52671 + Interview Questions in Chemistry in Class 11

1.	CO_(2)is isostructural with
Answer» `HgCl_(2)` `SnCl_(2)` `C_(2) H_(2)` `NO_(2)` Solution :Like `CO_(2),HgCl_(2) and C_(2)H_(2)` have sp hybridization and hence are isostructural ,i.e.,linear

Discussion

2.	CO_(2) is isostructural with
Answer» `C_(2)H_(2)` `SnCl_(2)` `NO_(2)` `SO_(2)` ANSWER :a

Discussion

3.	CO_(2) is heavler than n_(2) and O_(2) gases present in the atmosphere, yet it does not form the lower layer of the atmosphere. Why ?
Answer» SOLUTION :Gases prossess the PROPERTY of diffusion which is INDEPENDENT of the FORCE of gravitation. As a result of diffusion, gases MIX with each other and remain almost uniformly distributed in the atmosphere.

Discussion

4.	CO_(2) is a gaswhile SaO_(2)is a solid. Explain.
Answer» Solution :Carbon becauseof its small size and higher electronegativity than siliconforms `ppi-ppi`double bondswith O-ATOMS to form `CO_(2)`molecule. These molecules of `CO_(2)` are heldtogether byweak van der WAALS forces of attractswhich can be easilyovercome by collision of the molecules at room temperature. Consequently, `CO_(2)` isa gas. Silicon,on the otherhand,because of its bigger size andlowereletronegativitythan carbonhas little tendency to form `ppi-ppi`double BONDS with `O`-atoms . Instead, each silicon atom forms four single, covalentbonds withO-atomswhich aretetrahedrally arrangedaroundit leading to the FORMATIONOF a three-dimensionalnetworkstructure . To breakthese covalentbonds, a large amount of energyis needed and hence `SiO_(2)` is a high meltingsolid.

Discussion

5.	CO_2 has the same geometry as :
Answer» A ANDC B and D A and D C and D Solution :Both ` HgCl_2` and `C_2H_2` are LINEAR like ` CO_2` because of sp-hybridisation .

Discussion

6.	CO_(2) gas is not produced when the following is storngly heated
Answer» `CaCO_(3)` `CA(HCO_(3))_(2)` `K_(2)CO_(3)` `MgCI_(2)` Solution :Generally thermal stability of carbonaets increases from `BeCO_(3)` to `BaCO_(3)`. So `K_(2)CO_(3)` does not give `CO_(2)` on decomposition. All bicarbonates can decompose and give `CO_(2)`.

Discussion

7.	CO_(2) gas is cannot be liquefied at room temperature. Give reason.
Answer» SOLUTION :Only below the CRITICAL temperature,by the application of PRESSURE, a gas can be liquefied. `CO_(2)` has critical temperature as 303K. At room temperature critical temperature even by applying large AMOUNT of pressue `CO_(2)` cannot be liquefied only below the critical temperatures, it can be liquefied. At room temperature `CO_(2)` remains as gas.

Discussion

8.	CO_(2) can be easily absorbed by
Answer» `Na_(2)CO_(3)` `NaHCO_(3)` `NaOH` `Na_(2)SO_(4)` Answer :C

Discussion

9.	CO_(2) at 600 bar and a temperature above T_(c) (T_(c) = 304.15 K) is called
Answer» super cooled gas super critical fluid super critical liquid super cooled liquid Solution :Gases above critical temperature and under very HIGH pressure are CALLED SUPERCRITICAL fluides

Discussion

10.	CO_(2) and N_(2)are non-supporters of combustion. However, for putting out fires CO_(2)is preferred to N_(2)because CO_(2)
Answer» Does not BURN FORMS non-combustible products with burningsubstances Is DENSER than nitrogen Is more reactive gas. Answer :A

Discussion

11.	CO_(2) " and " H_(2) Oboth are triatomic molecule but their dipole moment values are different. Why ?
Answer» Solution :In `CO_(2)`, the dipole moments of TWO POLAR bonds (CO) are equal in MAGNITUDE but have OPPOSITE direction . Hence , the net dipole moment of the`CO_(2)` is , `mu = mu_(1) + mu_(2) = mu_(1) + (-mu_(1)) = 0 `. `{:(O = C = O),(" "to leftarrow),(" "mu_(1)"" mu_(2)):}` In this case `mu = mu_(1) + mu_(2)` ` = vec(mu_(1))+ (vec(-mu_(1))) = 0 ` Incase of water net dipole moment is the vector sum of`mu_(1) + mu_(2)` as shown. Dipole moment in water is found to be `1.85` D

Discussion

12.	Co-ordination number of central metal ion in filtrate (soluble complex) 'D' is:
Answer» 4 2 6 5 Answer :B

Discussion

13.	Co-ordination number and oxidation state of Cr in K_(3) [Cr(C_(2)O_(4))_(3)]are, respectively
Answer» 3 and + 3 3 and 0 6 and + 3 4 and + 2 Solution :`C_(2)O_(4)^(2-)`oxalate ion `implies` BIDENTATE ligand , So C.No = 6 `K_3[ulCr(C_2O_4)_3],+3+x+6=,x=+3`

Discussion

14.	Co-ordinate covalent compounds dissolvemore in
Answer» Polar solvents Non-polar solvents Both 1 and 2 WATER only Answer :B

Discussion

15.	CO is isostructural with
Answer» `SnCl_(2)` `HgCl_(2)` `SCl_(2)` `ZnCl_(2)` Solution :Both `HgCl_(2)andZnI_(2)` are LINEAR MOLECULES, WHEREAS `SnCl_(2)andSCl_(2)` are bent moelcules

Discussion

16.	CO is an unsaturated compound because all the valencies of carbon are not satisfied and forms addition compounds e.g. It forms carbonyl sulphide (COS ) with sulphur, carbonyl chloride ( phosgene COCl_(2) ) with chlorine, sodium formate with NaOH , methyl alcoholwith H_(2) in the presence of ZnO//Cr_(2)O as catalyst) A overset("Red hot coke")(rarr)CO overset(Cl_(2))(rarr)C overset(H_(2)O)(rarr)2HCl+A. The compounds A and C are
Answer» `CO_(2),COCl_(2)` `CO,COCl_(2)` `C,CO_(2)` `CO_(2),CO` Solution :`CO_(2) overset("Red hot coke")(RARR)CO overset(Cl_(2))(rarr)` `COCl_(2)overset(H_(2)O)(rarr) 2HCl + CO_(2)UARR` `A = CO_(2)` carbodioxide `C = COCl_(2)` PHOSGENE

Discussion

17.	CO is isoelectronic with
Answer» `NO^(+)` `N_(2)` `SnCl_(2)` `NO_(2)^(-)` Solution :(A,B) Electrons present in CO = 6 + 8 = 14 Then, In `NO^(+) ` = 7 + 8- 1 = 14 In `N_(2) ` = 7 + 7 = 14 In `SnCl_(2) ` = (very high) 50 + 17 `xx` 2 = 50 + 34 = 84 In `NO_(2)^(-) = 7 + 16 + 1 = 24 `

Discussion

18.	CO is heating with sulphur gives
Answer» `SO_(2)` `SO_(3)` `COS` `CO_(2)` SOLUTION :`CO+S rarr COS`

Discussion

19.	CO is an unsaturated compound because all the valancies of carbon are not satisfied and forms addition compounds e.g It forms carbonyl sulphide (COS) with sulphur, carbonyl chloride (phosgene COCl_(2)) with chlorine, sodium formate with NaOH, methyl alcohol with H_(2), in the presence of ZnO //Cr_(2)O as catalyst) The dehydration of malonic acid CH_(2)(COOH)_(2) with P_(4)O_(10) and heat give
Answer» Carbon monoxide Carbon sub oxide Carbon dioxide All three Solution :`underset("Malonic ACID")(o=underset(OH)underset(\|)(C ) -underset(H)underset(\|)OVERSET(H)overset(\|)(C)-overset(OH)overset(\|)(C)=O) underset(150^(@)C)overset(P_(4)O_(10))(rarr)C_(2) O_(3)[O=C=C=C=O]+2H_(2)O`

Discussion

20.	CO is a pollutant produced due to incomplete combustion of butane . One mole of butane requires 6.5 moles of O_2 for complete combustion. If 6 moles of oxygen are available , then number of moles of CO produced will be
Answer» SOLUTION :`C_4H_10+6.5O_2 to 4CO_2+ 5H_2O` `C_4H_10+6O_2 to 3CO_2 + CO+ 5H_2O`

Discussion

21.	CO in the atmosphere is due to
Answer» incomplete combustion of petorleum fuels Oxidation of carbon (coke) in metallurigical extraction of METAL THERMAL power plants all of the above SOLUTION :(a) PETROLEUM fuels are combusted incomplete to `CO`. `C_(x)H_(y)+((2x+y)/(4))O_(2) rarr xCO+y/2 H_(2)O` (b) `ZnO+C overset(Delta)(rarr) Zn+CO` (c ) `underset(coal)+1/2 O_(2) rarr CO`.

Discussion

22.	CO in monoxide is neutral while in dioxide CO_2 is acidic .
Answer» SOLUTION :TRUE STATEMENT

Discussion

23.	CO+H_(2)underset(Cu)overset(ZnO)rarr product. Identify the product formed in the given reaction.
Answer» `CH_(3)COOH` `CO_(2)` `C_(3)O_(2)` `CH_(3)OH` Answer :D

Discussion

24.	CO+H_(2) underset("catalyst")overset(300^(@)//300atm)(rarr)CH_(3)OH the catalyst is
Answer» FE `Cr_(2)O_(3)//ZnO` `V_(2)O_(5)` `Al_(2)O_(3)` Answer :B

Discussion

25.	CO forms a volatile compound with
Answer» Nickel Copper Sodium Aluminium Solution :`NI+4COtounderset("VOLATILE compound")(Ni(CO)_(4))overset(DELTA)toNi+4CO`

Discussion

26.	CO can be used as a fuel but not CO_(2) because
Answer» `CO` is a COMBUSTIBLE gas `CO` is neutral OXIDE `CO` can be OXIDIZED but not `CO_(2)` `CO_(2)` can be oxidized but not `CO` ANSWER :A

Discussion

27.	CO+2H_(2) rarr CH_(3)OH (all gases). An equilibrium mixture consists of 2.0 atm CH_(3)OH, 1 atm CO and 0.1 atm H_(2). The volume, at same T. Find new equilibrium pressures.
Answer» SOLUTION :`{:(CO(g),+,2H_(2)(g),hArr,CH_(3)OH(g)),(0.2 "mol",,-,,-),((0.2-0.1),,x,,0.1 "mol" larr "at equilibrium"):}` Total moles of equilibrium `=0.1+x+0.1=0.2+x` Also total moles `=(PV)/(RT)=(4.92xx5)/(0.0821xx600)=0.5` `rArr 0.5=0.2+x` `rArr x=0.3`="mol" of `H_(2)` at equilibrium `K_(C)=([CH_(3)OH])/([CO][H_(2)]^(2))=(0.1//5)/((0.1//5)(0.3//5)^(2))=277.8` If there is no catalyst, no REACTION occurs. `n=n_(CO)+n_(H_(2))=0.2+0.5=0.7` `P=n(RT)/V=6.88 "atm"`

Discussion

28.	CN^(-) ion is known but CP^(-) ion is not known.
Answer» Solution :Due to smaller SIZE and HIGHER electrongeativity of N than of P, N can form `ppi-ppi` multiple BONDS with C and hence forms `CN^(-)` ion. On the other hand, due to bigger size and lower electronegativity of P than that of N, P does not form multiple bonds with C and hence does not form `CP^(-)` ion. `underset(("Exists"))(""^(-):C-=N)""underset(("Does not EXIST"))(""^(-):C-=P)`

Discussion

29.	CN group is attached to cyclohexane ring. Should it be called as cyclohexanenitrile or cyclohexanecarbonitrile ? Explain.
Answer» SOLUTION :Since the suffix nitrile does not INCLUDE the CARBON atom of the CN group, therefore, this name is wrong. The correct name is cyclohexanecarbonitrile since the suffix carbonitrile INCLUDES the carbon atom of the CN group.

Discussion

30.	Cloud of fog is a colloidal system in which thedispersed phase and dispersed medium are
Answer» GAS, liquid liquid, gas liquid, liquid solid, liquid Answer :B

Discussion

31.	Clean water would have BOD value of less than
Answer» 17 ppm 5 ppm 200,000 ppm 10 ppm Answer :B

Discussion

32.	ClCH_(2)COOH is heated with fuming HNO_(3) in the presence of AgNO_(3) in carius tube. After filtration and washing a white precipitate is obtained. The precipitate is of
Answer» `Ag_(2)SO_(4)` `ClCH_(2)COOAG` `AgCl` `AGCN` Answer :C

Discussion

33.	Clean water should have BOD of 17 ppm or more.(T/F)
Answer» ANSWER :F

Discussion

34.	Cleaning action of detergent is due to
Answer» ADSORPTION Adsorption and EMULSIFICATION WASHING Only emulsification ANSWER :B

Discussion

35.	Clay contains aluminimum silicate to an extent of
Answer» 0.05 0.1 0.15 0.2 Solution :CLAY CONTAINS 10 % ALUMINIUM SILICATE

Discussion

36.	Classify the following transformation according to the reaction type. (a) H_(3)C-CH=CH-CH_(3) + Br_(2) rarr H_(3)C - CHBr - CHBr - CH_(3) (b) (H_(3)C)_(2)C = C(CH_(3))_(2) + Br_(2) rarr (H_(3)C)_(2)C = C(CH_(3))CH_(2)Br + HBr (c) H_(2)C = CH - CH_(2)CH_(3) rarr H_(3)C - CH = CH - CH_(3) (d) C_(6)H_(5)CHO + CH_(3)COCH_(3) rarr C_(6)H_(5)CH(OH)CH_(2)COCH_(3) (e) (CH_(3))_(3)C Cl + HO^(-) rarr (CH_(3))_(2)C = CH_(2) (f) {:(CH_(3)-C=N-OH+H_(3)O^(+) rarr CH_(3)-C=O),("\|""\|"),(""Ph""NHPh):}
Answer» Solution :(a) Electrophilic addition(b) Free RADICAL substitution (C) Isomerisation(d) CONDENSATION(e) `beta`-Elimination reaction(F) REARRANGEMENT

Discussion

37.	Classify the reactions in one of the reaction type studied in this unit. CH_3CH_2Br+hatSH rarr CH_3CH_2SH+Br^-
Answer» ANSWER :NUCLEOPHILIC SUBSTITUTION

Discussion

38.	which type of reactions is given below. CH_3CH_2Br+OH rarr CH_2=CH_2+H_2O+Br^-
Answer» ANSWER :ELIMINATION

Discussion

39.	Which type of reaction is given below. (CH_3)_3C-CH_2OH+HBr rarr (CH_3)_2C (Br) CH_2CH_3+H_2O
Answer» ANSWER :NUCLEOPHILIC SUBSTITUTION INVOLVING REARRANGEMENT

Discussion

40.	Which type of reaction is given below. (CH_3)_2C=CH_2+HCl rarr (CH_3)_3C Cl
Answer» ANSWER :Electrophillic ADDITION

Discussion

41.	Classify the oxides as neutral, acidic amphoteric and basic CO, B_2O_3, SiO_2, Al_2O_3, PbO_2, Tl_2O_3
Answer» ANSWER :ACIDIC - `B_2O_3, SiO_2, CO_2`; Basic - `Tl_2O_3`; NEUTRAL - CO; AMPHOTERIC - `PbO_2, Al_2O_3`

Discussion

42.	Classify the following substances into ionic, covalent, molecular or metallic. MgO, SO_(2), I_(2) , H_(2)O("ice"), SiO_(2) (quartz), brass.
Answer» Solution :MGO= Ionic, ` SO_(2)` = Polar molecular , ` I_(2)` = non-polar molecuar, ` H_(2)O` = HYDROGEN bonded molecualar , ` SiO_(2)` = Covalent or network, BRASS = metallic.

Discussion

43.	Classify the following substances into ionic, covalent, molecular or metallic : MgO, SO_2 , I_2, H_2O (ice) , SiO_2 (quartz), brass.
Answer» Solution :MgO=Ionic, `SO_2`=Polar MOLECULAR , `I_2`=non-polar molecular , `H_2O` = Hydrogen BONDED molecular , `SiO_2`=COVALENT or network, brass=metallic

Discussion

44.	Classify the following species into Lewis acids and Lewis bases and show how these act as such : (i)H_2O , (ii)HCO_3^- , (iii)HSO_4^- , (iv)NH_3
Answer» SOLUTION :`{:("Species +" H^+ "=conjugate acid","Species" - H^+ "=conjugate base"),(H_2O+H^(+) to H_3O^(+),H_2O-H^(+) to OH^(-)),(HCO_3^(-) + H^(+) to H_2CO_3,HCO_(3)^(-) -H^(+) to CO_3^(2-)),(HSO_4^(-)+H^(+) to H_2SO_4 , HSO_4^(-) -H^(+) to SO_4^(2-)),(NH_3 +H^(+) to NH_4^(+), NH_3 -H^(+) to NH_2^(-)):}`

Discussion

45.	Classify the following species into Lewis acids and Lewis bases and show how these act as such : (a)HO^- , (b)F^- , (c)H^+ , (d) BCl_3
Answer» Solution :(a)`HO^-` , (B)`F^-` , (C)`H^+` , (d) `BCl_3`

Discussion

46.	Classify the following species into lewis acid and lewis base. (i) OH^(-) (ii) BCl_(3)
Answer» ANSWER :(i)`HO^(-)to` LEWIS BASE

Discussion

47.	Classify the following species as Lewis acids and Lewis bases NH_(3), BF_(3), SnCl_(4), C_(5)H_(5)N, CO, Ni^(2+)
Answer» Solution :Lewis acids : `BF_(3), SnCl_(4), Ni^(2+)` Lewis BASES : `NH_(3), C_(5)H_(5)N, CO`.

Discussion

48.	Classifythe following species as Lewis acids and Lewis basesand show how these act as such : (a) HO^(-)(b) F^(-)(c) H^(+) (d) BCl_(3)
Answer» Solution :(a) : `OH^(-)` (hydroxyl ion ) is a Lewis base as it can DONATE an electron pair. (b) `:underset(..)overset(..)(F):^(-)` is a Lewis base as it has 4 lone pairs of ELECTRONS and can donate any one of these. (c) `H^(+)` is a Lewis base as it can accept electron pair from BASES LIKE `OH^(-), F^(-) ` ion etc. (d) `BCl_(3) ` is a Lewis acid as B is electron deficient (having only 6 electrons instead of complete octet). Hence, it can accept an electron pair from species like ammonia, amines etc.

Discussion

49.	Classifythe following species into Lewis acids and Lewis bases and show how theseact as Lewis acid/base : (a) OH^(-) " " (b) F^(-) " " (c)H^(+) " " (d)BCl_(3)
Answer» SOLUTION :(a) `OH^(-)` can donate electron pair. Hence, it is a Lewisbase (b) `F^(-)` can also donate electron pair. Hence, it is a Lewis base (c) `H^(+)` can accept electron pair. Hence, it is a Lewis acid (d) `BCl_(3)` is deficient is electrons. Hence, it can accept electron pair and is, therefore, a Lewis acid.

Discussion

50.	Classify the following solids in different categories based on the nature of intermolecular forces operating in them : Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer» Solution :Potassium sulphate = Ionic, Tin=Metallic, Benzene=Molecular (non-polar). Urea=Molecular (Polar), Ammonia= Molecular (Hydrogen bonded), Water=Molecular (Hydrogen bonded), Zinc sulphide = ionic, Graphite = Covalent or NETWORK, Rubidium = Metallic. ARGON = Molecular (Non-polar), Silicon carbide = Covalent or Network

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

CO_(2)is isostructural with

CO_(2) is isostructural with

CO_(2) is heavler than n_(2) and O_(2) gases present in the atmosphere, yet it does not form the lower layer of the atmosphere. Why ?

CO_(2) is a gaswhile SaO_(2)is a solid. Explain.

CO_2 has the same geometry as :

CO_(2) gas is not produced when the following is storngly heated

CO_(2) gas is cannot be liquefied at room temperature. Give reason.

CO_(2) can be easily absorbed by

CO_(2) at 600 bar and a temperature above T_(c) (T_(c) = 304.15 K) is called

CO_(2) and N_(2)are non-supporters of combustion. However, for putting out fires CO_(2)is preferred to N_(2)because CO_(2)

CO_(2) " and " H_(2) Oboth are triatomic molecule but their dipole moment values are different. Why ?

Co-ordination number of central metal ion in filtrate (soluble complex) 'D' is:

Co-ordination number and oxidation state of Cr in K_(3) [Cr(C_(2)O_(4))_(3)]are, respectively

Co-ordinate covalent compounds dissolvemore in

CO is isostructural with

CO is isoelectronic with

CO is heating with sulphur gives

CO is a pollutant produced due to incomplete combustion of butane . One mole of butane requires 6.5 moles of O_2 for complete combustion. If 6 moles of oxygen are available , then number of moles of CO produced will be

CO in the atmosphere is due to

CO in monoxide is neutral while in dioxide CO_2 is acidic .

CO+H_(2)underset(Cu)overset(ZnO)rarr product. Identify the product formed in the given reaction.

CO+H_(2) underset("catalyst")overset(300^(@)//300atm)(rarr)CH_(3)OH the catalyst is

CO forms a volatile compound with

CO can be used as a fuel but not CO_(2) because

CO+2H_(2) rarr CH_(3)OH (all gases). An equilibrium mixture consists of 2.0 atm CH_(3)OH, 1 atm CO and 0.1 atm H_(2). The volume, at same T. Find new equilibrium pressures.

CN^(-) ion is known but CP^(-) ion is not known.

CN group is attached to cyclohexane ring. Should it be called as cyclohexanenitrile or cyclohexanecarbonitrile ? Explain.

Cloud of fog is a colloidal system in which thedispersed phase and dispersed medium are

Clean water would have BOD value of less than

ClCH_(2)COOH is heated with fuming HNO_(3) in the presence of AgNO_(3) in carius tube. After filtration and washing a white precipitate is obtained. The precipitate is of

Clean water should have BOD of 17 ppm or more.(T/F)

Cleaning action of detergent is due to

Clay contains aluminimum silicate to an extent of

Classify the reactions in one of the reaction type studied in this unit. CH_3CH_2Br+hatSH rarr CH_3CH_2SH+Br^-

which type of reactions is given below. CH_3CH_2Br+OH rarr CH_2=CH_2+H_2O+Br^-

Which type of reaction is given below. (CH_3)_3C-CH_2OH+HBr rarr (CH_3)_2C (Br) CH_2CH_3+H_2O

Which type of reaction is given below. (CH_3)_2C=CH_2+HCl rarr (CH_3)_3C Cl

Classify the oxides as neutral, acidic amphoteric and basic CO, B_2O_3, SiO_2, Al_2O_3, PbO_2, Tl_2O_3

Classify the following substances into ionic, covalent, molecular or metallic. MgO, SO_(2), I_(2) , H_(2)O("ice"), SiO_(2) (quartz), brass.

Classify the following substances into ionic, covalent, molecular or metallic : MgO, SO_2 , I_2, H_2O (ice) , SiO_2 (quartz), brass.

Classify the following species into Lewis acids and Lewis bases and show how these act as such : (i)H_2O , (ii)HCO_3^- , (iii)HSO_4^- , (iv)NH_3

Classify the following species into Lewis acids and Lewis bases and show how these act as such : (a)HO^- , (b)F^- , (c)H^+ , (d) BCl_3

Classify the following species into lewis acid and lewis base. (i) OH^(-) (ii) BCl_(3)

Classify the following species as Lewis acids and Lewis bases NH_(3), BF_(3), SnCl_(4), C_(5)H_(5)N, CO, Ni^(2+)

Classifythe following species as Lewis acids and Lewis basesand show how these act as such : (a) HO^(-)(b) F^(-)(c) H^(+) (d) BCl_(3)

Classifythe following species into Lewis acids and Lewis bases and show how theseact as Lewis acid/base : (a) OH^(-) " " (b) F^(-) " " (c)H^(+) " " (d)BCl_(3)

Classify the following solids in different categories based on the nature of intermolecular forces operating in them : Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.