52671 + Interview Questions in Chemistry in Class 11

1.	The answer to each of the following questions is a single digit integer from 0 to 9. If the correct answer to the question number A,B, C and D say 3,0,8,5 respectively, then the correct darkening of bubbles should be as follow. Amonst the following, the ores which are roasted to convert them into their corresponding metal oxides are: Alumina, zinc, blende, iron pyrites, copper pyrites, galena, chalcorite
Answer» SOLUTION :N//A

Discussion

2.	The answer to each of the following question is a single interger, ranging from 0 to 9 If the correct answers to questions, numbers A,B,Cand D say are 4,0,9 and 2, then the correct darkening of bubbles should be as shown on the side (B) In haemoglobin , CO and not O_(2) links Fe (if both CO and O_(2) are present) beacuse CO is a stronger ligand than O_(2). The number of groups which Fe is coordinated other than vacant site for CO in haemoglobin is
Answer» ANSWER :B

Discussion

3.	The answer to each of the following question is a single interger, ranging from 0 to 9 If the correct answers to questions, numbers A,B,Cand D say are 4,0,9 and 2, then the correct darkening of bubbles should be as shown on the side (A) CO is a pollutant produced due to incomplete combindaation of butane. one mole of butane requires 6.5 moles of O_(2) for complete combustion .If 6 moles of oxygen are available , the number of CO produced will be
Answer» ANSWER :a

Discussion

4.	The answer to each of the following question is a single interger, ranging from 0 to 9 If the correct answers to questions, numbers A,B,Cand D say are 4,0,9 and 2, then the correct darkening of bubbles should be as shown on the side (C )In measurement of BOD_(n) ,n is generally taken as
Answer» ANSWER :C

Discussion

5.	The answer to each of the following questions is a single digit integer from 0 to 9. If the correct answer to the question number A,B, C and D say 3,0,8,5 respectively, then the correct darkening of bubbles should be as follow. How many of the following metals can be refined by the vapour phase refining? Zr, Zn, Cd, Hg, Ni, Co, Pt, Fe, Ti
Answer» SOLUTION :N//A

Discussion

6.	The answer of the calculation (2.568 xx 5.8)/(4.168) in significant figures will be
Answer» 3.57 3.6 3.57 3.579 Solution :B) `(2.568 XX 58)/(4.168)`=`3.5735` THUS, INSIGNIFICANT figures = 3.6

Discussion

7.	The anolic form of ethyl acetoacetate as shown below has
Answer» 9 SIGMA BONDS and 2 pi-bonds 9 sigma bonds and 1 pi-bond 18 sigma bonds and 2 pi-bonds 16 sigma bonds and 1 pi-bond. Solution :

Discussion

8.	The anolic form of acetone contains :
Answer» 9 sigma bonds, 1 pi bonds and two LONE pairs 8 sigma bonds, 2 pi bonds and two lone pairs 10 sigma bonds, 1 pi bonds and one lone pairs 9 sigma bonds, 2 pi bonds and one lone pairs Solution :`{:(""H""H),("\|""\|"),(H-C=C-C-H),("\|""\|"),(H-underset(..)(O: )""H),("ENOL FORM of acetone"):}``sigma`-Bonds = 9 `pi`-Bonds = 1 Lonepairs = 2

Discussion

9.	The anion nitrate is converted into ammonium ion. The equivalent mass of nitrate ion in the reactionwould be
Answer» `6.20` `7.75` `10.5` `21.0` SOLUTION :`OVERSET(+5)(NO_3^(-))rarroverset(-3)(NH_4^(+))` `E=M/8,(62)/8=7.75`

Discussion

10.	The anhydride of carbonic acid is
Answer» `CO` `CO_(2)` `C_(3)O_(2)` `C_(2)O` ANSWER :B

Discussion

11.	The angular part of the wave function depends on the quantum numbers are:
Answer» `N` `L` `m` `s` ANSWER :B::C

Discussion

12.	The angular shape of ozone molecule (O_(3)) consists of :
Answer» I sigma and 2 PI bonds 2 sigma and 2 pi bonds 1 sigma and 1 pi bonds 2 sigma and 1 pi bonds Solution :2 sigma and 1 pi bonds The SHAPE of ozone molecule is In it we FIND 2 `sigma and 1 pi` bond, i.e., option (D) is correct.

Discussion

13.	The angular momentum of the electron in the 4th energy shell in the hydrogen atom is ________.
Answer» SOLUTION :`(2H)/(PI)`

Discussion

14.	The angular momentum of electron in 'd' orbials is equal to:
Answer» `sqrt2 H` `sqrt(2/3)`h `0 h` `SQRT6 h` SOLUTION :`sqrt6 h` Orbital angularmoment= `sqrt(l(l +1)) h` dorbitals have ` l= 2` ` :.` orbital angular MOMENTUM `= sqrt92(2 +1)h0= sqrt(6) h`

Discussion

15.	The angular momentum of electron in 'd' orbital is equal to
Answer» `2 SQRT 3 H` `0 h` `sqrt6h` `sqrt2 h` Solution :Angular momentum `= sqrt(l (l + 1)) h` For d-orbital, l = 2 `:.` Angular momentum `= sqrt(2(2 + 1)) h = sqrt6 h`

Discussion

16.	The angular momentum of an electron present in the excited state of hydrogen is 1.5h//pi. The electron is present in
Answer» THIRD orbit Second orbit Fourth orbit FIFTH orbit Solution :`mvr =(NH)/(2pi) implies(3H)/(2pi) implies (1.5 h)/(PI)`

Discussion

17.	The angular momentum of an electron in Bohr's orbit of hydrogen atom is 4.22 xx 10^(-34) kg m^(2) s^(-1). Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level.
Answer» Solution :Angular momentum `(mvr) = n (h)/(2PI) = 4.22 xx 10^(-34) kg m^(2) s^(-1)` `:. n = 4.22 xx 10^(-34) xx (2pi)/(h) = (2 xx 4.22 xx 10^(-34) xx 3.14)/(6.626 xx 10^(-34)) = 4` (Given) When the electron jumps from n = 4 to n = 3, the wavelength of the SPECTRAL line can be calculated as FOLLOWS: `(1)/(lamda) = R_(H) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = 109, 677 cm^(-1) ((1)/(3^(2)) - (1)/(4^(2))) = 109677 xx ((1)/(9) - (1)/(16)) = 109677 xx (7)/(144) cm^(-1)` or `lamda = (144)/(109677 xx 7) cm = 1.88 xx 10^(-4) cm`

Discussion

18.	The angular momentum of an electron is zero. In which orbital may ti be present ?
Answer» 2s 2p 3d 4f Answer :A

Discussion

19.	The angular momentum of a revolving electron in an orbit is equal to
Answer» `(nh)/(2pi)` `(H)/(2pi)` `((nh)/(2pi))^(2)` `(n pi)/(2H)` ANSWER :A

Discussion

20.	The angular m om entum of electron in ‘d ’ orbital is equal to
Answer» 0h `SQRT(6)` h `sqrt(2)` h `2sqrt(2)` h Answer :B

Discussion

21.	The angle strain in cyclopentane is
Answer» `72^(@)` `1^(@)28'` `44'` `108^(@)` ANSWER :C::D

Discussion

22.	The angle between two covalent bonds is minimum in
Answer» `H_(2)O` is NON POLAR `CO_(2)` `NH_(3)` `CH_(4)` Answer :A

Discussion

23.	The angle between the planes of H_(2)O_(2) molecule in gaseous phase is
Answer» `101.5^(@)` `90^(@)` `111.5^(@)` `109^(@)28^(1)` ANSWER :C

Discussion

24.	The amphoteric oxide from among the following is
Answer» `B_(2)O_(3)` `In_(2)O_(3)` `Ga_(2)O_(3)` `Tl_(2)O_(3)` ANSWER :D

Discussion

25.	The amount of zinc required to produce 224 ml of H_(2) at STP on treatment with dilute H_(2)SO_(4) will be (Zn = 65)
Answer» 6.5 G 0.65 g 65 g 0.065 g ANSWER :B

Discussion

26.	The amount of zinc required to produce 224 ml of H_(2) at NTP on treatement with dilute H_(2)SO_(4) solution will be
Answer» `0.65 g` `0.065g` `65g` `6.5g` SOLUTION :`Zn +H_(2)SO_(4) rarr ZnSO_(4)+H_(2)`

Discussion

27.	The amount of oxygen used by the suitable microorganisms present in water during five days at 20^(@)C is called as BOD. For pure water BOD is about 1 ppm fairly pure water 3 ppm and doubtful purity 5 ppm the municipal sewage has BOD values 100-4000 ppm. BOD= Number of moles of O_2 required / Number of liters of water sample If water is polluted its BOD value is
Answer» 1 ppm 2 ppm 3 ppm 6 ppm ANSWER :D

Discussion

28.	The amount of oxygen used by the suitable microorganisms present in water during five days at 20^(@)C is called as BOD. For pure water BOD is about 1 ppm fairly pure water 3 ppm and doubtful purity 5 ppm the municipal sewage has BOD values 100-4000 ppm. BOD= Number of moles of O_2 required / Number of liters of water sample BOD is normally cxpressed for a time period of
Answer» 5 hours 5 minutes 5 days 5 seconds Answer :C

Discussion

29.	The amount of oxygen used by the suitable microorganisms present in water during five days at 20^(@)C is called as BOD. For pure water BOD is about 1 ppm fairly pure water 3 ppm and doubtful purity 5 ppm the municipal sewage has BOD values 100-4000 ppm. BOD= Number of moles of O_2 required / Number of liters of water sample The BOD value of municipal sewage is
Answer» 1 PPM 5 ppm 5-10 ppm 100-4000 ppm Answer :D

Discussion

30.	The amount of oxygen used by the suitable microorganisms present in water during five days at 20^(@)C is called as BOD. For pure water BOD is about 1 ppm fairly pure water 3 ppm and doubtful purity 5 ppm the municipal sewage has BOD values 100-4000 ppm. BOD= Number of moles of O_2 required / Number of liters of water sample The BOD values of three water sample x,y and z are 5 mg/litre, 16 mg/Litre and 400 mg/Litre respectively the more polluted sample of water is
Answer» x y z all are EQUALLY polluted Answer :C

Discussion

31.	The amount of oxygen used by micro organisms present in water for five days at 20^@Cis called
Answer» COD DO TLV BOD ANSWER :D

Discussion

32.	The amount of oxygen used by micro organisms present in water for five days at 20^@C is called
Answer» COD DO TLV BOD Solution :BOD

Discussion

33.	The amount of oxygen required to oxidise organic substance present in water is called
Answer» DO COD BOD TLV Answer :B

Discussion

34.	The amount of oxygen consumed by pollutants in water found by titration with acidified K_2Cr_2O_7 is called _____ of water.
Answer» ANSWER :CHEMICAL OXYGEN DEMAND (COD)

Discussion

35.	The amount of oxygen required for healthy growth of plants and animals in water is
Answer» 1-2 mg/mI 4-6g/lit 1-2g/100 ml 4-6mg/litre Solution :4-6m/litre

Discussion

36.	The amount of oxygen (in grams) present dissolved in 10^(x) gm of water is called dissolved oxygen. Then the value of x is ____.
Answer» ANSWER :6

Discussion

37.	The amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(2) to yield a solution og p^(H) = 9.35 (K_(b) for NH_(3) = 1.78 xx 10^(-5)) will be
Answer» `7.248 G` `6.248 g` `5.248 g` `52.48 g` SOLUTION :N//A

Discussion

38.	The amount of Mg in gms. to be dissolved in dilute H_(2)SO_(4) to liberate H_(2) which is just sufficient to reduce 160g of ferric oxide is
Answer» 24 48 72 96 Answer :C

Discussion

39.	The amount of Na_(2)S_(2)O_(3) to required for reducing iodine produced by the reaction of 1 mol of KI with H_(2)O_(2) in acid medium is-
Answer» 0.5 1mol 2 mol 2.5 mol Answer :C

Discussion

40.	The amount of heat released, when 20 ml of 0.5 M NaOH is mixed with 100 mL of 0.1M HCl, is x k cal. The heat of neutralisation is
Answer» `10x "K Cal mole"^(-1)` `100X "k Cal mole"^(-1)` `-100 x"k Cal mole"^(-1)` `-23.92x "k Cal mole"^(-1)` SOLUTION :EQ. NEUTRALISED = 10 m.eq 10 m.eq `rarr` - x kJ 1 eq `rarr` ? kJ `(-x)/(10 xx 10^(-3)) = - 100x kJ`/mole

Discussion

41.	The amount of heat exchanged with the surrounding at constant quantity ______
Answer» `DELTAE` `DeltaH` `DeltaS` `DELTAG` ANSWER :B

Discussion

42.	The amount of heat evolved when 500cm^(3) of0.1 MHCl is mixed with 200 cm^(3) of 0.2 M NaOH is "….................."
Answer» 2.292 Kj 1.292 kJ `0.292kJ` `3. 292 kJ` SOLUTION :`500 CM^(3)` of 0.1 HCl `=500 xx0.1` millimole `=50` millimole`=0.050`mole `200 cm^(3)` of0.2 M NaOH `= 200 xx0.2` millimole `=4` millimole `=40`millimole `=0.040 ` mole Thus, 0.04 mole of NaOH neturalizes 0.04 mole of HCl. Heat evolved from 1 mole of HCl`= 57.3 kJ` `:.` Heat evolved from0.04mole of HCl `= 57.3xx 0.04 kJ=2.292kJ`

Discussion

43.	The amount of heat evolved on combustion of 10 grams of benzoic acid is 10 K.Cals. The heat of combustion of benzoic acid is
Answer» `-122 K.Cals` `-112 K.Cals` `-132 K.Cals ` `-92 K.Cals ` ANSWER :A

Discussion

44.	The amount of gas adsorbed (x/m) is the function of temperatureand pressure, which can be shown as under (x)/(m)= f(p,T) The variation x/m and p, at constant temperature is known as adsorption isotherm i.e., x/m=f(p) at constant temperature Seven type of physisorption isotherms are known. Some of them are shown below Physisorption Type I. Where there is formation of monolayer on the surface of adsorbent, this typeof curve is obtained. Physisorption Type II. The transition point 'X' of this physisorption isotherm indicates, that the formation of mono layer is complete and that of multilayer is started. Physisorption Type III. This isotherm has no transition point. In this type of curve, the multimolecular starts before the completion of monomolecular layer. On the basis of this comprehension, answer teh following question from the physisorption isotherm of a gas on a solid shown as under. The point Z of the physisorption correctly represents
Answer» <P>Just beginning of formation of MULTILAYER Just beginning of formation of monolyer Just COMPLETION of multilayer and just formation of monolyer `(x)/(m) PROP p^(n)` where n `gt ` 1 Answer :A

Discussion

45.	The amount of gas adsorbed (x/m) is the function of temperatureand pressure, which can be shown as under (x)/(m)= f(p,T) The variation x/m and p, at constant temperature is known as adsorption isotherm i.e., x/m=f(p) at constant temperature Seven type of physisorption isotherms are known. Some of them are shown below Physisorption Type I. Where there is formation of monolayer on the surface of adsorbent, this typeof curve is obtained. Physisorption Type II. The transition point 'X' of this physisorption isotherm indicates, that the formation of mono layer is complete and that of multilayer is started. Physisorption Type III. This isotherm has no transition point. In this type of curve, the multimolecular starts before the completion of monomolecular layer. On the basis of this comprehension, answer teh following question from the physisorption isotherm of a gas on a solid shown as under. The point Y of physisorption isotherm correctly represents
Answer» FORMATION of MONOLAYER formation of multilayer first monolayer formation THEREAFTER the formation of multilayer none Answer :A

Discussion

46.	The amount of gas adsorbed (x/m) is the function of temperatureand pressure, which can be shown as under (x)/(m)= f(p,T) The variation x/m and p, at constant temperature is known as adsorption isotherm i.e., x/m=f(p) at constant temperature Seven type of physisorption isotherms are known. Some of them are shown below Physisorption Type I. Where there is formation of monolayer on the surface of adsorbent, this typeof curve is obtained. Physisorption Type II. The transition point 'X' of this physisorption isotherm indicates, that the formation of mono layer is complete and that of multilayer is started. Physisorption Type III. This isotherm has no transition point. In this type of curve, the multimolecular starts before the completion of monomolecular layer. On the basis of this comprehension, answer teh following question from the physisorption isotherm of a gas on a solid shown as under. The portion XY of the curve correctly represents
Answer» <P>x/m is DIRECTLY PROPORTIONAL to `p^(n)`, where n is greater than 1 x/m is directly proportional to `p^(n)` where n`LE`1, monolayer formation Both A and C Answer :A

Discussion

47.	The amount of gas adsorbed (x/m) is the function of temperatureand pressure, which can be shown as under (x)/(m)= f(p,T) The variation x/m and p, at constant temperature is known as adsorption isotherm i.e., x/m=f(p) at constant temperature Seven type of physisorption isotherms are known. Some of them are shown below Physisorption Type I. Where there is formation of monolayer on the surface of adsorbent, this typeof curve is obtained. Physisorption Type II. The transition point 'X' of this physisorption isotherm indicates, that the formation of mono layer is complete and that of multilayer is started. Physisorption Type III. This isotherm has no transition point. In this type of curve, the multimolecular starts before the completion of monomolecular layer. On the basis of this comprehension, answer teh following question from the physisorption isotherm of a gas on a solid shown as under. The portion ''OX'' of the physisorption isotherm represents
Answer» x/m is directly proportional to p x/m is linear FUNCTION of pressure adsorption isotherm has no transition POINT, the formation of multi-molecular layer STARTS before the completion of monomolecular layer Both B and C Answer :D

Discussion

48.	The amount of energy required to remove the electron from a Li^(+2) ion in its ground state is x times that of the amount of energy required to remove the electron from an H-atom in its ground state find .x.
Answer» ANSWER :9

Discussion

49.	The amount of energy released when an electron is added to an isolated gaseous atom to produce a monovalent anion is called electron affinity or first electron affinity or electron gain enthalpy. The first electron is given a negative sign as the addition of an electron to a neutral atom is an exoergic process. The addition of electron to A^(-1) requires energy to overcome the force of repulsion. Thus, the second electron affinity is an endoergic process. The magnitude of electron affinity depends on a number of factors such as (i) atomic size (ii) effective nuclear charge (iii) screening effect (iv) half and fully filled orbitals and (v) shape of orbital. In general, electron affinity increases as the atomic radii decrease in a period. However, there are exceptions when the atoms have stable configurations. In a group, electron affinity decreases as the size increases. However, the members of 3rd period have somewhat higher values than the members in the 2nd period of the same subgroups. Which of the following species has the highest electron affinity?
Answer» `F^(-1)` `O^-` `NA` `O` ANSWER :D

Discussion

50.	The amount of energyreleasedwhenone millionatomsof iodine are completely converted into I^(-) ionsin thevapour stateaccordingto theequationI(g)+ e^(-)to I^(-) ( g) " is" 4.9 xx 10^(-13) J Calculatethe electronaffinity ofiodine in(i) Kj//mol and (ii) in eVper atom.
Answer» Solution :Requiredenergy `=0.1 XX 3.36 xx 23.06 =7.748 kcal` Electronaffinity `=6.023 xx 10^(23) xx 4.9 xx 10^(-13) xx 10^(-6)` `=29 .5 xx 10^(4) xx 10^(-3) KJ =295 kJ mol^(-1)` `= 295// 96.49 eV //atom =3.06 eV //atom`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The answer to each of the following question is a single interger, ranging from 0 to 9 If the correct answers to questions, numbers A,B,Cand D say are 4,0,9 and 2, then the correct darkening of bubbles should be as shown on the side (C )In measurement of BOD_(n) ,n is generally taken as

The answer of the calculation (2.568 xx 5.8)/(4.168) in significant figures will be

The anolic form of ethyl acetoacetate as shown below has

The anolic form of acetone contains :

The anion nitrate is converted into ammonium ion. The equivalent mass of nitrate ion in the reactionwould be

The anhydride of carbonic acid is

The angular part of the wave function depends on the quantum numbers are:

The angular shape of ozone molecule (O_(3)) consists of :

The angular momentum of the electron in the 4th energy shell in the hydrogen atom is ________.

The angular momentum of electron in 'd' orbials is equal to:

The angular momentum of electron in 'd' orbital is equal to

The angular momentum of an electron present in the excited state of hydrogen is 1.5h//pi. The electron is present in

The angular momentum of an electron in Bohr's orbit of hydrogen atom is 4.22 xx 10^(-34) kg m^(2) s^(-1). Calculate the wavelength of the spectral line when the electron falls from this level to the next lower level.

The angular momentum of an electron is zero. In which orbital may ti be present ?

The angular momentum of a revolving electron in an orbit is equal to

The angular m om entum of electron in ‘d ’ orbital is equal to

The angle strain in cyclopentane is

The angle between two covalent bonds is minimum in

The angle between the planes of H_(2)O_(2) molecule in gaseous phase is

The amphoteric oxide from among the following is

The amount of zinc required to produce 224 ml of H_(2) at STP on treatment with dilute H_(2)SO_(4) will be (Zn = 65)

The amount of zinc required to produce 224 ml of H_(2) at NTP on treatement with dilute H_(2)SO_(4) solution will be

The amount of oxygen used by micro organisms present in water for five days at 20^@Cis called

The amount of oxygen used by micro organisms present in water for five days at 20^@C is called

The amount of oxygen required to oxidise organic substance present in water is called

The amount of oxygen consumed by pollutants in water found by titration with acidified K_2Cr_2O_7 is called _____ of water.

The amount of oxygen required for healthy growth of plants and animals in water is

The amount of oxygen (in grams) present dissolved in 10^(x) gm of water is called dissolved oxygen. Then the value of x is ____.

The amount of (NH_(4))_(2)SO_(4) in grams which must be added to 500 ml of 0.2 M NH_(2) to yield a solution og p^(H) = 9.35 (K_(b) for NH_(3) = 1.78 xx 10^(-5)) will be

The amount of Mg in gms. to be dissolved in dilute H_(2)SO_(4) to liberate H_(2) which is just sufficient to reduce 160g of ferric oxide is

The amount of Na_(2)S_(2)O_(3) to required for reducing iodine produced by the reaction of 1 mol of KI with H_(2)O_(2) in acid medium is-

The amount of heat released, when 20 ml of 0.5 M NaOH is mixed with 100 mL of 0.1M HCl, is x k cal. The heat of neutralisation is

The amount of heat exchanged with the surrounding at constant quantity ______

The amount of heat evolved when 500cm^(3) of0.1 MHCl is mixed with 200 cm^(3) of 0.2 M NaOH is "….................."

The amount of heat evolved on combustion of 10 grams of benzoic acid is 10 K.Cals. The heat of combustion of benzoic acid is

The amount of energy required to remove the electron from a Li^(+2) ion in its ground state is x times that of the amount of energy required to remove the electron from an H-atom in its ground state find .x.

The amount of energyreleasedwhenone millionatomsof iodine are completely converted into I^(-) ionsin thevapour stateaccordingto theequationI(g)+ e^(-)to I^(-) ( g) " is" 4.9 xx 10^(-13) J Calculatethe electronaffinity ofiodine in(i) Kj//mol and (ii) in eVper atom.