52671 + Interview Questions in Chemistry in Class 11

1.	Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Answer» SOLUTION :

Discussion

2.	Suggest a list of substacnes where carbon can exhibit oxidation state from -4 to +4 and nitrogen from -3 to +5
Answer» SOLUTION :

Discussion

3.	Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from –3 to +5.
Answer» SOLUTION :

Discussion

4.	Sugar is obtained from its saturated solution by
Answer» SUBLIMATION SIMPLE DISTILLATION CRYSTALLISATION Chromatograpy Answer :C

Discussion

5.	Sugar containing an impurity of common salt can be purified by crystallisation from
Answer» Benzene Ether Alcohol Water ANSWER :A

Discussion

6.	Sufficient amount of a solid X is taken in a rigid vessel at T^(@)C where it attained the equilibrium: X(s)hArrY_((g))+2Z(g) Total pressure was measured. Now the vessel is evcuated and filled with sufficient amount of another solid V under same conditions where it attained theequilibrium: Total pressure measured now is found to be double that of previous value. Now, if both X "&" V solids are allowed to attain their respective equilibrium together in the same vessel at same temperature, select the correct staement(s):
Answer» `K_(P)` for decompostion reaction of `V(s)=8xxK_(P)` for decomposition reaction of `X (s)`. In the`3^(RD)` case (when both solids are simultaneously establishing their equilibrium), `P_(Y)=(1)/(8)P_(W)`. `P_(Y) "in case"=(1)/(3sqrt3)xxP_(Y) "in" 1^(ST)` case. In the `3^(rd)` case,`P_(W),P_(Z)=4.9` Solution :N//A

Discussion

7.	Successive ionisation potentials of an element M are 8.3, 25.1, 37.9, 259.3 and 340.1ev. The formula of its bromide is
Answer» `MBr_5` `MBr_4` `MBr_3` `MBr_2` ANSWER :C

Discussion

8.	Substrate that show E1 reaction
Answer» `CH_3CH_2I` `CH_3-undersetunderset(CH_3)(\|)oversetoverset(CH_3)(\|)C-CH_2-I` ANSWER :C

Discussion

9.	Subtract : (i) 0.015 from 5.16 (ii) 16.92 from 237.5 (iii) 0.0604 from 0.08638
Answer» Solution :(i) Subtract : `{:(5.16),(0.015),(BAR(5.145)):}` The correct answer 5.14 (Digit 5 is to be DELETED) (ii) Subtract : `{:(237.5),(16.92),(bar(220.58)):}` The correct answer is 220.6 (Digit 5 has been rounded off to 6) (iii) Subtract : `{:(0.08638),(0.0604),(bar(0.02598)):}` The correct answer is 0.0260 (Digit 5 has been rounded off to 6)

Discussion

10.	Substract 24.5492 from 62.831 and report the answer in significant figures.
Answer» ANSWER :38.28

Discussion

11.	Substantiate lithium fluoride has the lowest solubility among group one metal fluorides
Answer» Solution :Lithium fluoride has high LATTICE enthalpy due to the small SIZE of `Li^(+)` and `F^(-)` So, DU high lattice enthalpy, LiF is less soluble in WATER

Discussion

12.	Substances which show permanent magnetism even in the absence of magnetic field are called ________
Answer» ANSWER :FERROMAGNETIC

Discussion

13.	Substances which readily undergo hydrolysis are
Answer» `AlCl_(3)` `BCl_(3)` `SiCl_(4)` `PbCl_(4)` Solution :All the chloridesbeing LEWISACIDS ACCEPT a pair of ELECTRONS from `H_(2)O`moleculehence undergohydrolysis.

Discussion

14.	Substances which can donate lone pairs of electrons to form a coordinate covalent bonds are called
Answer» Acids BASES Non-polar Amphiprotic Solution :LEWIS bases can DONATE ELECTRON PAIR.

Discussion

15.	Substance which is weakly repelled bya magnetic field is
Answer» `O_(2)` `H_(2)O` `CrO_(2)` `Fe_(3)O_(4)` SOLUTION :SUBSTANCE which is weakly repelled by a magnetic field is diamgagnetic , e.g, ` H_(2)O`

Discussion

16.	Substance which is weakly repelled by a magnetic field is
Answer» `O_2` `H_2O` `CrO_2` `Fe_3O_4` SOLUTION :Substance which is weakly repelled by a magnetic field is DIAMAGNETIC , e.g., `H_2O`.

Discussion

17.	Substance which is weakly rapelled by a magnetic field is
Answer» `O_(2)` `H_(2)O` `CrO_(2)` `Fe_(3)O_(4)` SOLUTION :SUBSTANCE which is WEAKLY repelled by magnetic FIELD is diamagnetic i.e. `H_(2)O`

Discussion

18.	Substance A_(2)B(g) can undergo decomposition to form of set of products : if the molar ratio of A_(2)(g) to A(g) is 5 : 3 in a set of product gases, then energy involved in the decomposition of 1 mole of A_(2)B is :
Answer» 43.75 KJ`//`mol 18.73 kJ`//`mol 46.25 kJ`//`mol None of these Answer :A

Discussion

19.	Sublimation is a process in which a solid :
Answer» CHANGES into ANOTHER ALLOTROPIC form changes into liquid form changes into vapour form None of the above Answer :C

Discussion

20.	SubQue : Whodiscoveredproton?
Answer» SOLUTION :Thesmallestand lightestpositiveion wasobtainedfromhydrogenand was calledproton `(H^(+))`

Discussion

21.	Sub. Que : Give number, bond angle and bond length in ethyne.
Answer» SOLUTION : In ethyne bond LENGTH, angle, shape and type of bonds : Two C -H sigma bond in it. In triple bond two `PI` and one `sigma`-bond. So total three bond. In it C - C bond length 120 pm and C - H bond length 106 pm. Both carbon have sp hybridization . So, H - C - C - H is linear and H- C - II and C - C - H bond angle `180^(@)`.

Discussion

22.	Sub. Que. : Explainpi -bond formation in ethyne,
Answer» Solution :Each of the two unhybridised p ORBITALS of both CARBON atoms overlap SIDEWISE to form two `pi`-bond between the carbon atoms. So the triple bond between the carbon atoms is MADE up of one sigma and two `pi` BONDS.

Discussion

23.	Sub Que : Give Bond length, Bond angle in ethene.
Answer» Solution :Bond length ANG angle in ETHENE : C - C Bond length is 134 pm C - H Bond length is 108 pm H- C-H : `117.6^(@) , "" H-C-C : " 121^(@)`

Discussion

24.	Sub Que. : Explain the formation pi-bond in ethene.
Answer» <P> Solution :`pi` BOND structured : In ETHENE the non HYBRID p and FORM `pi`-bond between two carbon atom. In ethene `sigmaand pi` bnd can be presented as follows.

Discussion

25.	Sub. Que. : Explain sigma-bond formation in ethene.
Answer» Solution :Now one sp ORBITAL of carbon combine with sp orbital of other carbon and form sigma BOND. Remaining TWO `sp^(2)` orbital. Combine with is orbital of H and form four C-H sigma bond. In this way four C-H sigma bond and two carbon & four Hydrogen are in one plane Now the each carbon has one non hybride 2p orbital which side by side OVERLAP and form `pi`-bond. in ethene both carbon has `sp^(2)` hybridization . So ethene is planar.

Discussion

26.	Sub que : Explain sigma-bond formation in ethyne .
Answer» Solution : In ethyne one sp hybrid orbital of one carbon atom OVERLAPS axially with sp hybrid orbital of the other carbon atom to form C-C SIGMA bond. While the other hybrids of orbital of each carbon atom overlaps axially with half filled s orbital of hydrogen atoms forming `sigma` honds. sp orbital are linear so all the four atoms H-C-C- H arrange in linear SHAPE figure of `sigma`-bond Still 2PX and 2py orbital are perpendicular to H-C-C-H bond AXIS.

Discussion

27.	Styrene (C_(6)H_(5)CH=CH_(2)) is treatedwith HBrpresenceof perxides.
Answer» Solution :Additionoceursaccording toanti- Markovnikov's rule . `C_(6)H_(5)CH =CH_(2)+ HBroverset("PEROXIDES ")UNDERSET("(Anti- Mark adition)") (to) underset(1-"Bromo-2- phenylethane ")(C_(6)H_(5)CH_(2)CH_(2)Br)`

Discussion

28.	Styphinic acid
Answer» 2,4,6-Trinitrophenol 2,4,6-Trinitroresorcinol 2,4-Dinitroresorcinol Phloroglucinol Solution :

Discussion

29.	Study the following reactions and mark the correct properties shown by water. (i) SO_(3) + H_(2)O to H_(2)SO_(4) (ii) CI_(2)O_(7) + H_(2)O to 2HCIO_(4) (iii)CaO + H_(2)O to Ca(OH)_(2)(iv) Na_(2)O + H_(2)O to 2NaOH
Answer» All oxides react with water to give hydroxides. Acidic oxides are formed by metals and basic oxides by non-metals. Non - metal oxides COMBINE with water to FORM acids while METALLIC oxides combine with water to form ALKALIES. Acidic oxides are stronger than basic oxides since they form STRONG acids. Solution :Acidic oxides form acids while basic oxides form alkalies with water.

Discussion

30.	Study the following reactants// product and choose the correct option of reagents for path Xand Y respectively.
Answer» SOLUTION :`(CH_(2)OH)_(2) & " " CH_(3)OH`are used to protect CARBONYL GROUP UNDESIRED reaction.

Discussion

31.	Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: At 21.5°C decrease in volume(B-C) doesn't result into increase in pressure because
Answer» The compression CAUSES increase in solidification At point B only all gases CONVERT into solid The compression causes further increase in condensation Carbon dioxide is real gas Solution :B-C ISTHE process of `L to G`

Discussion

32.	Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: Carbondioxide present at F can be liquified
Answer» At CRITICAL volume by compression At critical PRESSURE by compression By compression at any temperature By compression and lowering the temperature Solution :F is `T > T_C` To LIQUIFY, T should be dec and then should be inc.

Discussion

33.	Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: Choose the temperature above which CO_(2)cannot be liquified whatsoever applied pressure is
Answer» 290K 294.5 K 302K 304 K Solution :Above `T_C` GAS CANNONT be LIQUIFIED `T_C = 30.98^@C = 303.98 K`.

Discussion

34.	Studiedtheradioactivityand developed it ._(6) ^(14)C._(6)^(13) C ._(7) ^(14) N._(6)^(12) C
Answer» SOLUTION :ISOBARS`._(6)^(14) C._(7)^(14)N ` …. Samemassnumber Isotops `._(6)^(14)C ._(6)^(15) N ,_(6)^(12) C`….. Sameatomicnumber

Discussion

35.	Structures of molecules of two compounds are given below : (a) Which of the two compounds will have intennolecular hydrogen bonding and which compound is expected to show intramolecular hydrogen bonding. (b) The melting point of a cornpound depends on, among other things, the extent of hydrogen bonding. On this basis explain which of the above two cornpounds will show higher melting point. (c) Solubility of compounds in water depends on power to form hydrogen bonds with water. Which of the above compounds will form hydrogen bond with water easily and be more soluble in it.
Answer» Solution :(a) In compound (I) Intramolecular H-bonding is formed Here H-atom, in between the two HIGHLY ELECTRONEGATIVE atoms, is PRESENT within the same molecule. In ortho-nitrophenol (compound n. H-atom is in between the two oxygen atoms. Compound (II) forms intermolecular H- bonding. In para-nitrophenol cm there is a gap between `NO_(2)` and OH group. So, H-bond exists between H-atom of one molecule and O -atom of another molecule as shown below. (b) Compound (ID Possess higher melting point because large number of molecules are JOINED together by H-bonds. (c) Due to intramolecular H bonding, compound (I) is not able to form H-bond with water, so it is less soluble in water. While molecules of compound II form H-bonding with H20 easily, so it is soluble in water.

Discussion

36.	Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parenthesis are incorrect. (i) {:(CH_(3)-CH-CH_(2)-CH_(2)-CH-CH-CH_(2)-CH_(3)),("\|""\|""\|"),(""CH_(3)""CH_(3)" "CH_(3)),("2,5,6-Trimethyloctane"),("(and not 3,4,7-Trimethyloctane)"):} (ii) {:(CH_(3)-CH_(2)-CH-CH_(2)-CH-CH_(2)-CH_(3)),("\|""\|"),(""CH_(3)CH_(2)""CH_(3)),("3-Ethyl-5-methylheptane"),("(and not 5-Ethyl-3-methylheptane)"):}
Answer» Solution :(i) The set of locants (2,5,6) is lower than the set of locants (3,5,7). (ii) When substituents are at equivalent positions, lowest LOCANT is given to that substituents which comes first in the alphabetical ORDER. Therefore, lower locant 3 is assigned to ETHYL and HIGHER locant 5 is assigned to methyl GROUP.

Discussion

37.	Structures and IUPAC names of some hydrocarbons are given below. Explain why the names given in the parentheses are incorrect. (a) underset("2, 5,6-Trimethyloctane (and not 3,4,7-Trimethyloctane)")(CH_(3) - underset(underset(CH_(3))(\|))(CH)- CH_(2) - CH_(2) - underset(underset(CH_(3))(\|))(CH)-underset(underset(CH_(3))(\|))(CH)-CH_(2) - CH_(3)) (b) underset("3-Ethyl-5-methylheptane (and not 5-Ethyl-3-methylheptane)")(CH_(3) - CH_(2) - underset(underset(CH_(2)CH_(3))(\|))(CH)- CH_(2) - underset(underset(CH_(3))(\|))(CH)- CH_(2) - CH_(3))
Answer» Solution :(a) LOWEST locant number, 2, 5, 6 is lower than 3, 5, 7 (b) Substituents are in equivalent position, lower number is given to the one that comes FIRST in the name according to alphabetical order

Discussion

38.	Structure of X can be :
Answer» ANSWER :A::B::C::D

Discussion

39.	Structure of the compound whose IUPAC name is 5, 6-dimethylhept-2-ene is ........... .
Answer» NONE of these Answer :A

Discussion

40.	Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-ynoic acid is
Answer» SOLUTION :

Discussion

41.	Structure of the compound whose IUPAC name is 3-Ethyl -2-hydroxy -4-methylhex-3-en-5-ynoic acid is
Answer» ANSWER :D

Discussion

42.	Structure of product is
Answer» SOLUTION :

Discussion

43.	Structure of is :
Answer» SOLUTION :

Discussion

44.	Structure of diborane.
Answer» ANSWER :`B_(2)H_(6)`

Discussion

45.	Structure of benzene was proposed by
Answer» Mitcharlich FARADAY Kekule Berzelius ANSWER :C

Discussion

46.	Structure of a mixed oxide is cubic close-packed (c.c.p.) The cubic cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B . The formula of the oxide is :
Answer» `A_2B_3O_4` `AB_2O_2` `ABO_2` `A_2BO_2` Solution :SUPPOSE NUMBER of `O^(2-)` ions in close PACKING =n `therefore` No. of octahedral voids=n No. of TETRAHEDRAL voids=2n Divalent metal ions A present =`"2n"/4=n/2` Monovalent Metal ions B present =n `therefore` Ratio `A:B:O^(2-)=n/2:n:n=1/2:1:1` HENCE, formula =`AB_2O_2`

Discussion

47.	Structure of a mixed oxide is cubic close-packed (c.c.p)the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. the formula of the oxide is :
Answer» `A_(2)B_(3)O_(4)` `AB_(2)O_(2)` `ABO_(2)` `A_(2)BO_(2)` Solution :Suppose number of ` O^(2-)`ions in close PACKING = n No. of octahedral VOIDS = n No. of tetradral voids = 2n Divalent METAL ions A present = `(2n)/4 = n/2 ` Monovalent metal ions B present =n Ratio `A :B : O^(2-) = n/2 : n :n = 1/2 :1:1 ` Hence, formula = ` AB_(2)O_(2)`

Discussion

48.	Structural of Ascorbic acid is represented as follows How many of the following reagents can gives positive test with ascorbic acid. {:(Cu_(2)Cl_(2)+NH_(4)OH,"2,4"-DNP,"Na Metal",HCL+ZnCl_(2),FeCl_(3)),("(I)","(II)","(III)","(IV)","(V)"),("NAOH+Phenophthalein",dil. KMNO_(4),Br_(2)//H_(2)O,AgNO_(3)+NH_(4)OH,I_(2)+NaOH),("(VI)"," (VII)"," (VIII)","(IX)","(X)"):} Number of compound which gives positive Haloform test=(x) Number of compound which can gives positive Lucas reagent test=(y) Report your answer (x+y)
Answer» SOLUTION :Na Metal `HCl+ZnCl_(2), FeCl_(3), NaOH` +Phenolpthalein, DILL `KMnO_(4)`, `Br_(2)//H_(2)O` gives positive test with ascorbic acid.

Discussion

49.	Structural isomers have different covalent linkage of atoms. Stereoisomers are compounds that have same sequence of covalent bonds but differ in the relative dispositions of their atoms in space. Geometrical and optical isomers are the two important types of configurational isomers. The compound with double bonds or ring structure have restricted rotation, so exist in two geometrical forms. The double bonds in larger rings (ring size 10 carbon large) can also cause geometrical isomerism. The optical isomers rotate the plane of plane-polarised light. A sp^3 -hybridised carbon atom bearing four different types of substituents is called an asymmetric centre or chiral centre. A chiral object or molecule cannot be superimposed on its mirror image. Stereoisomers that are mirror images of each other are called enantiomers. The stereosomers that the pot mirror images of each other are called diastereomers. Diasteremers have different physical properties.A racemic mixture is optically inactive and contains equal amounts of both the enantiomers. Resolution refers to method of separating a racemic mixture. Into two pure enantiomers. A meso compound is an optically inactive stereoisomer, which is achiral due to the presence of an internal plane of symmetry or centre of symmetry within the molecule.The pair of showing identical species is
Answer» SOLUTION :

Discussion

50.	Structural isomers have different covalent linkage of atoms. Stereoisomers are compounds that have same sequence of covalent bonds but differ in the relative dispositions of their atoms in space. Geometri cal and optical isomers are the two important types of configurational isomers. The compound with double bonds or ring structure have restricted rotation, so exist in two geometrical forms. The double bonds in larger rings (ring size 10 carbon large) can also cause geometrical isomerism. The optical isomers rotate the plane of plane-polarised light. A sp^(3)-hybridised carbon atom bearing four different types of substituents is called an asymmetric centre or chiral centre. A chiral object or molecule cannot be superimposed on its mirror image. Stereoisomers that are mirror images of each other are called enantiomers. The stereosomers that the pot mirror images of each other are called diastereomers. Diasteremers have different physical properties. A racemic mixture is optically inactive and contains equal amounts of both the enantiomers. Resolution refers to method of separating a racemic mixture. Into two pure enantiomers. A meso compound is an optically inactive stereoisomer, which is achiral due to the presence of an internal plane of symmetry or centre of symmetry within the molecule. The pair of the following that showing identical species is
Answer» SOLUTION :

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Suggest a list of substacnes where carbon can exhibit oxidation state from -4 to +4 and nitrogen from -3 to +5

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from –3 to +5.

Sugar is obtained from its saturated solution by

Sugar containing an impurity of common salt can be purified by crystallisation from

Successive ionisation potentials of an element M are 8.3, 25.1, 37.9, 259.3 and 340.1ev. The formula of its bromide is

Substrate that show E1 reaction

Subtract : (i) 0.015 from 5.16 (ii) 16.92 from 237.5 (iii) 0.0604 from 0.08638

Substract 24.5492 from 62.831 and report the answer in significant figures.

Substantiate lithium fluoride has the lowest solubility among group one metal fluorides

Substances which show permanent magnetism even in the absence of magnetic field are called ________

Substances which readily undergo hydrolysis are

Substances which can donate lone pairs of electrons to form a coordinate covalent bonds are called

Substance which is weakly repelled bya magnetic field is

Substance which is weakly repelled by a magnetic field is

Substance which is weakly rapelled by a magnetic field is

Substance A_(2)B(g) can undergo decomposition to form of set of products : if the molar ratio of A_(2)(g) to A(g) is 5 : 3 in a set of product gases, then energy involved in the decomposition of 1 mole of A_(2)B is :

Sublimation is a process in which a solid :

SubQue : Whodiscoveredproton?

Sub. Que : Give number, bond angle and bond length in ethyne.

Sub. Que. : Explainpi -bond formation in ethyne,

Sub Que : Give Bond length, Bond angle in ethene.

Sub Que. : Explain the formation pi-bond in ethene.

Sub. Que. : Explain sigma-bond formation in ethene.

Sub que : Explain sigma-bond formation in ethyne .

Styrene (C_(6)H_(5)CH=CH_(2)) is treatedwith HBrpresenceof perxides.

Styphinic acid

Study the following reactions and mark the correct properties shown by water. (i) SO_(3) + H_(2)O to H_(2)SO_(4) (ii) CI_(2)O_(7) + H_(2)O to 2HCIO_(4) (iii)CaO + H_(2)O to Ca(OH)_(2)(iv) Na_(2)O + H_(2)O to 2NaOH

Study the following reactants// product and choose the correct option of reagents for path Xand Y respectively.

Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: At 21.5°C decrease in volume(B-C) doesn't result into increase in pressure because

Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: Carbondioxide present at F can be liquified

Study the following isotherms of carbondioxide at various temperature and answer the following questions based on that: Choose the temperature above which CO_(2)cannot be liquified whatsoever applied pressure is

Studiedtheradioactivityand developed it ._(6) ^(14)C._(6)^(13) C ._(7) ^(14) N._(6)^(12) C

Structure of X can be :

Structure of the compound whose IUPAC name is 5, 6-dimethylhept-2-ene is ........... .

Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-ynoic acid is

Structure of the compound whose IUPAC name is 3-Ethyl -2-hydroxy -4-methylhex-3-en-5-ynoic acid is

Structure of product is

Structure of is :

Structure of diborane.

Structure of benzene was proposed by

Structure of a mixed oxide is cubic close-packed (c.c.p.) The cubic cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B . The formula of the oxide is :

Structure of a mixed oxide is cubic close-packed (c.c.p)the cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. the formula of the oxide is :