94420 + Interview Questions in Chemistry in Class 12 Page 1749 InterviewSolution

87401.	A solid is made of two elements X and Z. The atoms Z are in C.C.P. arrangement while atoms X occupy all the tetrahedral sites. What is the formula of the compound-
Answer» XZ ` XZ_(2)` `X_(2)Z` Unpredictable Answer :C

Discussion

87402.	A solid is formed by two elements P and Q. The element Q forms cubic close packing and atoms of P occupy two-third of tetrahedral voids. The formula of the compound is
Answer» `PQ_(3)` `P_(3) Q` `P_(2)Q_(3)` `P_(3)Q_(2)` Solution :Let the no. of Q atoms = n No. of tetrahedral VOIDS = 2N No. of P atoms `=2n XX ( 1)/( 3)= ( 2n )/( 3)` FORMULA `: P : Q ` `( 2n )/( 3) : n ` or `2:3 ` `:. P _(2) Q_(3)`

Discussion

87403.	A solid is formed and it has three types of atoms X, Y, Z. X forms an FCC lattice with Y atoms occupying one-fourth of tetrahedral voids and Z atoms occupying half of the octahedral voids. The formula of the solid is
Answer» `X_4 Y_4 Z_2` `X_2 Y Z_2` `X_4 Y Z` `X_2 Y Z` ANSWER :D

Discussion

87404.	A solid having no definite geometrical shape with flat faces and sharp edges is :
Answer» AMORPHOUS solid Crystalline solid Isotropic solid None Answer :B

Discussion

87405.	A solid has a structure in which "W" atoms are located at the corner of the cubic lattice "O" atoms at the centre of edge and Na atom at the centre of cube. The formula of the compound is ...........
Answer» `NaWO_2` `Na_2WO_2` `NaWO_3` `Na_2WO_3` Solution :Number of W atoms = `( 8 xx 1/8) = 1` Number of Na atoms = 1 Number of O atoms `= (12 xx 1/4) = 3` `THEREFORE Na : W : O = 1 : 1 : 3 implies NaWO_3`.

Discussion

87406.	A solid has a strcture in which 'W' atoms are located at the corners of a cubic lattice 'O' atoms at the centre of edges and 'Na' atoms at the centre of the cube. The formula for the compound is
Answer» `NaWO_(2)` `NaWO_(3)` `Na_(2)WO_(3)` `NaWO_(4)` SOLUTION :In a unit cell, W atoms at the corner `=(1)/(8)xx8=1` O atoms at the centre of edges `=(1)/(4)xx12=3` Na atoms at the centre of the cube=1 `W:O:Na=1:3:1`, hence formula =`NaWO_(3)`

Discussion

87407.	A solid having density of 9 xx 10^3 "kg m"^(-3) form face centred cubic crystals of edge length 200sqrt(2) pm. What is the molar mass of solid ? (Avogadro's constant = 6 xx 10^23 "mol"^(-1), pi = 3)
Answer» `0.0216 "kg MOL"^(-1)` `0.0432 "kg mol"^(-1)` `0.4320 "kg mol"^(-1)` `0.0305 "kg mol"^(-1)` Solution :`d = (Z xx M)/(a^3 xx N_A)` `THEREFORE 9 xx 10^3 = (4 xx M)/((200 sqrt(2))^3 xx 10^(-36) xx 6 xx 10^(23))` `therefore M = 0.0305 "kg mol"^(-1)`.

Discussion

87408.	A solid has a structure in which W atoms are located at the corners of the cubic lattice, O atoms at the centre of the edges and Na atom at the centre of the cube. The formula of the compound is
Answer» `NawO_(2)` `Na_(2)WO_(3)` `NaW_(O_(3))` `NaWO_(4)` ANSWER :C

Discussion

87409.	A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73overset@A.The edge length of the cell is :
Answer» 200 pm142.2 pm Answer :A

Discussion

87410.	A solid has a b.c.c. structure . If the distance of closest approach between the two atoms is 1.73 Å. The edge length of the cell is :
Answer» 199 PM `SQRT(3//2)` pm 142.2 pm `SQRT2` pm ANSWER :A

Discussion

87411.	A solid element burns in oxygen without any change in volume (of gas) under similar conditions of temperature and pressure. If the vapour density of pure gaseous product is 32, what is the equivalent mass of the element?
Answer» SOLUTION :One vol. of oxide contains 1 VOLT. Of `O_(2)`. One MOLE of oxide contains one mole of `O_(2)`. Mol. Mass of oxide `=A=32=2V.D.=64` 32 PARTS of element combine with 32 parts of oxygen. So, equivalent mass of element`=(32)/(32)xx8=8`

Discussion

87412.	A solid dissolves in water exothermically, if its saturated solution at 20^@C is cooled to 0^@C then:-
Answer» Some solid separates out Some ICE separates out Both the solid and ice separate NEITHER the solid nor the ice separates out Solution :Due to DEPRESSION in P.P, WATER will freeze at a temperature LESS than `0^@C`.

Discussion

87413.	A solid crystal is composed of X, Y and Z atoms. Y atoms are occupying 50% of octahedral voids, whereas X atoms are occupying the 100 % tetrahedral void with Z atoms in ccp array arrangement, then the number of Y atoms in rational formula the given cell.
Answer» ANSWER :4

Discussion

87414.	A solid cube of edge length =25.32 mm of an ionic compound which has NaCl type lattice is added to 1kg of water.The boiling point of this solution is found to be 100.52^@C(assume 100% ionisation of ionic compound).If radius of anion of ionic solid is 200 pm then calculate radius of cation of solid in pm (picometer). (K_b of water =0.52 K kg "mole"^(-1),Avogadro's number,N_A=6xx10^(23),(root3(75))=4.22)
Answer» Solution :Effective molality of solution=1 Hence, number of moles of IONIC solid in GIVEN cube=0.5 So, number of formula units in given cube `=0.5xx6xx10^(23)` Number of units cells `=1/4xx0.5xx6xx10^23=7.5xx10^22` number of unit cells along ONE edge of cube `=root(3)(75)xx10^7=4.22xx10^7` edge length of unit cell =`a=(25.32xx10^(-3))/(4.22xx10^7)m=600xx10^(-12) m =600` pm forNaCl type unit cell, `a=2(r_(+)+r_(-)) " " IMPLIES " " r_(+)=100` pm

Discussion

87415.	A solid compound XY has NaCl structure . If the radius of the cation is 100 pm , the radius of the anion will be ……..
Answer» `((100)/(0.414))` `((0.732)/(100))` `100 xx 0.414` `((0.414)/(100))` Solution :`((100)/(0.414))` For an FCC structure `(r_(X))/(r_(Y)) = 0.414` , gives that `r_(X) = 100` PM `r_(Y) = (100 "pm")/(0.414)`

Discussion

87416.	A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion will be………………
Answer» `((100)/(0.414))` `((0.732)/(100))` `100xx0.414` `((0.414)/(100))` Solution :HINT : For an FCC STRUCTURE `(r_(X^(+)))/(r_(y^(-)))=0.414` GIVEN that `r_(X^(+))="100 PM "r_(y^(-))=("100 pm")/(0.414)`

Discussion

87417.	A solid compound 'X' on heating gives CO_(2) gas and a residue. The residue mixed with water forms 'Y'. On passing an excess of CO_(2) through Y in water, a clear solution 'Z' is obtained. On boiling 'Z' compound 'X' is formed. The compound 'X' is
Answer» `CaCO_(3)` `Na_(2)CO_(3)` `K_(2)CO_(3)` `Ca(HCO_(3))_(2)` ANSWER :A

Discussion

87418.	A solid compound contains X,Y and Z atoms in a cubic lattice with X atoms occupying the corners. Y atoms in the body centered positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound.
Answer» `XY_(2)Z_(1)` `XYZ_(3)` `X_(2)Y_(2)Z_(3)` `X_(8)YZ_(6)` Solution :Atoms of X PER UNIT cell `=8xx(1)/(8)=1` Atoms of Y per unit cell `=1` Atoms of Z per unit cell `=6xx(1)/(2)=3` Hence the formula is `XYZ_(3)`.

Discussion

87419.	Asolid compound contains XYZ atoms in a cubic lattice with X atoms occupying the corners, Y atomsin the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound
Answer» ANSWER :`XYZ_3`

Discussion

87420.	A solid catalyst is more effective in its finely divided form - Justify.
Answer» SOLUTION :Finely divided FORM of a catalyst has GREATER surface area when compared to their crystalline form. Greater the surface area, greater the ADSORPTION and HIGHER the catalytical activity.

Discussion

87421.	A solid AB has ZnS-type structure. The edge lenth of unit cell is 400 pm abd the radius of B^- ion is 0.130 nm. Then the radius of A^+ ion is
Answer» 35.8 pm 43.2 pm 60.5 pm 53.2 pm Answer :B

Discussion

87422.	A solid AB has the NaCl structure. If the radius of the cation A^(+) is 150 pm, calculate the maximum possible value of the radius of the anion B^(-).
Answer» Solution :For the NACL type structure, `(R^(+))/(r^(-))= 0.414 -0.732` `:. r^(+) = 150` PM and for MAXIMUM value of `r^(-), (r^(+))/(r^(-)) = 0.414` `:. (150)/(r^(-))= 0.414` or `r^(-) =(150)/(0.414) = 362.3` pm

Discussion

87423.	A solid AB has rock salt structure. If the radius of cation is 100 pm, the maximum radius of anion B^(ө)is ...........
Answer» 120.7 PM 68.3 pm. 136.61 pm. 241.54 pm SOLUTION :`(R^+)/(r^-) = 0.414 IMPLIES r^(-) = (100)/(0.414) = 241.54`

Discussion

87424.	A solid AB has NaCl structure. If the radius of cation A^(+) is 160 pm, calculate the maximum possible value of radius of the anion B^(-)?
Answer» Solution :The RADIUS ratio of `A^(+)` and `B^(-)` having `Na^(+)Cl^(-)` structure is = 0.414 to 0.732 `:. ("Radius of cation, "A^(+))/("Radius of anion, "B^(-))` = 0.414 to 0.732 `:. "Radius of anion "B^(-)=("Radius of cation, "A^(+))/("0.414 to 0.732") = (160)/("0.414 to 0.732")` = 386.47 to 218.58 PM `:.` Maximum POSSIBLE value of the radius of the anion `B^(-) = 386.47` pm

Discussion

87425.	A solid (A) which has photographic effect reacts with the solution of a sodium salt (B) to give a pale yellow ppt. (C). Sodium salt on heating gives brown vapour. Identify A, B and C
Answer» `AgNO_(3),NaBr,AgBr` `AgNO_(3),NACL,AgCl_(2)` `AgNO_(3),NaBr,AgCl_(2)` `AgCl,NaBr,AgBr_(2)` ANSWER :A

Discussion

87426.	Asolid A^+B^-has the B^-ions arranged as below .If the A^+ ions occupy half of the tetrahedral sites in the structure. The formula of solid is :
Answer» AB `AB_2` `A_2B` `A_3B_4` ANSWER :C

Discussion

87427.	A sol of egg albumin can be coagulated by addition of :
Answer» ELECTROLYTES Alcohols Both electrolytes and alcohols None of these Answer :C

Discussion

87428.	A sol may be prepared by a precipitation reaction. Give one such example. How can we find the nature of electric charge on the sol particles ?
Answer» Solution :`As_2S_3` SOL MAY be prepared by precipitation reaction as per the following reaction : `As_2O_3 + 3H_2S to As_2S_3 + 3H_2O` We can FIND the nature of charge on colloidal solution by electrophoresis. The colloidal PARTICLES move towards the oppositely charged electrode in an electric field. For example, the colloidal solution of `As_2O_3` is negatively charged and moves towards positive electrode.

Discussion

87429.	A soild, A which has photographic effect reacts with the solution of a sodium salt B to give a pale yellow precipitate C. Sodium salt on heatinggives brown vapours. Identity A,B and C.
Answer» `AgNO_(3),NaBr,AgBr` `AgNo_(3),NaCl,AgNO_(3)` `AgNO_(3),NaBr,AgCl_(2)` `AGCL,NaBr,AgBr_(2)` Solution :A solid silver nitrate `[AgNO_(3)(A)]` which has photographic effect reacts with the solution of NaBr (B) to give a pale YELLOW ppt. of AgBr which experience difficulty to be SOLUBLE in `NH_(4)OH NaBr` (B) on heating brown vapours of bromine. `underset(A)(AgNO_(3))+underset(B)(NaBr)rarrunderset(underset("Light yellow ppt.")(C))(AgBr+NaNO_(3))`

Discussion

87430.	A sodium salt X of an unknown anion when treated with MgCl_(2) gives precipate Y only on boiling. It gives lime water test positive but doesn't gives KMnO_(4) test. Compound X and Y can't be seperated by using.
Answer» Phenolphtalein TEST `AgNO_(3)` test `CaCl_(2) ` test `Hg_(2)(NO_(3))_(2)` test Answer :A

Discussion

87431.	A sodium salt on treatment with MgCl_2 gives white precipitate only on heating. The anion of sodium salt is:
Answer» `HCO_3^(-) ` `CO_3^(2-)` `NO_3^(-) ` `SO_4^(2-)` Solution :`MgCl_2 + NaHCO_3to UNDERSET("SOLUBLE")(Mg(HCO_3)_2)+2NaCl` `Mg(HCO_3)_2 overset("HEAT")to MgCO_3 + H_2O + CO_2 UARR `

Discussion

87432.	A sodium salt on treatment with MgCl_(2) gives white precipitate only on heating. The anion of sodium salt is
Answer» `HCO_(3)^(-)` `CO_(3)^(2-)` `NO_(3)^(-)` `SO_(4)^(2-)` Solution :N//A

Discussion

87433.	A sodium salt (A) when reacts with dilute HCI produces a salt and a colouless, rotten egg smelling gas (B). (B) when passed through lead acetate solution produces a black precipitate (C). Futher (C) on treatment with H_(2) O_(2) solution yields a white precipitate (D). Identify (A).
Answer» `H_(2) S` `Na_(2) S` `Na_(2) SO_(4)` `PbSO_(4)` Solution :As the GAS has rotten egg smell, so it may be `H_2S`. Hence the salf is sodium sulphide . Thus , `A = Na_2S, B= H_2S, C = PbS, D = PbSO_4` `underset((A))(Na_2S)+ 2HCL to 2NaCl + underset((B))(H_2S)` `Pb(CH_3COO)_(2) + H_2S to underset(("BLACK PPT."))underset((C))(PbS darr) + 2CH_3COOH` `PbS + 4H_2O_2 to underset(("WHITE ppt."))underset((D))(PbSO_4) darr + 4H_2O` .

Discussion

87434.	A sodium salt of an unknown anionwhentreates with MgCI_(2)gives whiteprecipitate onlyon boling .The anion is
Answer» `SO_(4)^(2-)` `HCO_(3)^(Θ)` `CO_(3)^(2-)` `NO_(3)^(Θ)` Solution :It must be a bicarbonate ionbecausefirst magnesiumbeicarbonate isformed which is soluble , then onheating , magnesiumcarbonate is formed which is a insoluble and FORMS a precipitate `MgCI_(2) + 2NaHCO_(3) RARR underset("soluble")((Mg(HCO_(3))_(2)) + 2NaCI` `Mg(HCO_(3))_(2) OVERSET(Delta)(to) underset(white)(MgCO_(3) darr) + H_(2)o + co_(2)`

Discussion

87435.	(a) Sodium metal crystallizes in bcc structure. Its unit cell edge length is 420 pm. Calculate its density. (Atomic mass of sodium = 23 u, N_(A) = 6.022 xx 10^(23) mol^(-)). (b) What is Frenkel defect ? How does it affect the density of a crystal ?
Answer» Solution :(a) Density `= (Z.M)/(a^(2) N_(A))` `= (2 XX 23 xx 10^(-3))/(420 xx 10^(-12) xx 6.022 xx 10^(23))` `= 1.031 xx 10^(3) kg//m^(3)` (b) This is a defect caused by the DISLOCATION of a smaller ION from its normal site to an intersitial site. Density does not change.

Discussion

87436.	(A) Sodium can't be used for drying ethyl alcohol.(R) Sodium displaces hydrogen from ethyl alcohol.
Answer» Both A & Rare true, R is the correct EXPLANATION of A Both A & Rare true, R is not correct explanation of A A is true, R is FALSE A is false, R is true Answer :A

Discussion

87437.	A + SOCl_2 to B + SO_2 + HCl X + Na to C + H_2B + C to (C_2H_5)_2O+ NaClThen A and X are respectively
Answer» ` C_2H_5CI and C_2H_5ONa` ` C_2H_5ONa and C_2H_5Cl` ` C_2H_5OH and C_2H_5OH` ` C_2H_5OH and C_2H_5ONa` ANSWER :C

Discussion

87438.	A soap can be obtained by the saponificattion of
Answer» Liquid paraffin COCONUT OIL LEMONGRASS oil SANDAL wool Answer :B

Discussion

87439.	(A) SO_(2) gas is easily liquified while H_(2) isdifficult to liquify . (R ) SO_(2) has low critical temperature while H_(2) has high critical temperature.
Answer» Both (A) and (R) are true and (R) is the correct explanation of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE (A) is false but (R) is true ANSWER :C

Discussion

87440.	(A): SO_(3) molecule has a planar structure (R) : S atom in SO_(3) is sp^(2)- hybridized and O-S-O bond angle is 120^(@)
Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R )are false ANSWER :A

Discussion

87441.	(A) SO _(2) acts as good reducing agent in aqueous solutions (R ) SO _(2) + 2 H _(O to H _(2) SO _(4) + 2 (H)
Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R )are false Answer :A

Discussion

87442.	(A) SO _(2) is reducing agent and TeO _(2) is oxidising agent (R )From SO _(2) to TeO _(2) acidic nature of dioxides increases.
Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R )are false Answer :C

Discussion

87443.	(A) SO _(2) is angular molecule (R ) In SO _(2) sulphur exhibit SP ^(2) hybridisation
Answer» Both (A) and (R ) are true and (R ) is the correct explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is FALSE Both (A) and (R )are false ANSWER :B

Discussion

87444.	(A) SnCl_(4) is more covalent than SnCl_(2) (R ) Compound with higher oxidation state of metal is more covalent than with lower oxidation state
Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is FALSE Both (A) and (R ) are false Answer :A

Discussion

87445.	(A) S_N2 reactions are exothermic (R) S_N2 reactions are thermochemically favored by stronger nucleophiles
Answer» Both (A) and (R) are correct and (R) is the correct EXPLANATION of (A) Both (A) and (R) are correct but (R) is not the correct explanation of (A) (A) is TRUE but (R) is False Both (A) and (R) are false Answer :A

Discussion

87446.	(A): S_(N^(2)) reaction takes place in single step.(R): S_(N^(2)) reaction involves the reactivity order of alkyl halides as 1^@ gt 2^@ gt 3^@ halides.
Answer» Both A & R are TRUE, R is the correct explanation of A Both A & R are true, R is not correct explanation of A A is true, R is false A is false, R is true Answer :B

Discussion

87447.	A smuggler could not carry gold by depositing iron on gold surface since
Answer» gold has higher standard reduction potential than iron gold has lower standard reduction potential than iron gold is denser iron rusts Solution :A SMUGGLER could not carry gold by deposting iron on gold surfase. We KNOW that GREATER is the standard ELECTRODE potential of an electrode, the more is the TENDENCY of its ion to get induced. Since the standard reduction potential of `Au^(3+)//Au` is greater than `Fe^(2+)//Fe`, iron cannot be deposited on gold by electroplating.

Discussion

87448.	A smuggler could not carry gold by depositing iron on the gold surface since
Answer» Gold is denser Iron rusts Gold has higher REDUCTION POTENTIAL than iron Gold has lower reduction potential than iron Answer :C

Discussion

87449.	A small sample of Uranium metal (0.119 gm) is heated to 900^(@) C in air to give 0.135 gm of a dark oxide, U_(x) O_(y). Then [U = 238]
Answer» Moles of `U_(X) O_(y)` formed `= 5 xx 10^(-3)` Empirical formula is `UO_(2)` Moles of Uranium metal used `5 xx 10^(-4)` Moles of oxygen gas used `1 xx 10^(-3)` Solution :`{:(U,+,O_(2),rarr,U_(x)O_(y),):}` Moles of `U = (0.119)/(238) = 0.5 xx 10^(-3) = 5 xx 10^(-4)` Mass of oxygen in oxide `= (0.135 - 0.119)` `= 0.016 gm` Moles of oxgyen ATOM in oxide `= (0.016)/(16) = 1 xx 10^(-3)` `{:(U,,O,),(5 xx 10^(-4),,1 xx 10^(-3),),(0.5,:,1,),(1,:,2,):}` Emprical formula `= UO_(2)`

Discussion

87450.	A small iodine crystal is added to each of the following aq solutions: I. sodium sulphate II. Sodiumthiosulphate III. Sodium tetrathionate IV. Sulphuric acid In which solutions the purple colour disappers
Answer» only in I only in I and II only in III and IV only in II Answer :D

Discussion

Explore topic-wise InterviewSolutions in .

A solid is made of two elements X and Z. The atoms Z are in C.C.P. arrangement while atoms X occupy all the tetrahedral sites. What is the formula of the compound-

A solid is formed by two elements P and Q. The element Q forms cubic close packing and atoms of P occupy two-third of tetrahedral voids. The formula of the compound is

A solid is formed and it has three types of atoms X, Y, Z. X forms an FCC lattice with Y atoms occupying one-fourth of tetrahedral voids and Z atoms occupying half of the octahedral voids. The formula of the solid is

A solid having no definite geometrical shape with flat faces and sharp edges is :

A solid has a structure in which "W" atoms are located at the corner of the cubic lattice "O" atoms at the centre of edge and Na atom at the centre of cube. The formula of the compound is ...........

A solid has a strcture in which 'W' atoms are located at the corners of a cubic lattice 'O' atoms at the centre of edges and 'Na' atoms at the centre of the cube. The formula for the compound is

A solid having density of 9 xx 10^3 "kg m"^(-3) form face centred cubic crystals of edge length 200sqrt(2) pm. What is the molar mass of solid ? (Avogadro's constant = 6 xx 10^23 "mol"^(-1), pi = 3)

A solid has a structure in which W atoms are located at the corners of the cubic lattice, O atoms at the centre of the edges and Na atom at the centre of the cube. The formula of the compound is

A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73overset@A.The edge length of the cell is :

A solid has a b.c.c. structure . If the distance of closest approach between the two atoms is 1.73 Å. The edge length of the cell is :

A solid element burns in oxygen without any change in volume (of gas) under similar conditions of temperature and pressure. If the vapour density of pure gaseous product is 32, what is the equivalent mass of the element?

A solid dissolves in water exothermically, if its saturated solution at 20^@C is cooled to 0^@C then:-

A solid crystal is composed of X, Y and Z atoms. Y atoms are occupying 50% of octahedral voids, whereas X atoms are occupying the 100 % tetrahedral void with Z atoms in ccp array arrangement, then the number of Y atoms in rational formula the given cell.

A solid compound XY has NaCl structure . If the radius of the cation is 100 pm , the radius of the anion will be ……..

A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion will be………………

A solid compound 'X' on heating gives CO_(2) gas and a residue. The residue mixed with water forms 'Y'. On passing an excess of CO_(2) through Y in water, a clear solution 'Z' is obtained. On boiling 'Z' compound 'X' is formed. The compound 'X' is

A solid compound contains X,Y and Z atoms in a cubic lattice with X atoms occupying the corners. Y atoms in the body centered positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound.

Asolid compound contains XYZ atoms in a cubic lattice with X atoms occupying the corners, Y atomsin the body centred positions and Z atoms at the centres of faces of the unit cell. What is the empirical formula of the compound

A solid catalyst is more effective in its finely divided form - Justify.

A solid AB has ZnS-type structure. The edge lenth of unit cell is 400 pm abd the radius of B^- ion is 0.130 nm. Then the radius of A^+ ion is

A solid AB has the NaCl structure. If the radius of the cation A^(+) is 150 pm, calculate the maximum possible value of the radius of the anion B^(-).

A solid AB has rock salt structure. If the radius of cation is 100 pm, the maximum radius of anion B^(ө)is ...........

A solid AB has NaCl structure. If the radius of cation A^(+) is 160 pm, calculate the maximum possible value of radius of the anion B^(-)?

A solid (A) which has photographic effect reacts with the solution of a sodium salt (B) to give a pale yellow ppt. (C). Sodium salt on heating gives brown vapour. Identify A, B and C

Asolid A^+B^-has the B^-ions arranged as below .If the A^+ ions occupy half of the tetrahedral sites in the structure. The formula of solid is :

A sol of egg albumin can be coagulated by addition of :

A sol may be prepared by a precipitation reaction. Give one such example. How can we find the nature of electric charge on the sol particles ?

A soild, A which has photographic effect reacts with the solution of a sodium salt B to give a pale yellow precipitate C. Sodium salt on heatinggives brown vapours. Identity A,B and C.

A sodium salt X of an unknown anion when treated with MgCl_(2) gives precipate Y only on boiling. It gives lime water test positive but doesn't gives KMnO_(4) test. Compound X and Y can't be seperated by using.

A sodium salt on treatment with MgCl_2 gives white precipitate only on heating. The anion of sodium salt is:

A sodium salt on treatment with MgCl_(2) gives white precipitate only on heating. The anion of sodium salt is

A sodium salt (A) when reacts with dilute HCI produces a salt and a colouless, rotten egg smelling gas (B). (B) when passed through lead acetate solution produces a black precipitate (C). Futher (C) on treatment with H_(2) O_(2) solution yields a white precipitate (D). Identify (A).

A sodium salt of an unknown anionwhentreates with MgCI_(2)gives whiteprecipitate onlyon boling .The anion is

(a) Sodium metal crystallizes in bcc structure. Its unit cell edge length is 420 pm. Calculate its density. (Atomic mass of sodium = 23 u, N_(A) = 6.022 xx 10^(23) mol^(-)). (b) What is Frenkel defect ? How does it affect the density of a crystal ?

(A) Sodium can't be used for drying ethyl alcohol.(R) Sodium displaces hydrogen from ethyl alcohol.

A + SOCl_2 to B + SO_2 + HCl X + Na to C + H_2B + C to (C_2H_5)_2O+ NaClThen A and X are respectively

A soap can be obtained by the saponificattion of

(A) SO_(2) gas is easily liquified while H_(2) isdifficult to liquify . (R ) SO_(2) has low critical temperature while H_(2) has high critical temperature.

(A): SO_(3) molecule has a planar structure (R) : S atom in SO_(3) is sp^(2)- hybridized and O-S-O bond angle is 120^(@)

(A) SO _(2) acts as good reducing agent in aqueous solutions (R ) SO _(2) + 2 H _(O to H _(2) SO _(4) + 2 (H)

(A) SO _(2) is reducing agent and TeO _(2) is oxidising agent (R )From SO _(2) to TeO _(2) acidic nature of dioxides increases.

(A) SO _(2) is angular molecule (R ) In SO _(2) sulphur exhibit SP ^(2) hybridisation

(A) SnCl_(4) is more covalent than SnCl_(2) (R ) Compound with higher oxidation state of metal is more covalent than with lower oxidation state

(A) S_N2 reactions are exothermic (R) S_N2 reactions are thermochemically favored by stronger nucleophiles

(A): S_(N^(2)) reaction takes place in single step.(R): S_(N^(2)) reaction involves the reactivity order of alkyl halides as 1^@ gt 2^@ gt 3^@ halides.

A smuggler could not carry gold by depositing iron on gold surface since

A smuggler could not carry gold by depositing iron on the gold surface since

A small sample of Uranium metal (0.119 gm) is heated to 900^(@) C in air to give 0.135 gm of a dark oxide, U_(x) O_(y). Then [U = 238]

A small iodine crystal is added to each of the following aq solutions: I. sodium sulphate II. Sodiumthiosulphate III. Sodium tetrathionate IV. Sulphuric acid In which solutions the purple colour disappers