94420 + Interview Questions in Chemistry in Class 12

1.	CuSO_(4) reacts with Kl in acidic medium to liberate l_(2) 2CuSO_(4) + 34Kl to Cu_(2)l_(2) + 2K_(2)SO_(4)+l_(2), Mercuric periodate Hg_(5)(lO_(6))_(2) reacts with a mixture of Kl and HCl according to the following equation : Hg_(5)(lO_(6))_(2) + 4Kl + 24HCl to 5K_(2)Hgl_(4) + 8l_(2) + 24KCl + 12H_(2)O The liberated iodine is titrated against Na_(2)S_(2)O_(3)solution, 1 ml of which is equivalent to 0.0499 gm of CuSO_(4). 5H_(2)O. What volume in ml of Na_(2)S_(2)O_(3) solution will be required to react with l_(2) liberated from 0.7245 gm ofHg_(5)(lO_(6))_(2) ? Molecular weight of Hg_(5)(lO_(6))_(2) ? Molecular weight of CuSO_(4).5H_(2)O - 249 . 5
Answer» SOLUTION :N/A

Discussion

2.	CuSO_4 solution reacts with excess of KCN solutionto from:
Answer» `Cu(CN)_2` `K_2[Cu(CN)_4]` `K_3[Cu(CN)_4]` `K[Cu(CN)_2]` ANSWER :C

Discussion

3.	{:(CuSO_(4)+NH_(4)OH to "Deep blue soluble"),("excesscomplex salt"):} The correct statement regarding the complex is/are :
Answer» It has square planar geometry. It is paramagnetic in nature . It GIVES black PRECIPITATE of `CuS` when treated with `H_(2)S`. All of these Solution :`[overset(+2)Cu(NH_(3))_(4)]SO_(4) overset(H_(2)S)to CuSdarr (Black)implies dsp^(2),"square planar (one unparied ELECTRON)"`

Discussion

4.	CuSO_4 on mixing with NH_3 (1:4 ratio) does not give test for Cu^(2+) ions but gives test for SO_(4)^(2-) ions. Why?
Answer» SOLUTION :It is because when `NH_3` COORDINATES to `Cu^(2+)` IONS and it forms the COMPLEX `[Cu(NH_3)_4]SO_4` copper ions are PRESENT in coordination sphere, therefore they are non-ionisable whereas `SO_(4)^(2-)` ions are counter ions which are ionisable.

Discussion

5.	CuSO_(4) solution +lime is called:
Answer» LUCA's reagent Befoed's reagent Fihling SOLUTION A Bordeaux mixture Answer :D

Discussion

6.	CuSO_(4) +NH_(4)OH gives deep blue complex of
Answer» `[Cu(NH_(3))_(4)]SO_(4)` `[Cu(NH_(3))_(2)]SO_(4)` `[Cu(NH_(3))_(6)]SO_(4)` `[Cu_(2)(NH_(3))_(4)]SO_(4)` ANSWER :A

Discussion

7.	CuSO_(4) is not stored in aluminium bottles because
Answer» Cu gets oxidised Cu gets REDUCED Al gets reduced `CuSO_(4)` gets DECOMPOSED Solution :Al is more REACTION than `Cu` i.e., `Cu^(2+)` gets reduced .

Discussion

8.	CuSO_(4) is colourless while CuSO_(4).5H_(2)O is coloured. Why?
Answer» Solution : `CuSO_(4)`does not has any LIGAND, so SPLITTING of d-orbital take place while`CuSO_(4).5H_(2)O`has WATER ligand.

Discussion

9.	CuSO_(4) is colourless while CuSO_(4) 5H_(2)O is coloured. Why? [Ti(H_(2)O)_(6)]^(3+)is coloured while [Sc(H_(2)O)_(6)]^(3+) is colourless, why?
Answer» SOLUTION :`CuSO_(4)` does not have any ligand, so splitting of d-orbitals take place while `CuSO_(4), 5H_(2)O` has water as ligand. `TI^(3)` has one UNPAIRED electron while `SC^(3+)` does not have any unpaired `e^(-)`, hence no `d-d` transition in `SC^(3+)`.

Discussion

10.	CuSO_(4) decolourises onaddition KCN , the produce is
Answer» `[CU(CN)_(4)]^(2-)` `Cu^(2+)` get redoced is FORM `[Cu(CN)_(4)]^(3-)` `Cu(CN)_(2)` `CUCN` SOLUTION :N//A

Discussion

11.	CuSO_(4) decolourises on addition of KCN, the product is
Answer» `[Cu(CN)_(4)]^(2-)` `Cu^(2+)` get reduced to FORM `[Cu(CN)_(4)]^(3-)` (C) `Cu(CN)_(2)` `CUCN` ANSWER :A

Discussion

12.	CuSO_4 crystallise in rock salt structure.Its cell parametercan be determined by various experimental methods like electrical conductivity measurement, colligative properties measurement, pHmeasurement etc. A cubic crystal of CuSO_4 of edge length 17.1 mm, is dissolved in water to make 500 ml solution of pH 5.root3(1.5)=1.14) Given : Cu(H_2O)_6^(2+)+H_2O hArr [Cu(H_2O)_5(OH)]^(+)+H_3O^(+) , K=10^(-5) In the given solution is made 1 M with respect to [Cu^(2+)] becomes 10^(-15) M therefore K_f for the formation of Cu(NH_3)_4^(2+) is
Answer» `10^10` `10^8` `2xx10^8` `2xx10^10` Solution :`{:(Cu(H_2O)_6^(2+)+H_2OhArr,[CA(H_2O)_5(OH)]^(+),+H_3O^(+)),(C, , ),(C-x,x,x):}` `10^(-5)=x^2/(C-x)=((10^(-5))^2)/(C-10^(-5))` `C=2xx10^(-5)` mol/lt `:.` MOLES of `CuSO_4` DISSOLVED =`2xx10^(-5)xx0.5 =10^(-5)` Number of units cells `=(10^(-5)xxN_A)/4=6/4xx10^(18)=1.5xx10^(18)` number of unit cells along one edge of the cube =`root3(1.5xx10^(18))=1.14xx10^8` If edge LENGTH of F C C unit cell is a Now`{:(Cu^(2+)+,4NH_3hArr,Cu(NH_3)_4^(2+)),(2xx10^(-5), , ),(10^(-15),1,2xx10^(-5)):}` `K_f=([Cu(NH_3)_4^(2+)])/([Cu^(2+)][NH_3]^(4))=(2xx10^(-5))/(10^(-15)xx1)=2xx10^10`

Discussion

13.	CuSO_4 crystallise in rock salt structure.Its cell parametercan be determined by various experimental methods like electrical conductivity measurement, colligative properties measurement, pHmeasurement etc. A cubic crystal of CuSO_4 of edge length 17.1 mm, is dissolved in water to make 500 ml solution of pH 5.root3(1.5)=1.14) Given : Cu(H_2O)_6^(2+)+H_2O hArr [Cu(H_2O)_5(OH)]^(+)+H_3O^(+) , K=10^(-5) Edge length of FCC unit cell of copper sulphate is
Answer» 120 Å 170 Å 150 Å 200 Å Solution :NA

Discussion

14.	CuSO_4. 5H_2O(s)hArrCuSO_4(s)+5H_2O(g)K_P=10^(-10)("atm").10^(-2)moles of CuSO_4. 5H_2O(s) is taken in a 2.5 L container at 27^@C then at equilibrium [Take: R=1/12litre atm "mol"^(-1)K^(-1)]
Answer» Moles of `CuSO_4. 5H_2O` LEFT in the container is `9xx10^(-3)` Moles of `CuSO_4. 5H_2O` left in the container is `9.8xx10^(-3)` Moles of `CuSO_4` left in the container is `10^(-3)` Moles of `CuSO_4.` left in the container is `2XX10^(-4)` Solution :`10^(-10)"atm"^5=P_(H_2O)^5 " " implies P_(H_2O)=10^(-2) "atm" "" n=(PV)/(RT)=(10^(-2)xx2.5)/(1/12xx300)=10^(-3)`

Discussion

15.	CuSO_4 , 5 H_2O is known as _______.
Answer» SOLUTION :BLUE VITRIOL

Discussion

16.	CuSO_4 5H_2O is commonly known as
Answer» SOLUTION :BLUE VITRIOL

Discussion

17.	CuSo_(4). 5 H_(2)O is a
Answer» solution of a solid in a LIQUID. solution of liquid in a solid salt only and cannot be called a solution coordination compound of copper with water molecules as the ligands. Solution :Liquid `H_(2)O` is UNIFORMLY mixed into solid `CuSo_(4)`.

Discussion

18.	Current is passed through two cells connected in series,the first cell contains X(NO_3)_(3(aq)) and the second cell contains Y(NO_(3))_(2(aq)). The relative atomic masses of X and Y are in the ratio 1 : 2. What is the ratio of the liberated mass of X to that of Y?
Answer» `3:2` `1:2` `1:3` `3:1` Solution :If atomic mass of X = a, then that of Y =2A As EQ.wt. = Atomic mass/Valency Eq. wt. of `X=a/3`, Eq. wt. of `Y=(2a)/2` =a According to Faraday.s second LAW : `"Liberated Mass of X"/"Liberated Mass of Y"="Eq. wt. of X"/"Eq. wt. of Y"=(a//3)/a=1/3`=1:3

Discussion

19.	Current market prices of Al, Zn and Fe scraps per kg are Rs. 20, Rs. 16 and Rs. 3 respectively. If H_(2) is to be prepared by the reaction of one of these metals with H_(2)SO_(4), which would be the cheapest metal to use ? Which would be most expensive ?
Answer» Solution :The various chemical reactions INVOLVED are given below : (i) `underset(2xx27=54 G)(2Al)+3H_(2)SO_(4)rarr Al_(2)(SO_(4))_(3)+underset(3xx2=6 g)(3H_(2))` (ii) `underset(65 g)(Zn)+H_(2)SO rarr ZnSO_(4) + underset(2g)(H_(2))` (iii)`underset(56 g)(Fe)+H_(2)SO_(4)rarr FeSO_(4)+underset(2 g)(H_(2))` Step 1. To calculate the cost of preparation of 100 g of `H_(2)` from Al 6 g of `H_(2)` prepared from Al = 54 g `THEREFORE` 100 g of `H_(2)` will be obtained from `Al=(54)/(6)xx100=900 g` Cost of 1000 g of Al = Rs. 20 `therefore` Cost of 900 g of `Al=(20)/(1000)xx900=Rs. 18` Step 2. To calculate the cost of preparation of 100 g of `H_(2)` from Zn. 2 g of `H_(2)` is peoduced from Zn = 65 g `therefore` 100 g of `H_(2)` will be obtained from `Zn=(65)/(2)xx100=3250 g` Cost of 1000 g of Zn Rs. 16 `therefore` Cost of 3250 g of `Zn=(16)/(1000)xx3250=Rs.52`. Step 3. To calculate the cost of preparation of 100 g of `H_(2)` from Fe. 2 g of `H_(2)` is produced from Fe = 56 g `therefore` 100 of `H_(2)` will be obtained from `Fe=(56)/(2)xx100=2800 g` Cost of 1000 g of Fe = Rs. 3 `therefore` Cost of 2800 g of `Fe=(3)/(1000)xx2800=Rs. 8.40`. Thus, Fe is the cheapest and Zn is the most expensive metal to use for the preparation of `H_(2)`.

Discussion

20.	Curves with the shape shown are often observed for reactions involving catalysts. The level portion of the curve is best attributed to the fact that:
Answer» PRODUCT is no LONGER being formed the REACTION has reached equilibrium all the CATALYTIC sites are occupied all the reactant has been consumed ANSWER :C

Discussion

21.	Current efficiency is defined as the extent of a desired electrochemical reaction, divided by the theoretical extent of the reaction times 100%. What is the current efficiency of an electrodeposition of Cy metal in which 9.8g Cu is deposited by passage of 3A current for 10000 seconds.
Answer» 0.009 0.499 0.991 0.511 Solution :`9.8g Cu=(9.8)/(63.5) XX 2=0.308" mol E"^(-1)` `1000 xx 3 xx 1/(96500)=0.311" mol e"^(-1)` If 0.311 mol electrons PROVIDED by the current 0.308 mol was used to DEPOSIT copper. The current efficiency is `(0.308)/(0.311) xx 100 =99.1%`

Discussion

22.	Curdling of milk is
Answer» naturation of protein denaturation of protein folding of POLYPEPTIDE chain coiling of polypeptide chain Answer :B

Discussion

23.	Curd'contain the following vitamin
Answer» VITAMIN `B_(12)` Vitamin `B_6` Vitamin A Vitamin C Answer :A

Discussion

24.	Cuprous salts are generally colourless while cuprous oxide is:
Answer» Green Blue Red Yellow Answer :C

Discussion

25.	Cuprous chloride is obtained from cupric chloride :
Answer» By HEATING CAPRIC chloride with CHOLRINE By the electrolysis of cupric chloride con-taining HCL By heating cupric chloride with conc. HCl and copper turnings By passing `H_2` OVER `CuCI_2` Answer :C

Discussion

26.	Cuprite contains
Answer» `Cu_(2)S` `CUFES` `Cu(OH)_(2).CuCO_(3)` `Cu_(2)O` ANSWER :D

Discussion

27.	Cuprous chloride can not be prepared :
Answer» by PASSING `SO_2` through the solution containing `CuSO_4` and NaCl by HEATING EXCESS of copper with conc. HCL in presence of a LITTLE `KClO_3` by boiling copper sulphate solution with excess of copper turnings in presence of hydrochloric acid by dissolving cupric oxide or copper carbonate in conc. HCl Solution :(D)`CuO+2HCl toCuCl_2+H_2O,Cu_2Cl_2` is not formed. (A)`2CuSO_4 + 2NaCl+2H_2O+SO_2 to Cu_2Cl_2 + Na_2 SO_4 +2H_2SO_4` (B)`Cu+HCl+[O]toCuCl_2+H_2O,CuCl_2+CutoCu_2Cl_2` (C )`CuSO_4 + 2HCl to CuCl_2 +H_2SO_4 , CuCl_2 + Cu to Cu_2Cl_2`

Discussion

28.	Cupric compounds are more stable than their cuprous counterparts in solid state. This is because
Answer» <P>the endothermic character of the 2nd I P of Cu is not so high size of `Cu^(2+)` is less than `Cu^(+)` `Cu^(2+)` has stabler electronic configuration as COMPARED to `Cu^(+)` the lattice energy RELEASED for cupric compounds is much higher than `Cu^(+)`. SOLUTION :(b,d) These are both correct options.

Discussion

29.	Cupra-ammonium silk is
Answer» natural FIBRE semi SYNTHETIC fibre PROTEIN fibre true synthetic fibre Answer :B

Discussion

30.	Cupellation process is used in he metallurgy of :
Answer» Copper Silver Lead Iron Answer :B

Discussion

31.	Cupellation is a process used for the refining of
Answer» Silver Lead Copper Iron Answer :A

Discussion

32.	"Cumene"underset((ii)H_(2)O,H^(+))overset((i)O_(2))to (X) and (Y) What is X and Y?
Answer» TOLUENE, Propene Toluene, PROPYL chloride PHENOL, ACETONE Phenol, Acetaldehyde Solution :Phenol, Acetone

Discussion

33.	Cumene on air oxidation give
Answer» CARBONIC acid carbolic acid carboxylic acid oxalic acid Answer :B

Discussion

34.	Cumene process is the most important commercial method for the manufacture of phenol. Cumene is
Answer» 1-methyl ethyl benzene ethyl benzene vinyl benzene PROPYL benzene. Solution :CUMENE is ISOPROPYL benzene or 1-methylethyl benzene which can be PREPARED as follows.

Discussion

35.	Cumene is isopropyl benzene.
Answer» ANSWER :T

Discussion

36.	Cumene is converted in phenol by
Answer» REDUCTION and DECOMPOSITION by acid oxidation and decomposition by acid reduction and decomposition by alkali oxidation and decomposition by alkali Answer :B

Discussion

37.	Cumene is
Answer» PHENYL n-propane 2-propyl benzene chlorobenzene benzene Answer :B

Discussion

38.	CuFeS_(2) mineral was analysed for Cu and Fe percentage : 10 g of it was boiledwith dil . H_(2)SO_(4)and diluted to 1 L. 10 ml ofthis solution required 2 mL of 0.01 M MnO_(4)^(-) in acidic medium . In another titration, 25 ml of the same solutionrequired . 5 ml of 0.01MMnO_(4)^(-) solution iodometrically . Calculate the percentage of Cu and Fe in the mineral.
Answer» SOLUTION :N/A

Discussion

39.	CuCO_(3) was strongly heated to obtain a residue A and gas B. The residue obtained was treated with a salt of sodium 'X' and oxide Y, which produced a blue colored glassy compound C on heating in oxidising flame. The same combination of x and Y gave a green colored glassy compound D when Cr_(2)(SO_(4))_(3) was heated with them in oxidizing flame. Match the following accordingly :
Answer» SOLUTION :`CuCO_(3) overset(Delta)to underset((A))(CuO)darr+underset((B))(CO_(2))uarr` `CuO+underset("Borax bead")ubrace(overset((X))(NaBO_(2))+B_(2)O_(3))to underset((C )"Blue bead")ubrace(CU(BO_(2))_(2))+NaBO_(2)` `Cr_(2)(SO_(4))_(3) underset("Bead test")overset("Borax")to ` GREEN COLORED metaborate

Discussion

40.	Cu(CN)_(4)^(2-) is colourless as it absorbs light in
Answer» VISIBLE region Ultraviolet region Infrared region All above are wrong Answer :D

Discussion

41.	CuCN reacts with FeCl_3 to give
Answer» `CuCl_2` `Cu(CN)_2` `(CN)_2` `All` Solution :`CUCN + FeCl_3 RARR (CN)_2 + CuCl + FeCl_2`

Discussion

42.	[CuCl_(4)]^(2-) exists while [CyI_(4)]^(2-) does not exist why ?
Answer» Solution :`(i)` In `[CuCl_(4)]^(2-)` complex, the size of chloride ion is less hence exist. But in `[CuI_(4)]^(2-)` the bigger iodide ion makes the compound unstable. `(ii)` When copper cation comes in contact with iodide anion, iodide get oxidised to iodine molecule hence the formation of the above complex ion does not take place. Hence `[CuCl_(4)]^(2-)` EXISTS while `[CuI_(4)]^(2-)` does not exist.

Discussion

43.	CuCl_(2) is used as catalyst in
Answer» the conversion of `(CH_(3))_(2)SiCl_(2)` to linear silicones making `Cl_(2)` fom HCl in Deacon's process the oxidation of the following metals does not DISSOLVE in aqua-regia ? in oxidation of alcohols ANSWER :B

Discussion

44.	CuCl is sparingly soluble in water but it dissolves in KCl solution due to the formation of
Answer» `K_(2)[CuCl_(4)]` `K_(2)[CuCl_(4)]` `K[CuCl_(2)]` NONE of the above. Solution :Hybrid state of NI in `Ni(CO)_(4)` is `sp^(3)` Ni(0) has CONFIGURATION `3d^(8)4s^(2)`. At the time of bonding two electrons of 4s-level go to 3d-orditals the 4s and 4p-orbitals hybridise to assume `sp^(3)` hybrid state.

Discussion

45.	CuCl is sparingly soluble in H_2O but it dissolves in KCI solution due to the formation of:
Answer» `K_2[CuCl_4]` `K_3[CuCl_4]` `K[CuCl_2]` None Answer :B

Discussion

46.	CuCl is insoluble in water but it dissolves in KCl solution. This is due to the formation of the complex.
Answer» `K_(2)[CuCl_(3)]` `K_(3)[CuCl_(4)]` `K[CuCl_(2)]` All of these Solution :All the given COMPLEXES are FORMED which are soluble in WATER.

Discussion

47.	CaCl is colourless, while CuSO_(4).5H_(2)O is coloured. Explain?
Answer» Solution :Electronic configuration of `Cu^+` ion is [Ar] `4s^0 3d^9` In `CUCL, Cu^(+)` ion has no unpaired d- electron . THUS d-d transition is no POSSIBLE in CuCl and HENCE it is colourless. In `CuSO_4. 5H_2O` the `Cu^(2+)` ion has one unpaired d-electron in 3d -subshell . Thus, d-d transition is possible and hence it EXHIBITS colour.

Discussion

48.	CuCl dissolves in ammonia forming a complex. The co-ordination number of copper in the complex is:
Answer» 1 2 4 6 Answer :B

Discussion

49.	Cubic unit cells are divided into primitive, bcc and fcc. Calculate the number of atoms in unit cell of each of the following : *fcc
Answer» SOLUTION :FCC, number of atoms = `(8xxfrac{1}{8})+ (6xxfrac{1}{2})=4`

Discussion

50.	Cubic unit cells are divided into primitive, bcc and fcc. Write two examples for covalent solids.
Answer» SOLUTION :GRAPHITE,DIAMOND

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

CuSO_4 solution reacts with excess of KCN solutionto from:

{:(CuSO_(4)+NH_(4)OH to "Deep blue soluble"),("excesscomplex salt"):} The correct statement regarding the complex is/are :

CuSO_4 on mixing with NH_3 (1:4 ratio) does not give test for Cu^(2+) ions but gives test for SO_(4)^(2-) ions. Why?

CuSO_(4) solution +lime is called:

CuSO_(4) +NH_(4)OH gives deep blue complex of

CuSO_(4) is not stored in aluminium bottles because

CuSO_(4) is colourless while CuSO_(4).5H_(2)O is coloured. Why?

CuSO_(4) is colourless while CuSO_(4) 5H_(2)O is coloured. Why? [Ti(H_(2)O)_(6)]^(3+)is coloured while [Sc(H_(2)O)_(6)]^(3+) is colourless, why?

CuSO_(4) decolourises onaddition KCN , the produce is

CuSO_(4) decolourises on addition of KCN, the product is

CuSO_4. 5H_2O(s)hArrCuSO_4(s)+5H_2O(g)K_P=10^(-10)("atm").10^(-2)moles of CuSO_4. 5H_2O(s) is taken in a 2.5 L container at 27^@C then at equilibrium [Take: R=1/12litre atm "mol"^(-1)K^(-1)]

CuSO_4 , 5 H_2O is known as _______.

CuSO_4 5H_2O is commonly known as

CuSo_(4). 5 H_(2)O is a

Current is passed through two cells connected in series,the first cell contains X(NO_3)_(3(aq)) and the second cell contains Y(NO_(3))_(2(aq)). The relative atomic masses of X and Y are in the ratio 1 : 2. What is the ratio of the liberated mass of X to that of Y?

Current market prices of Al, Zn and Fe scraps per kg are Rs. 20, Rs. 16 and Rs. 3 respectively. If H_(2) is to be prepared by the reaction of one of these metals with H_(2)SO_(4), which would be the cheapest metal to use ? Which would be most expensive ?

Curves with the shape shown are often observed for reactions involving catalysts. The level portion of the curve is best attributed to the fact that:

Current efficiency is defined as the extent of a desired electrochemical reaction, divided by the theoretical extent of the reaction times 100%. What is the current efficiency of an electrodeposition of Cy metal in which 9.8g Cu is deposited by passage of 3A current for 10000 seconds.

Curdling of milk is

Curd'contain the following vitamin

Cuprous salts are generally colourless while cuprous oxide is:

Cuprous chloride is obtained from cupric chloride :

Cuprite contains

Cuprous chloride can not be prepared :

Cupric compounds are more stable than their cuprous counterparts in solid state. This is because

Cupra-ammonium silk is

Cupellation process is used in he metallurgy of :

Cupellation is a process used for the refining of

"Cumene"underset((ii)H_(2)O,H^(+))overset((i)O_(2))to (X) and (Y) What is X and Y?

Cumene on air oxidation give

Cumene process is the most important commercial method for the manufacture of phenol. Cumene is

Cumene is isopropyl benzene.

Cumene is converted in phenol by

Cumene is

Cu(CN)_(4)^(2-) is colourless as it absorbs light in

CuCN reacts with FeCl_3 to give

[CuCl_(4)]^(2-) exists while [CyI_(4)]^(2-) does not exist why ?

CuCl_(2) is used as catalyst in

CuCl is sparingly soluble in water but it dissolves in KCl solution due to the formation of

CuCl is sparingly soluble in H_2O but it dissolves in KCI solution due to the formation of:

CuCl is insoluble in water but it dissolves in KCl solution. This is due to the formation of the complex.

CaCl is colourless, while CuSO_(4).5H_(2)O is coloured. Explain?

CuCl dissolves in ammonia forming a complex. The co-ordination number of copper in the complex is:

Cubic unit cells are divided into primitive, bcc and fcc. Calculate the number of atoms in unit cell of each of the following : *fcc

Cubic unit cells are divided into primitive, bcc and fcc. Write two examples for covalent solids.