ne c, `alpha = beta ne gamma = 90^(@)` ANSWER :A

Discussion

23.	In a cubic structure, atoms of 'X' occupy the comers, atoms of 'Y' occupy the centre of the body and atoms of 'Z' occupy the centres of all six faces. Write the composition of the unit cell.
Answer» Solution :Position of .X. is comers of CUBE. Number of comers `= 8 ` Contribution of ATOM as .X. to unit cell `= 1//8`. Number of atoms of `.X. = 8 xx 1//8 = 1` Position of .Y. is CENTRE of the body. Number of body CENTRES `= 1` Contribution of atom of .Y. to unit cell = 1. Number of atoms of `.Y. = 1 xx 1 = 1` Position of .Z. is centre of the face. Number of face centres ` = 6` Contribution of atom of .Z. to unit cell `= 1//2`. Number of atoms of `.Z. = 6 xx 1//2 = 3` The composition of unit cell containing atoms of X, Y and `Z = XYZ_3`

Discussion

24.	In a cubic type unit cell, atoms of A are at centre and corners of the cube. Atoms of B are at one half faces of the cube. What is the simplest formula of the compound?
Answer» SOLUTION :NUMBER of atoms of A = `8 XX 1/8+1xx1=1+1=2` Number of atoms of B = `3 xx 1/2 = 3//2` `:.` Simple formula of the COMPOUND `A_2B_(3//2) = A_4B_3`

Discussion

25.	In a cubic type unit cell, A atoms are at one half faces, while B atoms are at the corners of the cube. Calculate the formula of the compound.
Answer» SOLUTION :NUMBER of atoms of A : (3 faces, `1/2` fraction), i.e., `3 xx 1/2=3/2` Number of atoms of B : (8 corners, `1/8` fraction), i.e., `8 xx 1/8 =1` So, the formula of the compound = `A_(3//2) B = A_3B_2`

Discussion

26.	In a cubic structure, atoms of 'X' occupy the corners, atoms of 'Y' occupy the centre of the body and atoms of 'Z' occupy the centres of all six faces. Write the composition of the unit cell.
Answer» Solution : Position of ‘X’ is CORNERS of cube. Number of corners = 8 Contribution of atom as ‘X’ to unit cell = 1/8 Number of atoms of ‘X’ = 8 X 1/8 = 1 Position of ‘Y’ is centre of the body. Number of body centres = 1 Contribution of atom of ‘Y’ to unit cell = 1 Number of atoms of ‘ Y’ = 1 X 1 = 1 Position of ‘Z’ is centre of the face. Number of face centres = 6 Contribution of atom of ‘Z’ to unit cell = 1/2 Number of atoms of ‘Z’ = 6 X 1/2 = 3 The composition of unit cell CONTAINING atoms of X, Y

Discussion

27.	In a cubic packed structure of mixed oxides , the lattice is made up of oxide lois one fifth of retrahedral voids are occupied by cation of a while one half of the formula of the oxide is
Answer» `X_(5)Y_(4)O_(10)` `X_(4)Y_(5)O_(10)` `XY_(2)O_(4)` `X_(2)Y_(4)` ANSWER :B

Discussion

28.	In a cubic crystal of CsCl (density = 3.97 gm//cm^(3)) the eight corners are occupied by Cl^(–) ions with Cs+ ions at the centre. Calculate the distance between the neighbouring Cs^(+) and Cl^(–) ions.
Answer» Solution :`3.94=(1XX(M)/(N_(A)))/(a^(3))=(168.5)/(6.022xx10xxa^(3))` `impliesa^(3)=(168.5)/(6.022xx10^(-23)xx3.97)` `a^(3)=7.05xx10^(-23)CM^(3)` `a=4.13xx10^(-8)` cm `r_(+)+r_(-)=(asqrt(3))/(2)=3.577xx10^(-8)` 3.577xx10^(-10) cm` =3.577 Å

Discussion

29.	In a cubic crystal of CsCl (density = 3.97 gm/cm^(3)) the eight corners are occupied by Cl^(–) ions with Cs^(+) ions at the centre. Calculate the distance between the neighbouring Cs^(+) and Cl^(–) ions.
Answer» ANSWER :3.57 Å

Discussion

30.	In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A^(2+)) while one half of the octahedral voids occupied trivalent ions (B^(3+)). What is the formula of the oxide ?
Answer» `A_(2)BO_(4)` `AB_(2)O_(3)` `A_(2)BO_(3)` `AB_(2)O_(4)` ANSWER :D

Discussion

31.	In a cubic crystal anions are arranged in fcc arrangement and the cations occupy all the octahedral voids and half the tetrahederal voids. The ratio of the cations and anions in the crystal is
Answer» `1:1` `2:1` `1:2` `3:2` Solution :For fcc arrangement No. of anion = 4 Fro each atom, there is one octahedral site and 2 TETRAHEDRAL sites. THEREFORE, for 4 anions, there will be 4 octahedral sites and 8 tetrahedral sites. No. of CATIONS `= 4 XX 1 + 8 xx ( 1)/( 2) = 8` Ratio of cation and anion `= 8 : 4 ` or `2:1`

Discussion

32.	In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A2^(+)) while one half of the octahedral voids occupied trivalent ions (B3^(+)). What is the formula of the oxide?
Answer» ANSWER :`AB_(2)O_(4)`

Discussion

33.	In a cubic closed packed structure of mixed oxide, the oxide ions are in CCP arrangement 1/6 of tetrahedral voids are occupied by cations A and 1/2 of octahedral voids are occupied by cations B. The formula of the oxide is
Answer» `abo_(2)` `a_(2)bo_(2)` `a_(2)b_(2)O` `a_(2)b_(3)o_(6)` ANSWER :D

Discussion

34.	In a cubic close packing of spheres in there dimensions the coordination number of each sphere is:
Answer» 6 9 3 12 Answer :D

Discussion

35.	In a cubic close packed structure the number of nearest neighbours for a given lattice point is
Answer» 6 8 12 14 Answer :C

Discussion

36.	In acrystallinesubstanceatoms A occupycornersof a cubewhile atom B occupiesbodycentres .Whatis the formula of thesubstance?
Answer» ANSWER :AB

Discussion

37.	In a crystalline solid, anions B are arranged in ccp lattice and cations A occupy 50% of the octahedral voids and 50% of the tetrahedral voids. What is the formula of the solid
Answer» AB `A_(3)B_(2)` `A_(2)B_(2)` `A_(2)B_(3)` ANSWER :B

Discussion

38.	In a crystal of compound having molecular formula X_2 Y_3, Y atoms are arranged in CCP, then what fraction of tetrahedral voids will be covered by X atoms ?
Answer» `1/3` `1/4` `2/3` `3/4` ANSWER :A

Discussion

39.	In a crystal, the atoms are located at the position of……potential energy.
Answer» Zero Infinite Minimum Maximum Answer :C

Discussion

40.	In a crystal some ions are missing from normal sites This is an example of:
Answer» F-centres Interstitial DEFECT Frenkel defect SCHOTTKY defect Answer :D

Discussion

41.	In a crystal AB, which of the following crystal systems will have parameters, a ne b ne c and alpha = beta = gamma = 90^(@) ?
Answer» CUBIC Orthorhombic Monoclinic TRICLINIC ANSWER :B

Discussion

42.	In a coordination compound, if the metal ion has a secondary valence of six, it has an…………. Geometry.
Answer» ANSWER :OCTAHEDRAL

Discussion

43.	In a covalent bond, a difference of 1.7 in the electronegativities of the combing atoms produces an ionic character of about __________.
Answer» `50%` `30%` `90%` `70%` ANSWER :A

Discussion

44.	In a container of constant volume at a particular temperature N_2 and H_2 are mixed in the molar ratio of 9:13. The following two equilibria are found to be coexisting in the container N_2(g)+3H_2(g)hArr2NH_3(g) N_2(g)+2H_2(g)hArrN_2H_4(g) The total equilibrium pressure is found to be 3.5 atm while partial pressure of NH_3(g) and H_2(g) are 0.5 atm. and 1 atm respectively.Calculate of equilibrium constants of the two reactions given above.Give your answer as (Kp_1 + Kp_2)xx10
Answer» <P> SOLUTION :Let the initial partial pressures of `N_2` and `H_2` be 9P and 13 P respectively. `{:(N_2(g)+,3H_2(g)hArr,2NH_3(g)),(2P-y-x,13P-3x-2y,2x):}` `{:(N_2(g)+,2H_2(g)hArr,N_2H_4(g)),(2P-y-x,13P-3x-2y,y):}` Total PRESSURE=`P_(N_2)+P_(H_2)+P_(NH_3)+P_(N_2H_4)`=3.5 atm =(9P-x-y)+(13 P-3x-2y)+2x+y=3.5 atm ...(i) `P_(NH_3)=2 x=0.5` atm ...(2) `P_(H_2)=(13P-3x-2y)=1` atm from (1)`implies` (9P-x-y)+1 atm +0.5 +y=3.5 `implies` (9P-x)=2 atm so 9P=2.25 P=0.25 atm from (3)equation 2y=1.5 y=0.75 atm so `P_(N_2)`=9P-x-y=1.25 atm `P_(H_2)`=9P-x-y=1.25 atm `P_(H_2)`=1atm `P_(NH_3)`=0.5 atm `P_(H_2H_4)=`0.75 atm So, `K_(P_1)=(P_(NH_3)^2)/(P_(H_2)^3.P_(N_2^2))=(0.5xx0.5)/(1xx1xx1xx1.25)=0.2 atm^(-2)` `K_(P_2)=(P_(N_2H_2))/(P_(N_2).P_(H_2^2))=(0.75)/(1xx1xx1.25)=0.6 atm^(-2)` `P_(A_2C_2)`=y-z=1-`1/4=3/4` atm

Discussion

45.	In a container of 10 L volume, 20 g O_2(g) and 2.8 g CO(g) are taken at a total pressure of 2 atm and (27 ^0 C). Partial pressure and partial volume occupied by oxygen gas in container respectively are
Answer» `[50/29 ATM]` ,`[250/29]L` `[17/50 atm]` ,`[250/29]L` `[14/50 atm]` ,`[40/29]L` `[50/29 atm]` ,`[100/29]L` ANSWER :4

Discussion

46.	In a constant volume calorimeter 5g of a gas with molecular weight 40 was burnt in excess of oxygen at 298 K.the temperature of the calorimeter was found to increase from 298 K to 298.75 K due to combustion process.Given that the heat capacity of the calorimeter is 2.5 kJ K^(-1),a numerical value for the DeltaU of combustion of the gas in kJ mol^(-1) is
Answer» 15 12 90 8 Answer :A

Discussion

47.	In a constant volume calorimeter, 3.5g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0K to 298.45 K due to the combustion of the gas in kJ mol^(-1) is
Answer» <P> Solution :`n=(3.5)/(28)` `DeltaT=T_(2)-T_(1)=298.45-298=0.45` `C_(P)=C_(V)+R=2500+8.314=2508.314 JK^(-1)` `Q_(P)=C_(P)DeltaT=1128.74 J` `DeltaH=(Q_(P))/(n)=(1128.74)/(3.5//28)` implies 9030 J `mol^(-1)`=9.030 kJ `mol^(-1)` = 9 kJ `mol^(-1)`.

Discussion

48.	In a constant volume calorimeter, 3.5 g of a gas with moleular weight 28 was burnt in excess oxygen at 298.0 K . The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 KJK^(-1), the numerical value for the enthalphy of combustion of the gas in KJ mol^(-1) is .
Answer» Solution :`N=(3.5)/(28)` `DeltaT = T_(2)- T_(1) = 298.45 - 298 = 0.45` `C_(V)=2.5KJ K^(-1) = 2500 JK^(-1)` `C_(p) = C_(v) + R = 2500 + 8.314 = 2508.314 JK^(-1)` `Q_(p) = C_(p)DeltaT = 1128.74 J` `DELTAH = (Q_(p))/(n) = (1128.74 )/(3.5//28) = 9030 J mol^(-1) = 9.030 KJ mol^(-1) = 9 KJ mol^(-1).`

Discussion

49.	In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature process. Given that the heat capacity of the calorimeter is was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K^(-1), the numerical value for the enthalpy of combustion of the gas in kJ "mol"^(-1) is
Answer» Solution :`q=ms DeltaT=2.5xx0.45=1.125 KJ rArr DeltaE=-1.125/3.5xx28=-9" kJ/mole"` `\|DeltaE\|=9" kJ/mole"` [Since `DELTAH=DeltaE+Deltan_(g) RT` to determine `DeltaH` from `DeltaE` we MUST know `Deltan_(g)` (i.e. Chemical reaction or Compound) which is not given therefore having no other option answer is 9]

Discussion

50.	In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the heat capacity of the calorimeter is 2.5 kJ K^(-1), find the numerical value for the enthalpy of combustion of the gas in kJ mol^(-1)
Answer» Solution :Heat of combustion of 3.5 G of the gas = Heat TAKEN by the calorimeter= 2.5 (298.45 – 298.0) kJ `9kJ "MOL"^(-1)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In a face centred lattice with all the positions occupied by A atoms the body centred octahedral hole in it is occupied by an atom B of an appropriate size for such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound.

In a face centred cubic unit cell of close packed atoms, the radius of atom ( r ) is related to the edge length (a) of the unit cell by the expression

In a face centred cubic lattice the number of nearest neighbours for a given lattice point are:

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is :

In a face centred cubic lattice, a unit cell is shared equally by low many unit cells ?

In a face centred cubic lattice, a unit cell is shared equally by how many unit cells ?

In a face-centred cubic cell, an atom at the face centre is shared by:

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: One of the A atom is missing from one corner in unit cell . The simplest formula of compund is :

In a face centred cubic arrangement of A and B atoms in which ‘A’ atoms are at the corners of the unit cell and .B. atoms are at the face centers, one of the ‘A’ atoms is missing from one corner in the unit cell. The simplest formula of the compound is

In a face centred cubic arrangement of A and B atoms, A atoms are at the corners of the unit cell and B atoms at theface centres. One of the A atom is missing from one corner in the unit cell. The simplest formula of the compound is

In a face centered unit cell (fcc) the number of atoms present

In a Face centered Cubic lattice, the packing fraction is

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 centred. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is :

In a dry cell , zinc container is protected from the atmosphere by covering it with

In a Duma's method 0.3g of an organic compound gave 50ml of nitrogen collected at 27^(@)C and 715mm pressure . The percentage of N is (aqueous tension at 27^(@)C=15mm) :

In a dry cell the depolarizer is :

In a double sulphate salt K_(2)SO_(4)Al_(2)(SO_(4))_(3)4xx xH_(2)O. The value of x is

In a dilute aqueous H_(2)SO_(4), the complex disaquadioxalato ferrate (II) is oxidised by MnO_(4)^(-). For this reaction, the ratio of the rate of change of [H^(+)]. For this reaction, the ratio of the rate of change of [H^(+)] to the rate of change on [MnO_(4)^(-)] is:

In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present

In a cyclotrimetaphosphoric acid molecule, how many P-Osingle and double bonds are present ?

In a cubic unit cell:

In a cubic structure, atoms of 'X' occupy the comers, atoms of 'Y' occupy the centre of the body and atoms of 'Z' occupy the centres of all six faces. Write the composition of the unit cell.

In a cubic type unit cell, atoms of A are at centre and corners of the cube. Atoms of B are at one half faces of the cube. What is the simplest formula of the compound?

In a cubic type unit cell, A atoms are at one half faces, while B atoms are at the corners of the cube. Calculate the formula of the compound.

In a cubic structure, atoms of 'X' occupy the corners, atoms of 'Y' occupy the centre of the body and atoms of 'Z' occupy the centres of all six faces. Write the composition of the unit cell.

In a cubic packed structure of mixed oxides , the lattice is made up of oxide lois one fifth of retrahedral voids are occupied by cation of a while one half of the formula of the oxide is

In a cubic crystal of CsCl (density = 3.97 gm//cm^(3)) the eight corners are occupied by Cl^(–) ions with Cs+ ions at the centre. Calculate the distance between the neighbouring Cs^(+) and Cl^(–) ions.

In a cubic crystal of CsCl (density = 3.97 gm/cm^(3)) the eight corners are occupied by Cl^(–) ions with Cs^(+) ions at the centre. Calculate the distance between the neighbouring Cs^(+) and Cl^(–) ions.

In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A^(2+)) while one half of the octahedral voids occupied trivalent ions (B^(3+)). What is the formula of the oxide ?

In a cubic crystal anions are arranged in fcc arrangement and the cations occupy all the octahedral voids and half the tetrahederal voids. The ratio of the cations and anions in the crystal is

In a cubic closed packed structure of mixed oxides the lattice is made up of oxide ions, one eighth of tetrahedral voids are occupied by divalent ions (A2^(+)) while one half of the octahedral voids occupied trivalent ions (B3^(+)). What is the formula of the oxide?

In a cubic closed packed structure of mixed oxide, the oxide ions are in CCP arrangement 1/6 of tetrahedral voids are occupied by cations A and 1/2 of octahedral voids are occupied by cations B. The formula of the oxide is

In a cubic close packing of spheres in there dimensions the coordination number of each sphere is:

In a cubic close packed structure the number of nearest neighbours for a given lattice point is

In acrystallinesubstanceatoms A occupycornersof a cubewhile atom B occupiesbodycentres .Whatis the formula of thesubstance?

In a crystalline solid, anions B are arranged in ccp lattice and cations A occupy 50% of the octahedral voids and 50% of the tetrahedral voids. What is the formula of the solid

In a crystal of compound having molecular formula X_2 Y_3, Y atoms are arranged in CCP, then what fraction of tetrahedral voids will be covered by X atoms ?

In a crystal, the atoms are located at the position of……potential energy.

In a crystal some ions are missing from normal sites This is an example of:

In a crystal AB, which of the following crystal systems will have parameters, a ne b ne c and alpha = beta = gamma = 90^(@) ?

In a coordination compound, if the metal ion has a secondary valence of six, it has an…………. Geometry.

In a covalent bond, a difference of 1.7 in the electronegativities of the combing atoms produces an ionic character of about __________.

In a container of 10 L volume, 20 g O_2(g) and 2.8 g CO(g) are taken at a total pressure of 2 atm and (27 ^0 C). Partial pressure and partial volume occupied by oxygen gas in container respectively are

In a constant volume calorimeter, 3.5g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0K to 298.45 K due to the combustion of the gas in kJ mol^(-1) is