52671 + Interview Questions in Chemistry in Class 11

1.	The radius of an atom of an element is 500 pm. If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell ?
Answer» SOLUTION :For FCC, `r=a/(2SQRT2)` or `a=2sqrt2r=2xx1.414xx500` PM =1414 pm

Discussion

2.	The radius of an atom of an element is 500 pm.If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell ?
Answer» Solution :For FCC ` , r = a / (2SQRT2) ORA = 2sqrt2 r = 2 xx 1.414 xx 500 "PM"= 1414 "pm"`

Discussion

3.	A thin wire has a length of 21.7 cm and radius 0.46 mm. Calculate the volume of the wire to correct significant figures.
Answer» ANSWER :`0.14 CM^(3)`

Discussion

4.	The radioactive isotope of hydrogen is
Answer» `""_(1)^(1)H` `""_(1)^(2)H` `""_(1)^(3)H` `""_(1)^(0)H` ANSWER :C

Discussion

5.	The radioactive isotope of hydrogen is called ……….. And its nucleus contains ……….. Proton and ……….. Neutrons.
Answer» ANSWER :TRITIUM, ONE, TWO

Discussion

6.	The radioactive element of group 2 element is_________
Answer» Strontium Radium Beryllium francium Answer :B

Discussion

7.	Theradii of the F, F^(-) , O " and " O^(2-) are in the order ….
Answer» `O^(2-) GT O gt F^(-) gt F` `F^(-) gt O^(2-) gt F gt O` `O^(2-) gt F^(-) gt F gt O` `O^(2-) gt F^(-) gt O gt F` ANSWER :d

Discussion

8.	The radii of Na^(+) and Cl^(-) ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is
Answer» 276 PM 138 pm 552 pm 45 pm Solution :As NaCl has fcc lattice, edgelength =`2times"Destance between NA^(+) and Cl^(-)ions` =`2(r_(Na+)+r_(Cl-))=2(95+181)`pm. ltbrge552 pm.

Discussion

9.	The radii of maximum probability for 3s, 3p and 3d electrons are in the order :
Answer» `(r_("MAX"))_(3d) gt (r_("max"))_(3p) gt (r_("max"))_(3s)` `(r_("max"))_(3d) gt (r_("max"))_(3s) gt (r_("max"))_(3p)` `(r_("max"))_(3s) gt (r_("max"))_(3p) gt (r_("max"))_(3d)` `(r_("max"))_(3p) gt (r_("max"))_(3s) gt (r_("max"))_(3d)` ANSWER :C

Discussion

10.	The radii of F, F^(-), O, O^(2-) are in the order
Answer» `O^(2-) GT F^(-) gt O gt F` `O^(2-) gt F^(-) gt F gt O` `F^(-) O^(2-) gt F gt O` `O^(2) gt O gt F^(-) gt F` ANSWER :A

Discussion

11.	The radii of F, F^(-) , O andO^(2-)are in the order
Answer» `O^(2-)gtOgtF^(-)GTF` `F^(-)GTO^(2-)gtFgtO` `O^(2-)gtF^(-)gtOgt^(+)F` `O^(2)gtF^(-)gtFgtO` Answer :C

Discussion

12.	The radical halogenation of 2-methylpropane gives two products: (CH_(3))_(2)CHCH_(2)X (minor) and (CH_(3))_(3)CX (major) Chlorination gives a larger amount of the minor product than does bromination, why?
Answer» Bromine is more reactive than chlorine and is able to ATTACK the LESS reactive `3^(@)C-H` Bromine atoms are less reactive (more selective) than chlorine, and PREFERENTIALLY attack the weaker `3^(@)C-H` bond The methyl groups are more hindered to attack by the LARGER bromine atom. Bromination is reversible and more STABLE `3^(@)-` alkyl bromide is formed exlusively. Answer :B

Discussion

13.	The radical halogenation of 2 - methyl propane gives two products (CH_(3))_(2)CHCH_(2)X_("(minor)") and (CH_(3))_(3)CX_("(major)"). Chlorination gives larger amount of the minor product than the bromination because
Answer» Bromine is more REACTIVE than chlorine and is able to attack the less reactive `3^(@)C-H` Bromine ATOMS are less reactive (more selective) than chlorine and preferentially attack the WEAKER `3^(@)C-H` bond. The METHYL groups are more hindered to attack by the larger bromine atom Bromination is reversible and more stable `3^(@)-` alkyl bromide is formed exclusively. Solution :Chlorine is more reactive, less selective Bromine is less reactive, more selective

Discussion

14.	The radical halogenation of 2-methyl propane gives two products (CH_(3))_(2)CHCH_(2)X_(("minor")) and (CH_(3))_(3)CX_(("major")). Chlorination givs larger amount of the minor product than the bromination because
Answer» BROMIDE is more reactive than chlorine and is able to attack the less reactive `3^(@)C-H` Brominie atoms are less reactive (more selective) than chlorine and preferentially atack the weaker `3^(@)C-H` bond. The METHYL GROUPS are more HINDERED to attack by the larger bromine atom. Bromination is REVERSIBLE and more stable `3^(@)-` alkyl bromide is formed exlusively. Solution :Chlorine is more reactive, less selective bromine is less reactive, more selective

Discussion

15.	The radiation with highest wave number
Answer» Microwaves X- rays I.R. - rays RADIOWAVES ANSWER :B

Discussion

16.	The radial probability distribution curve of an orbital of H has '4' local maxima . If orbital has 3 angular node then orbital will be :
Answer» 7 f 8f 7d 8d Answer :A

Discussion

17.	The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function correspond to ?
Answer» Cu Li K Na Answer :D

Discussion

18.	The radial distribution function [P(r)] is used to determine the most probable radius, which is used tofind the electron in a given orgital (dP(r))/(dr) for 1s-orbital of hydrogen like atom having atomic number Z, is (dP)/(dr) = (4Z^3)/(a_0^3) (2r-(2Zr^2))/(a_0) e^(-27 x//a_0) : Then which is the following statements is/are correct ?
Answer» At the point of maximum value of radial distribution function `(dP(r))/(dr)=0`, one anti-node is PRESENT Most probable RADIUS of `Li^(2+)` is `(a_0)/(3) ` pm Most probable radius of `He^+` is `(a_0)/(2)` Most probable radius of hydrogen atom is (a_0) SOLUTION :(a,b,c,d) At the point of maximum value of RDF `(dP)/(dr) = 0 ` `(2r - (2Zr^2)/(a_0))= 0,r =(a_0)/(Z)` Where Z = 3 for `Li^(2+)` and Z= 2FOR the `He^+` , Z=1 for hydrogen.

Discussion

19.	The radial distribution function [P(r)] is use to determine the most probable radius, which is used to find the electron in a given orbital (dP(r))/(dr) for 1s-orbital of hydrogen like atom having atomic number Z, is (dP)/(dr) = (4Z^(3))/(a_(0)^(3))(2r-(2Zr^(2))/(a_(0)))e^(-2Ze//a_(0)). Then which of the following statements is/are correct
Answer» At the point of maximum value of radial distribution function `(dP(R))/(dr) = 0`, ONE antinode is PRESENT Most probable radius of `LI^(2+)` is `(a_(0))/(3)` pm Most probable radius of `He^(+)` is `(a_(0))/(2)` pm Most probable radius of hydrogen atom is `a_(0)` pm Answer :A::B::C::D

Discussion

20.	The R/S conifiguration of these compounds are respectively
Answer» R,R,R R,S,R R,S,S S,S,S Answer :A

Discussion

21.	The questionsgiven belowcontainstatement -1 (Assertion ) andstatement -2 (Reason ) . Ithas fouroptions(a), (b) , ( c) and (d ) out ofwhich ONLYONE iscorrect . Choosethe correctoptionas.Statement -1 . Of all theelementsheliumhas thehighestvalue offirstionizationenthalpy Statement-2Heliumhas themostpositiveelectrongainenthalpyof all theelement s.
Answer» Statement -1 ISTRUE Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -5 Statement -1 isTrueStatement-2 isTrue , Statement-2 is nota correctexplanationfor Statement -5 Statement -1 isTrueStatement-2 is False Statement -1 is FalseStatement-2 isTrue SOLUTION :CORRECTSTATEMENT -2 . Heliumis thesmallestinertgas .

Discussion

22.	The questionsgiven belowcontainstatement -1 (Assertion ) andstatement -2 (Reason ) . Ithas fouroptions(a), (b) , ( c) and (d ) out ofwhich ONLYONE iscorrect . Choosethe correctoptionas. Statement - 1. The ionicsize ofO^(2-) is biggerthan that ofF^(-)ion. Statement -2. O^(2-) and F^(-) are isoelectronic ions.
Answer» Statement -1 ISTRUE Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -3 Statement -1 isTrueStatement-2 isTrue , Statement-2 is NOTA correctexplanationfor Statement -3 Statement -1 isTrueStatement-2 is False Statement -1 is FalseStatement-2 isTrue Solution :CORRECTEXPLANATION .` O^(2-)` hashigher negativechargethatn `F^(-)`

Discussion

23.	The questionsgiven belowcontainstatement -1 (Assertion ) andstatement -2 (Reason ) . Ithas fouroptions(a), (b) , ( c) and (d ) out ofwhich ONLYONE iscorrect . Choosethe correctoptionas. Statement -1. The ionicradiifollowsthe order : I^(-)lt I lt I^(+) Statement -2 . Smallerthe valueof z//elargerthe sizeof the species .
Answer» Statement -1 isTrue Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -4 Statement -1 isTrueStatement-2 isTrue , Statement-2 is nota correctexplanationfor Statement -4 Statement -1 isTrueStatement-2 is False Statement -1 is FalseStatement-2 isTrue Solution :Correctstatement -1. Theionicradii followsthe ORDER `I^(+)LG I LT I^(-)`

Discussion

24.	The R and S configuration for each sterogenic centre in this from top to bottom?
Answer» R.R.R R.S.S R.S.R S.S.R Answer :3

Discussion

25.	The questionsgiven belowcontainstatement -1 (Assertion ) andstatement -2 (Reason ) . Ithas fouroptions(a), (b) , ( c) and (d ) out ofwhich ONLYONE iscorrect . Choosethe correctoptionas. Statement - 1 . The elements having 1 s^(2) 2s^(2)2p^(6)3s^(2) and1s^(2)2s^(2) configurationbelongto the samegroup Statement-2 . Bothhave samenumberof electronselectronsin the valenceshell.
Answer» Statement -1 ISTRUE Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -2 Statement -1 isTrueStatement-2 isTrue , Statement-2 is NOTA correctexplanationfor Statement -2 Statement -1 isTrueStatement-2 is False Statement -1 is FalseStatement-2 isTrue Solution :Correctstatement -1 . THEELEMENTWITH E.C`1s^(2)2S^(2)2p^(6)3s^(2) ` belongs to group 2 but theelementwith E.C. `1s^(2)2s^(2)` (i.e.,helium ) is aninertgas andhencebelongsto group 18.

Discussion

26.	The questionsgiven belowcontainstatement -1 (Assertion ) andstatement -2 (Reason ) . Ithas fouroptions(a), (b) , ( c) and (d ) out ofwhich ONLYONE iscorrect . Choosethe correctoptionas.Statement -1 . Sixthperiodis the longestperiodin the periodictable. Statement -2 . Sixthperiodinvolvesthe fillingof all theorbitalsof the sixthenergylevel
Answer» Statement -1 isTrue Statement-2 isTrue, Statement-2 iscorrectexplanationfor Statement -1 Statement -1 isTrueStatement-2 isTrue , Statement-2 is NOTA correctexplanationfor Statement -1 Statement -1 isTrueStatement-2 is False Statement -1 is FalseStatement-2 isTrue Solution :CORRECT statement -2 Sixthperioddoes not INVOLVE thefillingof all theorbitalsof the sixthenergylevel(i.e.,6s 4f, 5d 6 p- orbitals are FILLED )

Discussion

27.	The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question. Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted. Reason(R): High biological oxygen demand means high activity of bacteria in water
Answer» Both (A) and R are CORRECT and (R) is the correct EXPLANATION of (A) Both (A) and R are correct and (R) is not the correct explanation of (A) Both (A) and R are not correct (A) is correct but( R) is not correct Answer :d

Discussion

28.	The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. List if any of these combination (s) has/have the same energy: (i) n = 4, l = 2, m_(l) = -2, m_(s) = -1//2 (ii) n = 3, l = 2, m_(l) = -1, m_(s) = +1//2 (iii) n = 4, l = 1, m_(l) = 0, m_(s) = +1//2 (iv) n = 3, l = 2, m_(l) = -2, m_(s) = -1//2 (v) n = 3, l = 1, m_(l) = -1, m_(s) = +1//2 (vi) n = 4, l = 1, m_(l) = 0, m_(s) = +1//2
Answer» Solution :The ORBITALS occupied by the ELECTRONS are (i) 4D (ii) 3d (iii) 4P (iv) 3d (v) 3p (VI) 4p Their energies will be in the order:(v) `lt (ii) = (iv) lt = (iii) lt (i)`

Discussion

29.	The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists. (i) n = 4, l = 2, m_l = -2, m_s = -1/2 (ii) n = 3, l = 2, m_l = 1, m_s = +1/2 (iii) n = 4, l = 1, m_l = 0, m_s = +1/2 (iv) n = 3, l = 2, m_l = -2, m_s = -1/2 (v) n = 3, l = 1, m_l = -1, m_s = +1/2 (vi) n = 4, l = 1, m_l =0 , m_s = +1/2.
Answer» Solution : `:. (V) < (ii) = (iv) < (III) = (VI) < (i)` `:. 3p < 3d = 3d < 4P = 4p < 4d.`

Discussion

30.	The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. Do any of these combination have same energy. {:(1.,n=4,l=2,m_(l) = -2,m_(s)=-1//2,),(2.,n=3,l=2,m_(l) = 1,m_(s) = +1//2,),(3.,n=4,l=1,m_(l) = 0,m_(s) = +1//2,),(4.,n=3, l=2,m_(l) = -2,m_(s) = -1//2,),(5.,n=3,l=1,m_(l) = -1,m_(s) = +1//2,),(6.,n=4,l=1,m_(l) = 0,m_(s) = +1//2,):}
Answer» SOLUTION :The order or increasing energies is : `5<2 =4<3 = 6<1 ` 2 and 4 electrons have the same ENERGY, similarly 3 and 6 electrons also have same energy

Discussion

31.	The quantum numbers of four electrons (e_(1) " to " e_(4)) are given below : {:(,n,l,m,s,,,n,l,m,s),(e_(1),3,0,0,+1//2,,e_(2),4,0,0,+1//2),(e_(3),3,2,2,-1//2,,e_(4),3,1,-1,+1//2):} Correct order of decreasing energy of these electrons is
Answer» `e_(4) gt e_(3) gt e_(2) gt e_(1)` `e_(2) gt e_(3) gt e_(4) gt e_(1)` `e_(3) gt e_(2) gt e_(4) gt e_(1)` `e_(1) gt e_(4) gt e_(2) gt e_(3)` Answer :C

Discussion

32.	The quantum numbers n=2, l=1 represent
Answer» <P>1S ORBITAL 2s orbital 2p orbital 3d orbital Solution :l=1 is for p orbital

Discussion

33.	The quantum number which tells about the orientation of different orbitals of an atom is called........
Answer» SOLUTION :MAGNETIC QUANTUM NUMBER

Discussion

34.	The values of quantum numbers n, 1 and m for the fifth electron of boron is
Answer» rotation of the electron in clockwise and anticlockwise direction RESPECTIVELY. rotation of the electron in anticlockwise and clockwise direction respectively. magnetic moment of the electron POINTING up and down respectively. two quantum MECHANICAL spin states which have no CLASSICAL analogue. Solution :spin quantum number indicates the POSSIBLE Elements

Discussion

35.	The quantum number which may be designated by s,p,d and f instead of number is
Answer» <P>n l `m_1` `m_s` SOLUTION :l=0 is s,l=1 is p and l=2 is d and so on hence, s,p,d may be USED instead of no

Discussion

36.	The quantum number which tells about the angular momenta of the different electrons present in an atom is called........
Answer» SOLUTION :AZIMUTHAL QUANTUM NUMBERS

Discussion

37.	The quantum number which explain the line spectra observed as doublets in case of hydrogen and alkali metals and doublets & triplets in case of alkaline earth metals is
Answer» Spin Azimuthal Magnetic Principle Answer :A

Discussion

38.	The quantum number which is equal for all the d-electrons in an atom is
Answer» l m s n Answer :A

Discussion

39.	The quantum number of sixelectrons are givenbelow,Arrangethemin orderof increasingenergies . If anyofthesecombinationhashave thesameenergylists:(1) n-4,l=2 , m_(1)= 2m_(1) = (1)/(2) (2)n=3, l = 2, m_(1)= 1 ,m_(s) = (1)/(2) (3 ) n=4, l = 1 , m_(1) = =0 m_(s ) = (1)/(2) (4 )n=3: l= 2 m_(1)= - 2 m_(s ) = - (1)/(2) (5 )n=3, l = 1,m_(1)= - 1m_(s ) = (1)/(2) (6)n=4, l = 1, m_(1)= 0m_(s ) = (1)/(2)
Answer» SOLUTION :(1)L =2so dand n= 4 so4dsubshell (2)l=2so dand n=3 so3dsubshell (3 )l =1so p andn=4so 4p (5 )l =1 sop andn= 3 so 3p (6)l =1so p andn=4so 4p This4d3d , 3d4parrange as perenergythan (5) `3p(4 ) 3d = 2 (3d)lt (6) 4p= (3) 4p lt (1)4D (5 )lt (4) = (2)lt (6)= (3) lt (1) to`energyincreases(2) ,(3) (4) and(6) has sameenergy .

Discussion

40.	The quantum number not obtained from the Schrodinger.s wave equation is
Answer» n l m s Answer :D

Discussion

41.	The quantium number of four electrons (e_1to e_4) are given below The correct order of decreasing energy of these electrons is
Answer» `e_4 gt e_3 gt e_2 gt e_1` `e_2 gt e_3 gt e_4 gt e_1` `e_3 gt e_2 gt e_4 gt e_1` `e_1 gt e_3 gt e_4 gt e_2` Solution :N= l VALUES DESCENDING order.

Discussion

42.	The quantity that does not change for a sample of a gas in a sealed rigid container when. It is cooled from 120^(@)C to 90^(@)C at constant volume is:
Answer» average energy of the molecule pressure of the gas density of the gas average SPEED of the MOLECULES ANSWER :c

Discussion

43.	The quantity of heat required to raise the temperature of 1 gm of water by 1^@C is ____
Answer» SOLUTION :1 CALORIE

Discussion

44.	The quantim mechanical treatment of the hydrogen atom gives the energy vale:E_n=(-13.6)/(n^(2))eV "atom"^(-1)(i) use thios expression to find /_\Ebetween=3 and n=4(ii) Calcuilate the wavelength corresponding to the above transition.
Answer» Solution :(i) When n=3 `E_3=(-13.6)/(3^(2))=(-13.6)/(9)=-1.511eV "atom"^(-1)` When n=5`E_4=(-13.6)/(4^(2))=(-13.6)/(16)=-0.85 eV "atom"^(-1)` `/_\E=E_4-E_3` `=-0.85-(-1.511)=+0.661eV "atom"^(-1)` `/_\E=E_3-E_4` `=-1.511-(-0.85)` `=-0.661 eV " atom"^(-1)` (ii) WAVE length=`gamma ` `/_\E=(hc)/(gamma)` `therefore gamma=(hc)/(/_\E)` `/_\E=0.661xx1.6xx10^(-19)J` `=1.06xx10^(-19)j` h=planck.s constant=`6.626xx10^(-34)JS^(-1)` `c=3xx10^(8)m/s` `therefore gamma=(6.626xx10^(-34)xx3xx10^(8))/(1.06xx10^(-19))` `=1875xx10^(-6)m.`

Discussion

45.	The quality of diesel is expressed in terms of
Answer» OCTANE NUMBER decane number IGNITION number cetane number ANSWER :D

Discussion

46.	The pyknometric density of sodium chloride crystal is 2.165xx10^3 " kg m"^(-3) while its X-ray density is 2.178xx10^3 " kg m"^(-3) . The fraction of the unoccupied sites in sodium chloride crystal is
Answer» 5.96 `5.96xx10^(-2)` `5.96xx10^(-1)` `5.96xx10^(-3)` Solution :MOLAR volume from pyknometric DENSITY =`M/(2.165xx10^3)m^3` Molar volume from X-ray density `=M/(2.178xx10^3)m^3` `THEREFORE` Volume unoccupied `=M/10^3(1/2.165-1/2.178)m^3=(0.013Mxx10^(-3))/(2.165xx2.178)` `therefore` Fraction unoccupied `=((0.013Mxx10^(-3))/(2.165xx2.178))//((Mxx10^(-3))/2.165)=5.96xx10^(-3)`

Discussion

47.	The pyknometric density of sodium chloride crystal is 2.165times10^(3)kg"" m^(-3)while its X-ray density is 2.178times10^(3)kg"" m^(-3). The fraction of unoccupiedsites in sodium chloride crystal is
Answer» `5.96times10^(-3)` 5.96 `5.96times10^(-2)` `5.96times10^(-1)` Solution :Pyknometric DENSITY = Observed density `=2.165 times10^(3)KG"" m^(-3)` X-ray density = Calculated density =`2.178 times10^(3) kg"" m^(-3)` DECREASE indensity `=(2.178times10^(3)-2.165times10^(3))kg"" m^(-3)` =`(0.013 times10^(3))kg"" m^(-3)=13kg"" m^(-3)` Fraction of unoccupied sites =`("Decrease in density")/("Calculated density")` =`(13kg"" m^(-3))/(2.178times10^(3)kg"" m^(-3))=5.96times10^(-3)`

Discussion

48.	The pyknometric density of sodium chloride crystals is2.165 xx 10^(3) kg m^(-3)while X -ray density is2.178 xx 10^(3)kg m^(-3) . Thefraction of the unoccupied sites in sodium chloride crystal is
Answer» 5.96 `5.96 xx 10^(-2)` `5.96 xx 10^(-1)` ` 5.96 xx 10^(-3)` Solution :MOLAR VOLUME from pyknometric density ` = M/(2.165 xx 10^(3)) m^(3)` Molar volume from X-ray density ` M/(2.178 xx 10^(3)) m^(3)` Volume unoccpied `M/10^(3) ( 1/(2.165) - 1/(2.178))m^(3) = (( 0.013 M xx10^(-3))/(2.165 xx 2.178))//((M xx 10^(-3))/2.165) = 5.96 xx 10^(-3)`

Discussion

49.	The pyknometer density of NaCl crystal is 2.165xx10^(3)kg m^(-3) while its X - rays density is 2.178xx10^(3)kg m^(-3). The fraction of the unoccupied sites in NaCl crystal is
Answer» `5.968` `5.96xx10^(-2)` `5.96xx10^(-3)` `5.96xx10^(-4)` SOLUTION :Molar volume from pyknometer density `= (M)/(2.165xx10^(3))m^(3)` Molar volume from X - RAY density `= (M)/(2.178xx10^(3))m^(3)` `therefore` Volume unoccupied `=((1)/(2.165)-(1)/(2.178))(M)/(10^(3))m^(3)` Fraction unoccupied `=((1)/(2.165)-(1)/(2.178))//(1)/(2.165)` `= 5.96xx10^(-3)`

Discussion

50.	The purpose of chemistry can be understood and described by...........
Answer» FUNDAMENTAL PARTICLES atoms molecules all of the above Answer :A::B::C::D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The radius of an atom of an element is 500 pm. If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell ?

The radius of an atom of an element is 500 pm.If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell ?

A thin wire has a length of 21.7 cm and radius 0.46 mm. Calculate the volume of the wire to correct significant figures.

The radioactive isotope of hydrogen is

The radioactive isotope of hydrogen is called ……….. And its nucleus contains ……….. Proton and ……….. Neutrons.

The radioactive element of group 2 element is_________

Theradii of the F, F^(-) , O " and " O^(2-) are in the order ….

The radii of Na^(+) and Cl^(-) ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is

The radii of maximum probability for 3s, 3p and 3d electrons are in the order :

The radii of F, F^(-), O, O^(2-) are in the order

The radii of F, F^(-) , O andO^(2-)are in the order

The radical halogenation of 2-methylpropane gives two products: (CH_(3))_(2)CHCH_(2)X (minor) and (CH_(3))_(3)CX (major) Chlorination gives a larger amount of the minor product than does bromination, why?

The radical halogenation of 2 - methyl propane gives two products (CH_(3))_(2)CHCH_(2)X_("(minor)") and (CH_(3))_(3)CX_("(major)"). Chlorination gives larger amount of the minor product than the bromination because

The radical halogenation of 2-methyl propane gives two products (CH_(3))_(2)CHCH_(2)X_(("minor")) and (CH_(3))_(3)CX_(("major")). Chlorination givs larger amount of the minor product than the bromination because

The radiation with highest wave number

The radial probability distribution curve of an orbital of H has '4' local maxima . If orbital has 3 angular node then orbital will be :

The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function correspond to ?

The R/S conifiguration of these compounds are respectively

The R and S configuration for each sterogenic centre in this from top to bottom?

The quantum numbers of four electrons (e_(1) " to " e_(4)) are given below : {:(,n,l,m,s,,,n,l,m,s),(e_(1),3,0,0,+1//2,,e_(2),4,0,0,+1//2),(e_(3),3,2,2,-1//2,,e_(4),3,1,-1,+1//2):} Correct order of decreasing energy of these electrons is

The quantum numbers n=2, l=1 represent

The quantum number which tells about the orientation of different orbitals of an atom is called........

The values of quantum numbers n, 1 and m for the fifth electron of boron is

The quantum number which may be designated by s,p,d and f instead of number is

The quantum number which tells about the angular momenta of the different electrons present in an atom is called........

The quantum number which explain the line spectra observed as doublets in case of hydrogen and alkali metals and doublets & triplets in case of alkaline earth metals is

The quantum number which is equal for all the d-electrons in an atom is

The quantum number not obtained from the Schrodinger.s wave equation is

The quantium number of four electrons (e_1to e_4) are given below The correct order of decreasing energy of these electrons is

The quantity that does not change for a sample of a gas in a sealed rigid container when. It is cooled from 120^(@)C to 90^(@)C at constant volume is:

The quantity of heat required to raise the temperature of 1 gm of water by 1^@C is ____

The quantim mechanical treatment of the hydrogen atom gives the energy vale:E_n=(-13.6)/(n^(2))eV "atom"^(-1)(i) use thios expression to find /_\Ebetween=3 and n=4(ii) Calcuilate the wavelength corresponding to the above transition.

The quality of diesel is expressed in terms of

The pyknometric density of sodium chloride crystal is 2.165xx10^3 " kg m"^(-3) while its X-ray density is 2.178xx10^3 " kg m"^(-3) . The fraction of the unoccupied sites in sodium chloride crystal is

The pyknometric density of sodium chloride crystal is 2.165times10^(3)kg"" m^(-3)while its X-ray density is 2.178times10^(3)kg"" m^(-3). The fraction of unoccupiedsites in sodium chloride crystal is

The pyknometric density of sodium chloride crystals is2.165 xx 10^(3) kg m^(-3)while X -ray density is2.178 xx 10^(3)kg m^(-3) . Thefraction of the unoccupied sites in sodium chloride crystal is

The pyknometer density of NaCl crystal is 2.165xx10^(3)kg m^(-3) while its X - rays density is 2.178xx10^(3)kg m^(-3). The fraction of the unoccupied sites in NaCl crystal is

The purpose of chemistry can be understood and described by...........