52671 + Interview Questions in Chemistry in Class 11

1.	In Huckel's (4n+2) rule of for aromaticity, 'n' represents
Answer» NUMBER of CARBON ATOMS Number of RINGS whole number Fractional number (or) integer (or) zero Solution :AROMATIC character

Discussion

2.	In HS^(-), I^(-), R-NH_(2), NH_(3), order of proton accepting tendency will be
Answer» `I^(-) gt NH_(3) gt R - NH_(2) gt HS^(-)` `NH_(3) gt R - NH_(2) gt HS^(-) gt I^(-)` `R-NH_(2) gt NH_(3) gt HS^(-) gt I^(-)` `HS^(-) gt R - NH_(2) gt NH_(3) gt I^(-)` Solution :In `R-NH_(2)`, electron pair on N is most easily AVAILABLE for protonation because R is eletron releasing group.

Discussion

3.	In how many reactions CO_(2) gas is released out after reaction with NaHCO_(3)
Answer» SOLUTION :

Discussion

4.	In how many of the folowinf reaction \|H^(@)\|gt\|DeltaE^(@)\|, assuming reacftion to be occurring at a constant temperature and pressure. H_(2)+(1)/(2)O_(2)rarrH_(2)O(l) Combustion of C_(2)H_(8)(g) at 400 K Dimerisation of NO_(2)(g) Fusion of water at 0^(@) C Vaporisation of any liquid at its boiling poing Dossociation of CaCO_(3)(s) Combustion of C_(2)H_(4) at 300m K
Answer» ANSWER :5

Discussion

5.	In how many reaction CH_(3)-CHO is obtained as major product ?
Answer» SOLUTION :All FOUR reaction GIVES `CH_(3)-CHO` as major PRODUCT.

Discussion

6.	In how many of the following the concept of resonance can be observed CH_(2)=CH-Cl, CH_(3)-O-overset(+)(C)H-CH_(3) CH_(3)-O-CH_(2)overset(+)(C)H, CH_(3)-NH-CH_(2)-CH-CH_(3) CH_(3)-CH=CH-NH_(2), CH_(3)-CH_(2)=CH_NH-CH_(3)
Answer» ANSWER :4

Discussion

7.	In how many of the following reactions, colored precipitate is obtained? (a). CuSO_(4)+NaoH("excess")to, (b) CuCl_(2)+NH_(3)("excess")to ltBrgt (c) ZnSO_(4)+NaOH("excess")to (d) ZnSO_(4)+NH_(3)("excess")to , (e) FeSO_(4)+NaOH("excess")underset(air)overset("in")to , (f) FeCl_(3)+NaoH("Excess")to (g) CuSO_(4)+Kl("excess")to, (h) ZnSO_(4)+NaHCO_(3)to (i) CuCl_(2)+NaNO_(3)to
Answer» Solution :(a) `Cu(OH)_(2)darr(blue)` (b) `[Cu(NH_(3))_(4)]^(2+)(aq.)` (c). `Na_(2)[ZN(OH)_(4)](aq.)` (d). `[Zn(NH_(3))_(4)]^(2+)(aq.)` (e) `Fe(OH)_(2)darr("greenish")` (f) `Fe(OH)_(3)darr("reddish brown")` (g) CUL (white) (H) `ZnCO_(3)darr ("white")` (i) `Cu(NO_(3))_(2)darr(aq.)`

Discussion

8.	In how many of the following processes, heat of neutralization released is less than 13.7 kcal (if 1gram equivalent of each is taken)? HCL(aq)+ NAOH(aq) HI(aq)+ NaOH(aq) HCN(aq)+KOH(aq) C_(6)H_(5)COOH(aq)+ NaOH(aq) HCl(aq)+ NH_(4)OH(aq)
Answer» ANSWER :5

Discussion

9.	In how many of the following entropy increases? {:((a) N_(2(g)) + 3H_(2(g)) rarr 2NH_(3(g)),(b)PCl_(5(g)) rarr PCl_(3(g)) + Cl_(2(g))),((c ) H_(2)O_((s)) rarr H_(2)O_((g)),(d)H_(2)O_((l)) rarr H_(2)O_((g))),((e ) 2NaHCO_(3(s)) rarr Na_(2)CO_(3(s)) + H_(2)O_((g)) + CO_(2(g)),(f ) NH_(2)CO_(2)NH_(4(s)) rarr 2NH_(3(g)) + CO_(2(g))),((g) H_(2)O_((l)) rarr H_(2)O_((s)),(h) CO_(2(s)) rarr CO_(2(g))):}
Answer» Solution :CASES in which entropies INCREASE are b,C,d,e,f,H

Discussion

10.	In how many of the following pairs will have higher enol content than second.
Answer» ANSWER :3

Discussion

11.	In how many of the following compounds Hyperconjugation effect is observed -
Answer» Answer :3(B, C & G).

Discussion

12.	In how many of the following cases, the negative charge is delocalised?
Answer» ANSWER :5(I, II, iv, VI, ix)

Discussion

13.	In hot alkaline solution, Br_2disproportionates to 3Br_(2)- 6OH^(-) rarr 5Br^(-) + BrO_3^(-) +3H_(2)O .The equivalent weight of Br_(2)is 3M/x (mol wt = M). Then X is?
Answer» <BR> Solution :For `3Br_(2)`MOLECULES change in. no = 5 Change in OXIDATION number for 1 MOLECULE = `5/3` =n-factor equivalent weight of `Br_(2)= M/(5//3) =(3M)/5`

Discussion

14.	in homologous series all the members have the same physical prperties
Answer» ANSWER :t

Discussion

15.	In hexogonal systems of crystals, a frequently encountered arrangement ofatoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexongonas and threeatoms are sandwiched inbetween them. A space filling model of this structure called haxagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The empty space in this HCP unit cell is
Answer» 0.74 0.476 0.32 0.26 Solution :Space OCCUPIED by SPHERES in HCP = 74 % Empty space = ( 100 -74) % = 26 %

Discussion

16.	In hexogonal systems of crystals, a frequently encountered arrangement ofatoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexongonas and threeatoms are sandwiched inbetween them. A space filling model of this structure called haxagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The number of atoms in this HCP unit cell is
Answer» 4 6 12 17 Solution :According to the given DESCRIPTION, the ARRANGEMENT of atoms in HCP will be as shownFig. 1.61. Effective number of atoms present in the unit CELL of HCP `12 xx1/6 + 3 + 2 xx 1/2 = 6`

Discussion

17.	In hexogonal systems of crystals, a frequently encountered arrangement ofatoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexongonas and threeatoms are sandwiched inbetween them. A space filling model of this structure called haxagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. The volume of this HCP unit cell is
Answer» `24sqrt2 r^(3)` ` 16 SQRT2 r^(3)` `12 sqrt2 r ^(3)` `64/(3sqrt3)r^(3)` Answer :d

Discussion

18.	In hexagonal systems of crystals, a frequently ecounted arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are reqular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. There spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three sphere touches three spheres of the bottom layer. Finally the second layer is covered with third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r' The volume of this HCP unit cell is
Answer» `24sqrt2 r^(3)` `16 SQRT2 r^(3)` `12 sqrt2 r^(3)` `(64)/(3 SQRT3)r^(3)` ANSWER :A

Discussion

19.	In hexagonal systems of crystals, a frequently ecounted arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are reqular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. There spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three sphere touches three spheres of the bottom layer. Finally the second layer is covered with third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r' The number of atoms in the HCP unit cell is
Answer» 4 6 12 17 Answer :B

Discussion

20.	In hexagonal systems of crystals , a frequenctly encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are regular hexagons and three atoms are sandwiched inbetween them. A space filling model of this structure called hexagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position . Assume radius of every sphere to be 'r'. The empty space in this HCP unit cell is
Answer» `74%` `47.6%` `32%` `26%` Solution :SPACE occupied by spheres in HCP =74% `therefore` EMPTY space =(100-74)%=26%

Discussion

21.	In hexagonal systems of crystals , a frequenctly encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are regular hexagons and three atoms are sandwiched inbetween them. A space filling model of this structure called hexagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position . Assume radius of every sphere to be 'r'. The volume of this HCP unit cell is
Answer» `24sqrt3r^3` `16sqrt2r^3` `12sqrt2r^3` `64/(3SQRT3)r^3` ANSWER :A

Discussion

22.	In hexagonal systems of crystals , a frequenctly encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are regular hexagons and three atoms are sandwiched inbetween them. A space filling model of this structure called hexagonal closed packed (HCP) is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible . Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position . Assume radius of every sphere to be 'r'. The number of atoms in this HCPunit cell is
Answer» 4 6 12 17 Solution :According to the given descripition, the ARRANGEMENT of ATOMS in HCP will. EFFECTIVE number of atoms PRESENT in the UNIT cell of HCP `=12xx1/6+3+2xx1/2`=6

Discussion

23.	In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagonand three atomsare sandwiched in between them. A space filling model of this structure, called hexagonal close packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical sphere as closely as possible. Three spheres are then placed over the first layer so that they touch other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be same The empty space in this HCP unit cell is
Answer» 0.74 0.476 0.32 0.26 Solution :Packing FRACTION =`("Volume of the atoms in one unit CELL")/("Volume of one unit cell")=74%` `"EMPTY space"=26%`

Discussion

24.	In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagonand three atomsare sandwiched in between them. A space filling model of this structure, called hexagonal close packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical sphere as closely as possible. Three spheres are then placed over the first layer so that they touch other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be same The volume of this HCP unit cell is
Answer» `24sqrt2r^(3)` `16sqrt2r^(3)` `12sqrt2r^(3)` `64/(3SQRT3)r^(3)` Solution :Height of unit cell =`sqrt(2/3)4r` `"Base AREA" = 6timessqrt3/4(2r^(2))` `"VOLUME" = (6sqrt3)/4(2r)^(2)*SQRT2/(3)4r=24sqrt2 r^(3)`

Discussion

25.	In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagonand three atomsare sandwiched in between them. A space filling model of this structure, called hexagonal close packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical sphere as closely as possible. Three spheres are then placed over the first layer so that they touch other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be same The number of atoms in this HCP unit cell is
Answer» 4 6 12 17 Solution :The EFFECTIVE NUMBER of ATOMS in the HCP is `=12times1/6+2times1/2+3=6`

Discussion

26.	In hexagonal close packing, the difference in the number of tetrahedral and octahedral voids per unit cell is
Answer» Solution :In one HCP unit cell, ATOMS in the packing = 6 `therefore" octahedral voids " =6,` tetrahedral voids =12, DIFFERENCE = 12-6=6

Discussion

27.	In hexa-1, 3-dien-5yne, the number of C-C, sigma, C-C pi and C-H sigma bonds respectively are
Answer» 5, 4 and 6 6, 3 and 5 5, 3 and 6 6, 4 and 5 Solution :`{:(H-overset(1)(C)=overset(2)(C)-overset(3)(C)=overset(4)(C)-overset(5)(C)-=overset(6)(C)-H),("\|""\|""\|""\|"),(""H""H" "H" "H),("Hexa-1,3-diene-5-yne"):}` C-C, `sigma`-BONDS=5 C=C, `PI`-bonds = 4 C-H, `sigma`-bonds = 6

Discussion

28.	In Heisenberg.s uncertainty experiment
Answer» Use of HIGH INTENSITY light increases accuracy in the measurement of distance of `E^(-)` from thenucleus Use of high FREQUENCY light increases accuracy in the measurement of distance of `e^(-)` from the nucleus If an experiment is designed to measure the distance of `e^(-)` from nucleus, the speed measured in the same experiment would be HIGHLY imprecise Use of low wavelength light increases accuravy in the measurement at distance of `e^(-)` form the nucleus Answer :D

Discussion

29.	In hevy water molecule
Answer» Both hydrogen ATOMS and OXYGEN atoms are heavier isotopes Hydrogen atoms are heavy hydrogens while oxygen atoms are lighter isotopes Both hydrogen and oxygen atoms are lighter isotopes none of these ANSWER :B

Discussion

30.	In heavy water molecualar
Answer» Both HYDROGEN atomsand oxygen atoms are heavier isotopes Hydrogen atoms are HEAVY HYDROGENS while oxygen are lighter isotopes Both hydrogen and oxygen atoms are lighter isotopes Hydrogen atoms are lighter isotopes and oxygen atoms are heavier isotopes Solution :Heavier hydrogen and lighter isotope FO oxygen

Discussion

31.	In HCOO- ion, the two carbon-oxygen bonds are found to be equal length, what is the reason for it?
Answer» The C = O BOND is weaker than the C - O bond The ANION `HCOO^(-)` has two resonating structures The ELECTRONIC orbital's of carbon atom are hybridized The anion is obtained by removal of a proton from the acid molecule Solution :

Discussion

32.	In halogens, which of the following decreases from flourine to Iodine?
Answer» BOND length Electronegativity Ionization energy Oxidizing power Answer :A

Discussion

33.	In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the species
Answer» X `X^(-)` `X^(+)` `X^(**)` ANSWER :C

Discussion

34.	In halogenationof aromaticcompounds, catalyst is
Answer» LEWIS BASE `FeCl_3` `AICI_3` `FeCl_3 or AICI_3` ANSWER :D

Discussion

35.	In Hall's process, tha main reagent is mixed with
Answer» NaF `Na_(3)AlF_(3)` `AlF_(3)` None of these Solution :In Hall's process `Al_(2)O_(3)` (ALUMINA) is mixed with `Na_(3)AlF_(6)` to reduce its melting POINT for electrolysis.

Discussion

36.	In haemoglobin, CO and not O_2 links to Fe (if both CO and O_2 are present ) because CO is a stronger ligand than O_2. The number of groups to which Fe is coordinated other than vacant site for CO in haemoglobin is
Answer» Solution :In haemoglobin, Fe is COORDINATED to FIVE groups and sixth SITE is free to which CO links if present.

Discussion

37.	In Haber's process 50.0 g of N_(2)(g) and 10.0 g of H_(2)(g) are mixed to produce NH_(3)(g). What is the number of moles of NH_(3)(g) formed?
Answer» 3.33 2.36 2.01 5.36 Answer :A

Discussion

38.	In Haber process for the synthesis of ammonia which of the following catalysts are used ?
Answer» `Fe, K_(2)O` `K_(2),Al_(2)O_(3)` `Fe,Al_(2)O_(3)` `Fe, K_(2)O, Al_(2)O_(3)` Answer :D

Discussion

39.	In H_2O_2the degree of hydrogen bonding is
Answer» Same as in WATER More than in water LESS than in water EQUAL to zero Answer :B

Discussion

40.	In H_2O_2molecule, the O-O bond length is
Answer» `1.34A^(@)` `1.48A^(@)` `1.54A^(@)` `1.20A^(@)` ANSWER :B

Discussion

41.	In H_2O_2molecule, the H-O-O angle is
Answer» `95^@` `106^@ 30' ` `111^@ 30'` `120^@ 18'` ANSWER :C

Discussion

42.	In H_(2)O and H_(2)O_(2)mass oxygen
Answer» `1:2` `2:1` `32:16` `1:8` ANSWER :A::B

Discussion

43.	In H_(2(g)) + I_(2(g)) (violet) hArr 2HI_((g))if H_2 is added than at new equilibrium concentration and colour will be ….
Answer» SOLUTION :(i) As the concentration of `H_2` increase, it more COMBINE with `l_2` so , `I_2` decrease and intensity of COLOUR decreases. (ii) The amount of `H_2` is less at equilibrium compare to initial. (III) The amount of `H_1` increases but `K_c` remain constant.

Discussion

44.	In H-atoms, the energy of electron in the nth orbit is given as E_(n) = - (13.6)/(n^(2)) eV Show that E_((n +1)) - E_(n) = (13.6 xx 2)/(n^(3)) eV for large value of n.
Answer» Solution :`E_((n +1)) - E_(n) = - (13.6)/((n +1)^(2)) - (-(13.6)/(n^(2))) = 13.6 [(1)/(n^(2)) - (1)/((n +1)^(2))] = 13.6 xx ((n +1)^(2)- n^(2))/(n^(2) (n +1)^(2))` `= (13.6 xx (2N +1))/(n^(2) (n +1)^(2)) = (13.6 xx 2n)/(n^(2) (n +1)^(2)) + (13.6)/(n(n + 1)^(2)) =(13.6 xx 2)/(n(n +1)^(2)) + (13.6)/(n^(2) (n + 1)^(2))` As n is LARGE, `n + 1 ~=n`. HENCE, `E_((n +1)) - E_(n) ~= (13.6 xx 2)/(n xx n^(2)) + (13.6)/(n^(2) xx n^(2)) = (13.6 xx 2)/(n^(3))` (Neglecting second term because `n^(4)` is very large)

Discussion

45.	In gylcerine,
Answer» One PRIMARY `-OH` group is present only One TERTIARY `-OH` group is present Two SECONDARY `-OH` GROUPS are present One secondary `-OH` group is present Answer :D

Discussion

46.	Due to inert pair effect, the stability of +2 oxidation state increases on moving down the group 14. Why?
Answer» SOLUTION :DECREASE,Increases , inert PAIR EFFECT

Discussion

47.	In group 14, the tendency for catenation decreases with increasing atomic number .
Answer» Solution :As we move down the group 14, the atomic SIZE increasesand HENCETHE STRENGTH of the element -elementbond, i.e, the bonddissociationenthalphy decreasessteadily. Consequentlythe tendencyfor catendation decreasesdown the group with increasingatomic NUMBER.

Discussion

48.	In group 14 of the periodic table, the oxidising power of tetravalent species decreases in the ordera. Ge gt Pb gt Snb. Ge gt Sn gt Pbc. Pb gt Ge gt Snd. Pb gt Sn gt Ge
Answer» `GE GT PB gt SN` `Ge gt Sn gt Pb` `Pb gt Ge gt Sn` `Pb gt Sn gt Ge` SOLUTION :`Pb gt Sn gt Ge`

Discussion

49.	In graphite, hybridization of carbon is
Answer» `SP` `sp^(3)d` `sp^(3)` `sp^(2)` ANSWER :B

Discussion

50.	In graphite, electrons are ………
Answer» localised on every third C-atom. present in anti-bonding orbital. localised on each C-atom. SPREAD out between the STRUCTURE. Solution :In graphite, each CARBON is `sp^2` -hybridized and the single occupied unhybridized p-orbitals of C-atoms OVERLAP side wise to GIVE `pi`-electron cloud which is delocalized and thus the electrons are spread out between the structure.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In Huckel's (4n+2) rule of for aromaticity, 'n' represents

In HS^(-), I^(-), R-NH_(2), NH_(3), order of proton accepting tendency will be

In how many reactions CO_(2) gas is released out after reaction with NaHCO_(3)

In how many reaction CH_(3)-CHO is obtained as major product ?

In how many of the following the concept of resonance can be observed CH_(2)=CH-Cl, CH_(3)-O-overset(+)(C)H-CH_(3) CH_(3)-O-CH_(2)overset(+)(C)H, CH_(3)-NH-CH_(2)-CH-CH_(3) CH_(3)-CH=CH-NH_(2), CH_(3)-CH_(2)=CH_NH-CH_(3)

In how many of the following processes, heat of neutralization released is less than 13.7 kcal (if 1gram equivalent of each is taken)? HCL(aq)+ NAOH(aq) HI(aq)+ NaOH(aq) HCN(aq)+KOH(aq) C_(6)H_(5)COOH(aq)+ NaOH(aq) HCl(aq)+ NH_(4)OH(aq)

In how many of the following pairs will have higher enol content than second.

In how many of the following compounds Hyperconjugation effect is observed -

In how many of the following cases, the negative charge is delocalised?

In hot alkaline solution, Br_2disproportionates to 3Br_(2)- 6OH^(-) rarr 5Br^(-) + BrO_3^(-) +3H_(2)O .The equivalent weight of Br_(2)is 3M/x (mol wt = M). Then X is?

in homologous series all the members have the same physical prperties

In hexagonal close packing, the difference in the number of tetrahedral and octahedral voids per unit cell is

In hexa-1, 3-dien-5yne, the number of C-C, sigma, C-C pi and C-H sigma bonds respectively are

In Heisenberg.s uncertainty experiment

In hevy water molecule

In heavy water molecualar

In HCOO- ion, the two carbon-oxygen bonds are found to be equal length, what is the reason for it?

In halogens, which of the following decreases from flourine to Iodine?

In halogenation of aromatic hydrocarbon, a halogen carrier is used which is generally a Lewis acid. The main function of this reagent is to generate the species

In halogenationof aromaticcompounds, catalyst is

In Hall's process, tha main reagent is mixed with

In haemoglobin, CO and not O_2 links to Fe (if both CO and O_2 are present ) because CO is a stronger ligand than O_2. The number of groups to which Fe is coordinated other than vacant site for CO in haemoglobin is

In Haber's process 50.0 g of N_(2)(g) and 10.0 g of H_(2)(g) are mixed to produce NH_(3)(g). What is the number of moles of NH_(3)(g) formed?

In Haber process for the synthesis of ammonia which of the following catalysts are used ?

In H_2O_2the degree of hydrogen bonding is

In H_2O_2molecule, the O-O bond length is

In H_2O_2molecule, the H-O-O angle is

In H_(2)O and H_(2)O_(2)mass oxygen

In H_(2(g)) + I_(2(g)) (violet) hArr 2HI_((g))if H_2 is added than at new equilibrium concentration and colour will be ….

In H-atoms, the energy of electron in the nth orbit is given as E_(n) = - (13.6)/(n^(2)) eV Show that E_((n +1)) - E_(n) = (13.6 xx 2)/(n^(3)) eV for large value of n.

In gylcerine,

Due to inert pair effect, the stability of +2 oxidation state increases on moving down the group 14. Why?

In group 14, the tendency for catenation decreases with increasing atomic number .

In group 14 of the periodic table, the oxidising power of tetravalent species decreases in the ordera. Ge gt Pb gt Snb. Ge gt Sn gt Pbc. Pb gt Ge gt Snd. Pb gt Sn gt Ge

In graphite, hybridization of carbon is

In graphite, electrons are ………