52671 + Interview Questions in Chemistry in Class 11

1.	The best combination of reagents for carrying out the conversion RCH_(2)CH_(2) OH rarr RCH_(2)CH_(2)COOH is
Answer» `PBr_(3), KCN, H_(3)O^(+)` `PBr_(3), KCN, H_(2)//PT` `KCN, H_(3)O^(+)` `PBr_(3), H_(3)O^(+)` Answer :A

Discussion

2.	The best and latest technique for isolation, purification and separation of organic compound is : (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
Answer» SOLUTION :CHROMATOGRAPHY. THUS, OPTION (d) is CORRECT.

Discussion

3.	The best and latest technique for isolation, purification, and separtion of organic compounds is
Answer» crystallisation distillation sublimation CHROMATOGRAPHY Solution :Chromatography is the BEST and LATEST TECHNIQUE`.

Discussion

4.	The best and latest technique for isolation, purification and separation of organic compound is:
Answer» CRYSTALLISATION DISTILLATION SUBLIMATION Chromatography Answer :D

Discussion

5.	The behaviour of temporary gases like CO_(2) approaches that of permanent gases like N_(2), O_(2), as we go
Answer» Below CRITICAL temperature Above critical temperature Above absolute zero Below absolute zero Solution :The gas cannot be liqueffied unit COOLED to critical temperature

Discussion

6.	The benzene molecule has two different carbon - carbon bond lengths, corresponding to alternate single and double bonds.
Answer» ANSWER :F

Discussion

7.	Thebelow configuration is not correct as it violates .
Answer» Only Hund's RULE Only PAULI's EXCLUSION principle `(N + 1)` rule (Hund + Pauli) rule Solution :(b) Only Pauli's exclusion principle.

Discussion

8.	The behaviour of temporary gases like CO_(2)approaches that of permanent gases like N_(2), O_(2)etc. as we go
Answer» Below critical temperature Above critical temperature Above ABSOLUTE zero Below absolute zero Solution :Above `T_C` TEMPERARY GAS follow gas LAWS.

Discussion

9.	The behaviour of matter in different states is governed by various physical laws. According to you what are he factors that determine the state of matter ?
Answer» Solution :TEMPERATURE, pressure, mass and VOLUME are the factors which FOLLOWS the different STATES of matter. i.e., solid, liquid and gas.

Discussion

10.	The behaviour of ideal gas is goverened by various gas laws which are described by mathematical statements as given below: (i) PV=k (constant) at constant n and T (ii) V//T=k_(2) (constant) at constant n and P (iii) V//n=k_(3) (constant) at constant T and P (iv) PV=nRT (v) P//T=k_(4) (constant) at constant n and V Answer the following The value of k_(2) is
Answer» <P>INDEPENDENT of nature and AMOUNT of gas Depends on temperature and PRESSURE conditions Depends on pressure and amount of gas Depends only on nature of gas Solution :Depends only on `T` and `P`

Discussion

11.	The behaviour of matter in different states is governed by various physical laws. According to what are the factors that determine the state of mater?
Answer» SOLUTION :The factors that determine the state of MATTER are pressure, TEMPERATURE, MASS and volume.

Discussion

12.	The behaviour of an electron in an atom is described mathematically by a wave function, or orbital.Spin of the electron produce angular momentum equal to S= sqrt(s(s+1)) (h)/(2pi)where S= +1/2. Total spin of an atom=+ n/2orh/2 Where n is the number of unpaired electron. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment the magnetic moment of an atom mu_s sqrt(s(s+1)) (eh)/(2pi mc) = sqrt((n/2)(n/2+1)) (eh)/(2pi mc) s = n/2 impliesmu_s = sqrt(n(n+2)) B.M n = number of unpaired electrons 1 B.M. (Bohr magneton) = (eh)/(4pi mc) If magnetic moment is zero the substances is di-magnetic. If an ion of ""_(25)Mn has a magnetic of 3.873 B.M. Then Mn is in which state.
Answer» `+2` `+3` `+4` `+5` SOLUTION :`sqrt(n(n+2)) = 3.813 B.M.` n= 3, no. of unpaired `E^-` =3

Discussion

13.	The behaviour of an electron in an atom is described mathematically by a wave function, or orbital.Spin of the electron produce angular momentum equal to S= sqrt(s(s+1)) (h)/(2pi)where S= +1/2. Total spin of an atom=+ n/2orh/2 Where n is the number of unpaired electron. The substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment the magnetic moment of an atom mu_s sqrt(s(s+1)) (eh)/(2pi mc) = sqrt((n/2)(n/2+1)) (eh)/(2pi mc) s = n/2 impliesmu_s = sqrt(n(n+2)) B.M n = number of unpaired electrons 1 B.M. (Bohr magneton) = (eh)/(4pi mc) If magnetic moment is zero the substances is di-magnetic. Which of the following ion has lowest magnetic moement.
Answer» `Fe^(2+)` `MN^(2+)` `Cr^(3+)` `V^(3+)` Solution :Lowest NUMBER of UNPAIRED `e^-` have lower magnetic MOMENT

Discussion

14.	The behaviour of an electron in an atom is described mathematically by a wave function, or orbital. It turns out that each wave function contains three variables, called quantum numbers, which are represented as n, l and m_1. These quantum numbers describe the energy level of an orbital and define the shape and orientation of the region in space where the electron will be found.Radial nodes are maximum in
Answer» 4S 4P 3D 5f Solution :RADIAL NODES = n-l-1

Discussion

15.	Which quantum number describes the orientation of the orbital in space ?
Answer» PRINCIPAL SECONDARY Magnetic SPIN Solution :ORIENTATION of `e^-` is given by `m_l`

Discussion

16.	The beauty of Tajmahal is getting destroyed due to
Answer» GLOBAL wanning PHOTOCHEMICAL REACTION presence of CO GAS in air ACID rain Answer :D

Discussion

17.	the beamof electronis differented ? Whatis itindicates ?
Answer» SOLUTION :ELECTRON is WAVE.

Discussion

18.	The basis of modern periodic law is
Answer» ATOMIC NUMBER atomic size atomic VOLUME atomic MASS ANSWER :A

Discussion

19.	The basis building unit of silicates is "………….." .
Answer» ANSWER :`SiO_(4)^(4-)` TETRAHEDRA

Discussion

20.	The basic structural unit of silicates is
Answer» `SiO_(3)^(2-)` `SiO_(4)^(2-)` `SiO^(-)` `SiO_(4)^(4-)` SOLUTION :Basic STRUCTURAL UNIT of silicatesis `SiO_(4)^(4-)`.

Discussion

21.	The basic structural unit in silicates is
Answer» `SiO_(2)` `Si_(2)O_(6)` `Si_(3)O_(2)` `SiO_(4)^(-4)` Answer :D

Discussion

22.	The basic reaction involved in the synthesis of linear silicones is
Answer» the HYDROLYSIS of TRIMETHYL CHLOROSILANE the hydrolysis of dimethyl dichlorosilane the hydrolysis of EHTYL chlorosilane the acid hydrolysis of dimenthyl silane Answer :B

Discussion

23.	The basic principle of Cottrel's precipitator is
Answer» neutralization of charge on COLLOIDAL particles scattering of light Lechateliar's principle Peptization Solution :In Cottrel's PRECIPITATOR, the CHARGED plates NEUTRALIZE the charge on the smoke particles which get precipitated.

Discussion

24.	The basic chartacter of the alkaline earth metal hydroxides is as follows:
Answer» `Mg(OH)_(2)gtBa(OH)_(2)gtCa(OH)_(2)gtSr(OH)_(2)` `Be(OH)_(2)gtSr(OH)_(2)gtCa(OH)_(2)gtMg(OH)_(2)` `Sr(OH)_(2)gtCa(OH)_(2)gtBa(OH)_(2)gtMg(OH)_(2)` `Mg(OH)_(2)gtBa(OH)_(2)gtSr(OH)_(2)gtCa(OH)_(2)` Solution :BASIC character increases down the GROUP due to increase in POLARITY of `M-OH` bond.

Discussion

25.	The basic character of oxides MgO, SrO, K_(2)O, NiO, Cs_(2)O increase in the order
Answer» `MGO gt SRO gt K_(2)O gt NIO gt Cs_(2)O` `Cs_(2)O lt K_(2)O lt MgO lt SrO lt NiO` `NiO lt MgO lt SrO lt K_(2)O lt Cs_(2)O` `K_(2)O lt NiO lt MgO lt SrO lt Cs_(2)O` Answer :C

Discussion

26.	The band formed from atomic orbitals of lower energy is called ___________ while that formed from atomic orbitals of higher energy is called ________
Answer» ANSWER :VALENCE BAND, CONDUCTION band

Discussion

27.	The Balmer series in the hydrogen spectrum corresponds to the transition from n_(1) = 2 " to " n_(2) = 3, 4... This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit (R_(H) = 109677 cm^(-1))
Answer» Solution :`bar(v) = 109677 cm^(-1) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) = 109677 XX ((1)/(2^(2)) - (1)/(4^(2))) = 109677 ((1)/(4) - (1)/(16)) cm^(-1) = 20564.4 cm^(-1)`

Discussion

28.	The Balm er series in the hydrogen sp ectru m corresponds to the transition from n_(1) = 2 to n_(2) = 3, 4,........This series lies in th e visible region. Calculate the wave num ber of line associated with the tran sitio n in Balm er series w hen the electron m oves to n = 4 orbit. (R_(H) = 109677 cm^(-1) )
Answer» Solution :`barv=109677 [(1)/(n_(i)^(2))-(1)/(n_(f)^(2))]cm^(-1)` so `n_(i) =2 to n_(f) =4` TRANSITION in BALMER series `=109677 (1/4-1/16)cm^(-1) =20564.44 cm^(-1)`

Discussion

29.	The balanced equation for a reaction is given below 2x + 3y rarr 41 +m When 8 moles of x react with 15 moles of y, then Which is the limiting reagent ? (b) Calculate the amount of products formed. Calculate the amount of excess reactant left at the end of the reaction.
Answer» Solution :2X + y `RARR` 41 + m (i) 2x reacts with 3y to give products. 8x reacts with 15Y means, y is the excess because 8 moles of x should react with `4 xx 3y = 12y` moles of y to give products. Therefore, 3 moles of y is excess and x is the limiting agent. When 8 moles of x react with 12 moles of y, the product formed will be `4 xx 4l` i.e. 161 and 4m as product. `8x + 2y rarr 161 + 4m` At the end of the REACTION, the excess reactant left is 3 moles of y.

Discussion

30.	The balanced equation for a reaction is given below 2x + 3y rarr 4l + m When 8 moles of x react with 15 moles of y, then(i) Which is the limiting reagent?(ii) Calculate the amount of products formed.(iii) Calculate the amount of excess reactant left at the end of the reaction.
Answer» Solution :`2x + 3y rarrt 41 + m` (i) 2x reacts with 3y to give products. 8X reacts with 15y means, y is the excess because 8 moles ofx should react with 4 `XX` 3y = I2y moles of y to give products. In this reaction 15y moles are used. Therefore, 3 moles of y is excess and x is the limiting agent. (II) When 8 moles of x react with 12 moles of y, the PRODUCT formed will be 4 `xx` 41 i.e. 161 and 4m as product. 8x + 12y `rarr` 161 + 4m (iii) At the end of the reaction, the excess reactant left is 3 moles of y

Discussion

31.	The azetrophic mixture of water (b.pt. 100^(@)C) and HCl (b.pt. 85^(@)C) boils at 108.5^(@)C. What this mixture is distilled, it is possible to obtain :
Answer» PURE HCL pure WATER pure water as WELL as HCl neither HCl nor `H_(2)O` in their pure states Answer :D

Discussion

32.	The azimuthal quantum number for the last electron in sodium atom is
Answer» 1 2 0 3 Solution :`3s^1`

Discussion

33.	The azimuthal quantum number indicates …..of the orbital
Answer» Size Shape Orientation Spin Answer :B

Discussion

34.	The average velocity of nitrogen at 27^@C is 0.3ms^(-1). At what temperature it will be 0.6 ms^(-1)?
Answer» Solution :Average VELOCITY is given as `bar(C) = sqrt((8RT)/(PI M))` The RATIO of velocities at two different temperatures for the same GAS is given as`(bar(C_1))/(bar(C_2)) = sqrt((T_1)/(T_2))` Substituting the values, the ratio is `= (0.3)/(0.6) = sqrt((300)/(T_2)) = 1/2` The TEMPERATURE, `T_2 = 2^2 xx 300 = 1200 K = 927^@C`.

Discussion

35.	Theaverage velocity ofgas molecules is 400 m//sec. Calculate its rms velocity at the same temperature .
Answer» Solution :Average SPEED , `v=sqrt((8 RT)/(pi M))` Root mean square speed, `u=sqrt((3 RT)/(M))` `:. (u)/(v)=sqrt((3 RT)/(M)XX(PIM)/(8 RT))=sqrt((3pi)/(8))=sqrt((3xx3.143)/(8))=1.085` `:. u=1.085xxv=1.085xx400 m s^(-1)=434 m s^(-1)`ordirectly, v=0.921 u, i.e., `u=1.085xxv`

Discussion

36.	The average velocity of gas is 2.9 xx 10^4 cms^(-1).Calculate the RMS and most probable velocities of the gas.
Answer» ANSWER :`2.9 XX 10^4cms^(-1)`

Discussion

37.	Theaverage velocity of an ideal gas molecule at 300 K is 2 m//s. The average velocity at 1200 Kwill be
Answer» 6 m/s 4 m/s 2 m/s 8 m/s Solution :(B) `v_(av)=sqrt((8RT)/(PIM))rArrv_(av)propsqrtT` `(2)/(V_(2))=sqrt((300)/(1200))=sqrt((1)/(4))=(1)/(2)rArrv_(2)=4 m//s`

Discussion

38.	Isotherms of carbon dioxide at various temperatures are repersented in figure. Answer the following questions based on this figures. (i) In which state will CO_(2) exist between the points a and b at temperature T_(1) ? (ii) At what point will CO_(2) start liquefyinh when temperature is T_(1) ? (iii) At what point will CO_(2) be completely liquefued when temperature is T_(2) ? (iv) Will condensation take place when the temperature is T_(3) ? (v) What portion of the isotherm at T_(1) represent liquid and gaseous CO_(2) at equilibrium ?
Answer» SOLUTION :The average VELOCITY of a gas is given by `BARU =sqrt((8RT)/(piM))` In the present case, `baru =9.0xx10^4 " cm " s^(-1)`, `R=8.314xx10^7 " erg " K^(-1) mol^(-1)` `T=T_1 (?) and M=44` `:. "" 9.0xx10^4=sqrt((8xx8.314xx10^7xxT_1)/(3.14xx44))` HENCE, `T_1 = 1682.5 K`. The most probable velocity is given by `ALPHA = sqrt((2RT)/M)` `:. "" 9.0xx10^4=sqrt((2xx8.314xx10^7xxT_2)/(44))` Which gives `T_2 = 2143.3 K`.

Discussion

39.	The average person can see the red colour imparted by the complex [Fe(SCN)]^(2+) to an aqueous solution if the concentration of the complex is 6xx10^(-6) M or greater.What minimum concentration of KSCN would be required to make it possible to detect 1 ppm (part per million) of Fe(III) in a natural water sample ? The instability constant for Fe(SCN)^(2+)hArrFe^(3+)+SCN^(-) "is"7.142xx10^(-3)
Answer» `0.0036 M` `0.0037 M` `0.0035 M` NONE of these Answer :A

Discussion

40.	The average speed at T_1 K and most probable speed at T_2K of CO_2 gas is 9 xx 10^4 cm/sec. The values of T_1 and T_2 are
Answer» 2143K, 1694K 2126K, 1726K 1684 K, 2143K 1684K, 3368K Answer :C

Discussion

41.	The average particle mass in colloidal solutions can be determined by
Answer» elevation in boiling point depression in FREEZING point relative lowering in vapour pressure osmotic pressure. Solution :The only MEASURABLE COLLIGATIVE property of colloidal solution is osmotic pressure. This is due to very large particle MASS.

Discussion

42.	The average oxidation number of iron in Fe_(3)O_(4) (ferrousferic oxide) is ____
Answer» `+2` `+3` `8/3` `2/3` SOLUTION :Ferrousferic OXIDE is `Fe_(3)O_(4)` `therefore3(FE)+4(O)=0` `therefore3(X)+4(-2)=0` `therefore3x-8=0` `therefore3x=8` `thereforex=8/3` OXIDATION of Fe

Discussion

43.	The average molecular velocity in a gas sample at 300K is 500m/s. The temperature of this gas is increased until the average velocity of its molecules is 1000m/s. what is the new temperature?
Answer» 420K 573k 600k 1200k Answer :d

Discussion

44.	The average life of a excited state of hydrogen atom is of the order of 10^(-8)sec. the number of revolutions made by an electron when it returns from n=2 to n=1 is:
Answer» `2.28xx10^(6)` `22.8xx10^(6)` `8.23xx10^(6)` `2.82xx10^(6)` ANSWER :C

Discussion

45.	The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure is:
Answer» `900 cal K^(-1) mol^(-1)` `6.21 xx 10^(-21) J K^(-1) "MOLECULE"^(-1)` `336.7 J K^(-1) "molecule"^(-1)` `3741.3 J K^(-1) mol^(-1)` Solution :`KE = (3)/(2) (R)/(N) T` `= (3)/(2) xx (8.314)/(6.023 xx 10^(23)) xx 300` `= 6.21 xx 10^(-21) J K^(-1) "molecule"^(-1)`

Discussion

46.	The average kinetic energy of gas molecules is directly proportional to the………………..
Answer» SOLUTION :KELVIN TEMPERATURE

Discussion

47.	The average kinetic energies of helium (mol.wt.4), methane (mol.wt.16) and sulphurdioxide (mol.wt.64), at certain temperature are respectively x, y and z kJ mol^(-1) . The ratio of x, y and z is
Answer» `1 :4: 16` `1 :2: 4` `1 :1:1` `4:2:1` Answer :C

Discussion

48.	The average kinetic energy of a gas molecule at 0^(@)C is 5.621xx10^(-27)J. Calculate Boltzman constant. Also calculate the number of molecules present in one mole of the gas.
Answer» Solution :Average kinetic ENERGY `KE=3/2kT` `:.k=2/3xx(KE)/^=2/3xx(5.621xx10^(-21)J"MOLECULE"^(-1))/(273K)=1.373xx10^(-23)JK^(-1)"molecule"^(-1)` No. of molecules in 1 mole of the gas (Avagadro.s no.) `=R/k` `:.R/k=(8.314JK^(-1)"MOL"^(-1))/(1.373xx10^(-23)JK^(-1)"molecule"^(-1))=6.05xx10^(23)"molecules mol"^(-1)`

Discussion

49.	The average kinetic energy of a gas molecule is given by
Answer» 3/2 RT 3/2 KT 2/3 RT 2/3 KT Answer :B

Discussion

50.	The average K.E. of one mole of an ideal gas in calories is equal to
Answer» 3 times of its ABSOLUTE TEMP 2 times of its absolute temp 4 times of its absolute temp 2/3 times of its absolute temp Solution :KE = `3/2 XX RT = 3/2 xx 2T = 3T`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The best combination of reagents for carrying out the conversion RCH_(2)CH_(2) OH rarr RCH_(2)CH_(2)COOH is

The best and latest technique for isolation, purification and separation of organic compound is : (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

The best and latest technique for isolation, purification, and separtion of organic compounds is

The best and latest technique for isolation, purification and separation of organic compound is:

The behaviour of temporary gases like CO_(2) approaches that of permanent gases like N_(2), O_(2), as we go

The benzene molecule has two different carbon - carbon bond lengths, corresponding to alternate single and double bonds.

Thebelow configuration is not correct as it violates .

The behaviour of temporary gases like CO_(2)approaches that of permanent gases like N_(2), O_(2)etc. as we go

The behaviour of matter in different states is governed by various physical laws. According to you what are he factors that determine the state of matter ?

The behaviour of matter in different states is governed by various physical laws. According to what are the factors that determine the state of mater?

Which quantum number describes the orientation of the orbital in space ?

The beauty of Tajmahal is getting destroyed due to

the beamof electronis differented ? Whatis itindicates ?

The basis of modern periodic law is

The basis building unit of silicates is "………….." .

The basic structural unit of silicates is

The basic structural unit in silicates is

The basic reaction involved in the synthesis of linear silicones is

The basic principle of Cottrel's precipitator is

The basic chartacter of the alkaline earth metal hydroxides is as follows:

The basic character of oxides MgO, SrO, K_(2)O, NiO, Cs_(2)O increase in the order

The band formed from atomic orbitals of lower energy is called ___________ while that formed from atomic orbitals of higher energy is called ________

The Balmer series in the hydrogen spectrum corresponds to the transition from n_(1) = 2 " to " n_(2) = 3, 4... This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit (R_(H) = 109677 cm^(-1))

The balanced equation for a reaction is given below 2x + 3y rarr 41 +m When 8 moles of x react with 15 moles of y, then Which is the limiting reagent ? (b) Calculate the amount of products formed. Calculate the amount of excess reactant left at the end of the reaction.

The balanced equation for a reaction is given below 2x + 3y rarr 4l + m When 8 moles of x react with 15 moles of y, then(i) Which is the limiting reagent?(ii) Calculate the amount of products formed.(iii) Calculate the amount of excess reactant left at the end of the reaction.

The azetrophic mixture of water (b.pt. 100^(@)C) and HCl (b.pt. 85^(@)C) boils at 108.5^(@)C. What this mixture is distilled, it is possible to obtain :

The azimuthal quantum number for the last electron in sodium atom is

The azimuthal quantum number indicates …..of the orbital

The average velocity of nitrogen at 27^@C is 0.3ms^(-1). At what temperature it will be 0.6 ms^(-1)?

Theaverage velocity ofgas molecules is 400 m//sec. Calculate its rms velocity at the same temperature .

The average velocity of gas is 2.9 xx 10^4 cms^(-1).Calculate the RMS and most probable velocities of the gas.

Theaverage velocity of an ideal gas molecule at 300 K is 2 m//s. The average velocity at 1200 Kwill be

The average speed at T_1 K and most probable speed at T_2K of CO_2 gas is 9 xx 10^4 cm/sec. The values of T_1 and T_2 are

The average particle mass in colloidal solutions can be determined by

The average oxidation number of iron in Fe_(3)O_(4) (ferrousferic oxide) is ____

The average molecular velocity in a gas sample at 300K is 500m/s. The temperature of this gas is increased until the average velocity of its molecules is 1000m/s. what is the new temperature?

The average life of a excited state of hydrogen atom is of the order of 10^(-8)sec. the number of revolutions made by an electron when it returns from n=2 to n=1 is:

The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure is:

The average kinetic energy of gas molecules is directly proportional to the………………..

The average kinetic energies of helium (mol.wt.4), methane (mol.wt.16) and sulphurdioxide (mol.wt.64), at certain temperature are respectively x, y and z kJ mol^(-1) . The ratio of x, y and z is

The average kinetic energy of a gas molecule at 0^(@)C is 5.621xx10^(-27)J. Calculate Boltzman constant. Also calculate the number of molecules present in one mole of the gas.

The average kinetic energy of a gas molecule is given by

The average K.E. of one mole of an ideal gas in calories is equal to

The band formed from atomic orbitals of lower energy is called ___ while that formed from atomic orbitals of higher energy is called