94420 + Interview Questions in Chemistry in Class 12

1.	Fill in the blanks EDTA is a …. ligand.
Answer» SOLUTION :HEXADENTATE LIGAND,

Discussion

2.	EDTA is a bidentate ligand . True or False?
Answer» SOLUTION :HEXADENTATE LIGAND

Discussion

3.	EDTA is a…..ligand:
Answer» Monodentate Hexadentate Bidentate Tridentate Answer :B

Discussion

4.	The edge length of a face centred cubic cell of an ionic substance is 508 pm .If the radius of the cation is 110 pm the radius of the anion is
Answer» ANSWER :144

Discussion

5.	Edge length of a cube is 400 pm. Then its bodydiagonal length would be
Answer» 50 PM 600pm 566 pm 639 pm ANSWER :D

Discussion

6.	Edge length of a cube is 400 pm. Its body diagonal would be :
Answer» 600 pm 566 pm 693 pm 500 pm Answer :C

Discussion

7.	E^@Cu^(2+)//Cuis + 0.34VAt which one of the following concentrations of Cu^(2+), reduction potential of copper electrode becomes zero
Answer» `1.76 xx 10^(-6)M` `3.12 xx 10^(-12)M` `2 xx 10^(-8) M` `3 xx 10^(-14) M` Answer :B

Discussion

8.	E^(@)(Cu^(2+)//Cu) and E^(@)(Ag^(+)//AG)" is "0.337V and +0.799 V respectively. Make a cell whose EMF is +ve. If the concentration of Cu^(2+)" is 0.01M and "E_("cell")" at "25^(@)C is zero, calculate the concentration of Ag^(+).
Answer» Solution :Cu is more reactive than silver, so that the cell is as `Cu//Cu_(2)^(+) (0.01M) \|\| Ag^(+) (C )//Ag` or cell reaction `Cu+2Ag^(+)rarrCu^(2+)+2Ag` `E_("cell")=E_("cell")^(@)-(0.0591)/(n)log.([Cu^(2+)][Ag]^(2))/([Cu][Ag^(+)]^(2))` `=E_("cell")^(@)-(0.0591)/(n)log.((0.01)xx1^(2))/(1xx[Ag^(+)]^(2))` `"Or"[Ag^(+)]=1.47xx10^(-9)M`

Discussion

9.	Edge length of a cube is 300 pm. Its body diagonal would be
Answer» 600 pm 423 pm 519.6 pm 450.5 pm Solution :LENGTH of BODY diagonal `=sqrt3 xx`EDGE length of the cube (a) `=sqrt3xx300` pm `= 1.732xx300` =519.6 pm

Discussion

10.	E_(Cu^(2+)\|Cu)^@ = +0.337 V, E_(Zn^(2+)\| Zn)^@ =- 0.762 V.The EMF of the cell ,Zn\|Zn^(2+) (0.01M)\|\|Cu^(2+) (0.01M)\|Cu is
Answer» +1.099 V - 1.099 V +1.069 V - 1.069 V Answer :C

Discussion

11.	E^(@)Cu=0.34V,E_(Zn)^(@)=-0.76V. A Daniell cell contains 0.1M ZnSO_(4) solution and 0.01 M CuSO_(4) solution at its electrodes. E.M.F. of the cell is
Answer» 1.10V 1.04V 1.16V 1.07V Solution :`ZN\|Zn^(2+)\|\|CU^(2+)\|Cu` `Zn+Cu^(2+)+rarrZn^(2+)+Cu` `E_("cell")^(@)=E_(Cu^(2+)._(//Cu))-E_(Zn^(2+)._(//Zn))^(@)` `=0.34-(-0.76)=1.10V` `E_("cell")=E_("cell")^(@)(0.0591)/(2)"LOG"([Zn^(2+)])/([Cu^(2+)])` `=1.10V-(0.0591V)/(2)"log"(0.1)/(0.01)` `=10.7045V`

Discussion

12.	E_(Cu^(+2)//Cu)^(@)=0.34V, E_(Zn//Zn^(+))^(@)=0.76V A cell formed by the conbination of Cu and Zn (a) when CuSO_(4) is added to Cu^(+2) compartment what is the effect on emf of cell (b) when ZNH_(3) is added to Cu^(+2) compartment what is the effect on emf of cell (c ) When ZnSO_(4) is added to Zn^(+2) compartment is the effect on emf of cell (d) When Zn^(+2) is diluted what is the effect on emf of cell ?
Answer» (a) INCREASES (B) DECREASES (C ) decreases (d) increases (a) decreases (b) decreases (c ) increases (d) increases (a) increases (b) decreases (c ) decreases (d) decreases (a) increases (b) increases (c ) decreases (d) increases SOLUTION :(a) increases (b) decreases (c ) decreases (d) increases

Discussion

13.	E_(Cu)^(0)=+0.34V and E_(Ag)^(0)=+0.8V" calculate "E_(cell")^(0).
Answer» SOLUTION :`E_("CELL")^(0)overset(oplus_(0))(E)-overset((-0)_(0))(E) =E_(Ag)-E_(Cu)` =+0.8-0.34 `E_("cell")^(0)=+0.46V`

Discussion

14.	Eclipsed and staggered forms of n-butane are called a pair of:
Answer» Diastereomers Conformers Isomers Enantiomers Answer :B

Discussion

15.	E_(Cell)^(Theta)=1.1V for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?
Answer» `1.1=K_(c)` `(2.303RT)/(2F)logK_(c)=1.1` `logK_(c)=(2.2)/(0.059)` `logK_(c)=1.1` SOLUTION :For Daniell cell `ZN+Cu^(2+)TOZN^(2+)+Cu,E_(cell)=E_(cell)^(@)-(2.303RT)/(nF)"log"([Zn^(2+)])/([Cu^(2+)])` At equilibrium `E_(cell)=0`. Hence, `E_(cell)^(@)=(2.303RT)/(2F)logK_(c)=1.1` (given) i.e.., (b) Putting `(2.303RT)/(F)=0.059` at 298K, `(0.059)/(2)logK_(c)=1.1` or log`K_(c)=(2.2)/(0.059)` i.e., (c).

Discussion

16.	E_(Cell)^(Theta) for some half cell reactions are given below. On the basis of these mark the correct anwer. (i) H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E_(Cell)^(Theta)=0.00V (ii) 2H_(2)O(l)toO_(2)(g)+4H^(+)(aq)+4e^(-),E_(Cell)^(Theta)=1.23V (iii) 2SO_(4)^(2-)(aq)toS_(2)O_(8)^(2-)(aq)+2e^(-),E_(Cell)^(Theta)=1.96V
Answer» In dilute SULPHURIC acid solution, HYDROGEN will be reduced at cathode. In concentrated sulphuric acid solution, WAER will be oxidised at anode. In dilute sulphuric acid solution, WATER will be oxidised at anode. in dilute sulphuric acid solution, `SO_(4)^(2-)` ion will be oxidised to tetrathionate ion at anode. Solution :Indilute `H_(2)SO_(4)`, (acidulated water), `H^(+)` ions are reduced to `H_(2)` at cathode, i.e., (a) and `H_(2)O` is oxidized to `O_(2)` at anode i.e., (c) reaction (III) occurs when `H_(2)SO_(4)` is concentrated

Discussion

17.	E_(cell)^(Theta) for some half-cell reactions are given below. On the basis of these mark the correct answer. (a) H_((aq))^(+)+e^(-) to (1)/(2) H_(2(g))E_(cell)^(Theta)=0.00V (b) 2H_(2)O_((l)) toO_(2(g))+4H_((aq))^(+)+4e^(-),E_(cell)^(Theta)=1.23V (c) 2SO_(4)^(2-) to S_(2)O_(8(aq))^(2-) +2e^(-),E_(cell)^(Theta)=1.96V
Answer» In dilute sulphuric acid SOLUTION, HYDROGEN will be reduced at cathode. In CONCENTRATED sulphuric acid solution, water will be oxidised at ANODE. In dilute sulphuric acid solution, water will be oxidised at anode. In dilute sulphuric acid solution, `SO_(4)^(2-)` ion will be oxidized to tetrathionate ion at anode. Solution :In dilute sulphuric acid solution, water will be oxidised at anode.

Discussion

18.	E_(cell)^(Theta)=1.1V for Daniel cell. Which of the following expression are correct description of state of equilibrium in this cell ?
Answer» `1.1=K_(C)` `(2.303RT)/(2F)log" "K_(C)=1.1` `logK_(C)=(2.2)/(0.059)` `logK_(C)=1.1` Solution :* According to NERNST formula `E_(cell)=E_(cell)^(THETA)=-(2.303RT)/(NF)log_(10)Q` At the TIME of equilibrium `E_(cell)=0 and Q=K,N=2` `therefore E_(cell)^(Theta)=(2.303RT)/(2F)logK_(C)=(0.059)/(2)logK_(C)` `therefore (2.303RT " log "K_(C))/(2F)=1.1` So, option (B) is correct. * `E_(cell)^(Theta)=(0.059)/(n)log" "K_(C)=0` `therefore 1.1=(0.059)/(2)log" "K_(C)` `therefore logK_(C)=(2.2)/(0.059)` So, option (C) is correct.

Discussion

19.	E_(cell)=0.78 volt for the following cell. underset(Fe_((s))\|Fe_((aq))^(2+))\|\|underset((0.01M)(Cu_((aq))^(2+)\|Cu_((s))) E_(Fe//Fe^(2+)(aq))^(@)=0.44V,E_(Cu//Cu^(2+)(aq))^(@)=-0.34V
Answer» x cannot be predicted x=0.01M `xgt0.01M` `xlt0.01M` SOLUTION :If a solution contains two or more catios, the cation to be deposited on the cathode is the one which has HIGHER reduction potential. Here the REACTION will be `Fe+Cu^(2+)TOFE^(2+)+Cu` `E_(cell)=E_(cell)^(o)-(0.0591)/(n)"log"([Fe^(2+)])/([Cu^(2+)])` `0.78=0.78-(0.591)/(n)"log"[(Fe^(2+))/(Cu^(2+))]` or `0="log"(x)/(0.01)implies(x)/(0.01)=1impliesx=0.01`.

Discussion

20.	E_(cell)^(@) is 1.89Vfor the reaction 2Ce^(4+) +Co rarr 2Ce^(3+)+Co^(2+). If E_(Co^(2+)//Co) is -0.28V.What is the value of E_(Ce^(4+)//Ce^(3+))^(@)?
Answer» `-1.61V` `+1.61V` `-1.89V` `2.17V` ANSWER :B

Discussion

21.	E_(cu)^(0)=0.34 and E_(zn)^(0)=-0.76V. Calculate E_("cell")^(0).
Answer» Solution :`E_("CELL")^(0)=OVERSET(oplus_(0))(E)-overset((-0)_(0))(E)` `E_("cell")^(0)=E_("cell")^(0)-E_(ZN)^(0)=+0.34-(0.76)=+1.1V`

Discussion

22.	E_("cell")^(@) for the redox reation 2Ag_((aq))^(+)Cu rarr Cu_((aq))^(2+) +2Ag is 0.46V. Calculate the equibilrium constant of the reaction.
Answer» Solution :`E_("cell")^(@)` for the given reaction `= (0.059)/(2) log K = 0.46 V` (or) `log K=(0.46xx2)/(0.059)=15.6` Equilibrium CONSTANT, `K=10^(15.6) =3.9xx10^(15)`.

Discussion

23.	E_("Cell") for the given redox reaction is 2.71 V Mg_((s))+Cu_((0.01 M))^(2+)to Mg_((0.011M))^(2+)+Cu_((s)) Calculate E_("cell") for the reaction. Write the direction of flow of current when an external opposite potential applied is: (i) less than 2.71 V and (ii) greater than 2.71 V
Answer» Solution :`E_("cell")=E_("Cell")^(@)-(0.059)/(N)log K_(C)` `=E_("Cell")^(@)-(0.059)/(2)"log"(10^(-3))/(10^(-2))` `=2.71+0.0295` `E_("cell")=2.7395V` (i) CU to Mg / Cathode to ANODE / Same DIRECTION (ii) Mg to Cu / Anode to cathode / Opposite direction

Discussion

24.	E_("cell")^(@) for the redox reaction 2Ag^(+)(aq) + Cu to Cu^(2+)(aq) + 2Ag is 0.46 V. Calculate the equilibrium constant of the reaction.
Answer» Solution :`E_("cell")^(@)` for the given REATION= `(0.059)/(2) "log" K = 0.46 V " (or) " log K = (0.46 xx 2)/(0.059) = 15.6` Equilibrium CONSTANT, `K = 10^(15.6) = 3.9 xx 10^(15)`.

Discussion

25.	E_(cell)^(@) for some half cell reaction are given below on the basis of these markthe correct answer I H^(+)(aq)+e^(-)rarr1/2 H_(2)(g),E_(cell)^(-)=0.00 V II2H_(2)O(l)rarrO_(2)(g)+4H^(+)(Aq)+4e^(-),E_(cell)^(-)=1.23 V III 2SO_(4)^(2-)(aq)rarrS_(2)O_(8)^(2-)(aq)+2e^(-)+2e^(-),E_(cell)^(-)=1.96 V
Answer» in dilute suplhuric acid solution HYDROGEN will be REDUCED at cathode in CONCENTRATED sulplhuric acid solution water will be oxidised at anode in dilutesulphuric acid solutionwater will be oxidised at cathode in dilutesulphuricacid solution`SO_(4)^(2-)` ION will beoxidisedto TETRATHIONATE ion anode Solution :In dilue `H_(2)SO_(4)`solutionhydrogen will be reduced at cathode

Discussion

26.	E_("cell")^(@) for some half - cell reactions are given below. On the basis of these mark the correct answer - (a) H^(+)(aq)+e rarr (1)/(2)H_(2)(g), E_("cell")^(@)=0.00V (b) 2H_(2)O(l) rarr O_(2)(g)+4H^(+)(aq)+4e , E_("cell")^(@)=+1.23V (c) 2SO_(4)^(2-)(aq) rarr S_(2)O_(8)^(2-)(aq)+2e , E_("cell")^(@)=+1.96V
Answer» in dilute `H_(2)SO_(4)` solution, hydrogen will be reduced at cathode in cocentrated `H_(2)SO_(4)` solution, water will be oxidised at anode in dilute `H_(2)SO_(4)` solution, water will be oxidised at anode in dilute `H_(2)SO_(4)` solution, `SO_(4)^(2-)` ION will be oxidised to tetrahionate ion at anode ANSWER :A::C

Discussion

27.	Ecamine structures a through j, shown below, with reaspect to their symmerty or lack of it. Assuming that the five-membered rings and the ring in compound g are plannr. The wedgephatched bonds in b,c,d & e designate specific configuration. Also for the acyclic compounds assume stable anti conformations. Answer each of the following questions by writing letters (a through j), corresponding to your selections, in each answer box. If there is no structure that fits the description enter an x in the answer box.
Answer» SOLUTION :(A) e,f,j `to` STR. Do not have P.O.S. & C.O.S. so optically ACTIVE. (B) a,b,c,d G,i-has P.O.S. so are O.I. (C ) No ONE str. Has C.O.S..

Discussion

28.	E_(cell) and triangleG^o are related as:triangleG^o=nFE^o,triangleG^o=-nFE^o,triangleG^o=-nFE_(cell)^o,triangleG^o=nFE_(cell)^o=0.
Answer» `triangleG^o=nFE^o` `triangleG^o=-nFE^o` `triangleG^o=-nFE_(CELL)^o` `triangleG^o=nFE_(cell)^o=0` ANSWER :C

Discussion

29.	E^(c-) of Cu is + 0.34 V while that of Zn is - 0.76V . Explain.
Answer» Solution :Positive `E^(@)` value for Cu is due to the fact THATSUM of SUBLIMATION enthalpy and ionization enthalpy to convert Cu(s) to `Cu^(2+)` (aq) is so high that it is not BALANCED by its hydratio enthalpy. Negative `E^(@)` for ZN ( more negative than expected value) is due to the fact that after removal of electrons from 4s orbital , stable `3d^(10)` configuration is ATTAINED.

Discussion

30.	Ebullioscopy is concerned with
Answer» osmotic PRESSURE lowering of VAPOUR pressure elevation of B. pt depression of F.pt Solution :It is a FACT.

Discussion

31.	Ebonite is a
Answer» NATURAL RUBBER synthetic rubber highly vulcanized rubber polypropene Solution :EBONITE is a hard highly vulcanized rubber, containing 20-25%, sulphur.

Discussion

32.	Ebullioscopy is concerned with:
Answer» Osmotic pressure of a SOLUTION Elevation of boiling point of a solution Depression in FREEZING point of a solution Relative lowering in VAPOUR pressure of a solution Answer :B

Discussion

33.	Ebonite is
Answer» NATURAL rubber synthetic rubber HIGHLY vulcanized rubber polypropene Solution :Ebonite is a HARD highly vulcanized rubber, CONTAINING 20-30%, rubber

Discussion

34.	Easily ___________ gases are readily adsorbed by the adsorbent.
Answer» SOLUTION :LIQUEFIABLE

Discussion

35.	Ease of nucleophilic addition in the given compounds is
Answer» `IgtIIIgtII` `IIgtIIIgtI` `IIgtIgtIII` `IIIgtIgtII` ANSWER :B

Discussion

36.	Easiest to perform in the following is
Answer» `F_(2(G))+2E^(-)to2F_((g))^(-)`<BR>`Cl_(2(g))+2e^(-)to2Cl_((g))^(-)` `Br_(2(g))+2e^(-)to2Br_((g))^(-)` `I_(2(g))+2e^(-)to2I_((g))^(-)` ANSWER :A

Discussion

37.	Ease of sulphonation of alkanes is:
Answer» `3^@gt 2^@gt1^@` `1^@gt2^@GT3^@` `2^@gt3^@gt1^@` `3^@gt1^@gt2^@` ANSWER :A

Discussion

38.	Ease of abstraction of hydrogen is greater when attached to:
Answer» `1^@` CARBON `2^@` carbon `3^@` carbon Neo-carbon Answer :C

Discussion

39.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The volume of the above solution required to make 1.0 L of 0.10 M HCl will be
Answer» 5.6 mL 6.6 mL 7.6 mL 8.6 mL Solution :`UNDERSET("(GIVEN vol.)")(M_(1)V_(1))=underset("(Required SOL.)")(M_(2)V_(2))` `13.15xxV_(1)=0.10xx1000mL or V_(1)=7.6mL`

Discussion

40.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The equivalent weight of Cu
Answer» will be same in CuO and `Cu_(2)O` will be double in `Cu_(2)O` than in CuO will be double in CuO than in `Cu_(2)O` DEPENDS upon whether COPPER is pure or impure. Solution :In CuO, 16 parts of oxygen combines with Cu = 63.5 parts by mass. Hence, 8 parts of oxygen will combine with Cu = 31.75 parts `therefore` EQ. wt. = 31.75 In `Cu_(2)O,` 16 parts of oxygen combine with Cu `=2xx63.5` parts by mass `therefore"8 parts of oxygen will combine with Cu"` `= 63.5 ` parts `therefore" Eq. wt. = 63.5"`

Discussion

41.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The mole fraction of hydrochloric acid in the solution will be
Answer» 0.25 0.3 0.35 0.4 Solution :`"Mass of solvent"=100-40=60g=0.060kg.` `"Moles of solvent (WATER)"=(60)/(18)=3.33` MOLE fraction of HCL in the solution `=(1.096)/(1.096+3.33)=0.25`

Discussion

42.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The molality of the above solution will be nearly
Answer» 15.3 m 16.3 m 17.3 m 18.3 m Solution :`"Molarity"=("Moles of SOLUTE")/("SOLVENT in KG")=("1.096 MOL")/("0.060 kg")=18.3m`

Discussion

43.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. The chloride of an element is found to contain 52.8% chlorine. The equivalent weight of the element is
Answer» 63.4 31.7 47.2 18.7 Solution :Element present = `100-52.8=47.2` PARTS `THEREFORE"52.8 parts by MASS of CHLORINE combine with element = 47.2 parts"` `therefore"35.5 parts by mass of chlorine will combine with element"=(47.2)/(52.8)xx35.5=31.7" parts"` HENCE, eq. wt. = 31.7.

Discussion

44.	Earlier the concept of equivalent weights was very common and the concentrations of the solutions were expressed in terms of normalities. The convenience was that the substances reacted in the ratio of their gram equivalents . So there was no need to write balanced equation to determine the amounts of the substraces reacted. However, determination or equivalent weights posed difficulty in certain cases. Moreover, the equivalent weight of the same substance is not same in different reactions, For example, KMnO_(4) has different equivalent weight in the basic medium than inthe acidic medium. Hence, now a days, mole concept is more common and the concentrations of the solutions are generally expressed in terms of molarities, though some other methods like molality. mole fraction etc. are also used. A 40% hydrochloric acid is found to have a density of 1.20"g mL"^(-1). The molarity of the solution is nearly
Answer» 11 M 12 M 13 M 14 M Solution :`40%HCL " solution means 40 g HCl in 100 g of the solution."` `"40 g HCl"=(40)/(36.5)"mol = 1.096 mol"` `"Volume of solution"=(100G)/(1.20"g mL"^(-1))` `=83.3mL=0.0833L` `"MOLARITY"=(1.096mol)/(0.0833L)` `=13.15"mol L"^(-1)or 13.15M`

Discussion

45.	EAN rule is followed by the complexes. (i)[Fe(pi-C_5H_5)_2](ii)[Mn_2(CO)_10](iii)[V(CO)_6]^(-)
Answer» only (i) only (II) only (III) all Solution :For all THREE EAN=36

Discussion

46.	EAN of the elements (*) are equla in :
Answer» `overset()(Ni)(CO)_(4), [overset()(FE)(CN)_(6)]^(4-)` `[overset()(Ni)(en)_(2)]^(2+), [overset()(Fe)(H_(2)O)_(6)]^(2+)` `[overset()(Co)(CN)_(6)]^(3-), [overset()(Ni)(CN)_(6)]^(3-)` `[overset()(Ni)(en)_(2)]^(2+), [overset()(Sc)(H_(2)O)_(6)]^(3+)` SOLUTION :EAN of `[Ni(CO)_(4)]=Fe(CN)_(6)]=36`

Discussion

47.	EAN of Fe in K_3[Fe(CN)_6] is.
Answer» 36 37 38 35 Answer :D

Discussion

48.	EAN of cobalt is 36 in [Co(NH_(3))_(2)O_(2)(en)Cl]. Knowing that atomic number of Co is 27, O_(2) is present is
Answer» PEROXIDE ION dioxide ion superoxide ion OXIDE ion Solution :Peroxide = `O_(2)^(2-)`, dioxide = `2 O^(2-)`, superoxide `=O_(2)^(-)`, oxide =`O^(2-)` EAN = Z - Oxidation no. +2 `xx` coordination no. If `O_(2)` is present as `O_(2)^(-)`, oxidation no. of Co = + 2 ltbr EAN = `27-2+2xx6=37` If `O_(2)` is present as `O_(2)^(2-)`, oxidation no. of Co=+3 EAN=`27-3+2xx6=36`.

Discussion

49.	EAN of a metal carbonyl M(CO)_(x) is 36. If atomic number of metal M is 26, what is the value of x?
Answer» 4 8 5 6 Solution :36 = 26 + 2X `RARR x=5`

Discussion

50.	EAN of Cr in [Cr(NH_3)_6]Cl_3 is:
Answer» 32 33 34 35 Answer :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Fill in the blanks EDTA is a …. ligand.

EDTA is a bidentate ligand . True or False?

EDTA is a…..ligand:

The edge length of a face centred cubic cell of an ionic substance is 508 pm .If the radius of the cation is 110 pm the radius of the anion is

Edge length of a cube is 400 pm. Then its bodydiagonal length would be

Edge length of a cube is 400 pm. Its body diagonal would be :

E^@Cu^(2+)//Cuis + 0.34VAt which one of the following concentrations of Cu^(2+), reduction potential of copper electrode becomes zero

E^(@)(Cu^(2+)//Cu) and E^(@)(Ag^(+)//AG)" is "0.337V and +0.799 V respectively. Make a cell whose EMF is +ve. If the concentration of Cu^(2+)" is 0.01M and "E_("cell")" at "25^(@)C is zero, calculate the concentration of Ag^(+).

Edge length of a cube is 300 pm. Its body diagonal would be

E_(Cu^(2+)|Cu)^@ = +0.337 V, E_(Zn^(2+)| Zn)^@ =- 0.762 V.The EMF of the cell ,Zn|Zn^(2+) (0.01M)||Cu^(2+) (0.01M)|Cu is

E^(@)Cu=0.34V,E_(Zn)^(@)=-0.76V. A Daniell cell contains 0.1M ZnSO_(4) solution and 0.01 M CuSO_(4) solution at its electrodes. E.M.F. of the cell is

E_(Cu)^(0)=+0.34V and E_(Ag)^(0)=+0.8V" calculate "E_(cell")^(0).

Eclipsed and staggered forms of n-butane are called a pair of:

E_(Cell)^(Theta)=1.1V for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?

E_(cell)^(Theta)=1.1V for Daniel cell. Which of the following expression are correct description of state of equilibrium in this cell ?

E_(cell)=0.78 volt for the following cell. underset(Fe_((s))|Fe_((aq))^(2+))||underset((0.01M)(Cu_((aq))^(2+)|Cu_((s))) E_(Fe//Fe^(2+)(aq))^(@)=0.44V,E_(Cu//Cu^(2+)(aq))^(@)=-0.34V

E_(cell)^(@) is 1.89Vfor the reaction 2Ce^(4+) +Co rarr 2Ce^(3+)+Co^(2+). If E_(Co^(2+)//Co) is -0.28V.What is the value of E_(Ce^(4+)//Ce^(3+))^(@)?

E_(cu)^(0)=0.34 and E_(zn)^(0)=-0.76V. Calculate E_("cell")^(0).

E_("cell")^(@) for the redox reation 2Ag_((aq))^(+)Cu rarr Cu_((aq))^(2+) +2Ag is 0.46V. Calculate the equibilrium constant of the reaction.

E_("Cell") for the given redox reaction is 2.71 V Mg_((s))+Cu_((0.01 M))^(2+)to Mg_((0.011M))^(2+)+Cu_((s)) Calculate E_("cell") for the reaction. Write the direction of flow of current when an external opposite potential applied is: (i) less than 2.71 V and (ii) greater than 2.71 V

E_("cell")^(@) for the redox reaction 2Ag^(+)(aq) + Cu to Cu^(2+)(aq) + 2Ag is 0.46 V. Calculate the equilibrium constant of the reaction.

E_(cell)^(@) for some half cell reaction are given below on the basis of these markthe correct answer I H^(+)(aq)+e^(-)rarr1/2 H_(2)(g),E_(cell)^(-)=0.00 V II2H_(2)O(l)rarrO_(2)(g)+4H^(+)(Aq)+4e^(-),E_(cell)^(-)=1.23 V III 2SO_(4)^(2-)(aq)rarrS_(2)O_(8)^(2-)(aq)+2e^(-)+2e^(-),E_(cell)^(-)=1.96 V

E_(cell) and triangleG^o are related as:triangleG^o=nFE^o,triangleG^o=-nFE^o,triangleG^o=-nFE_(cell)^o,triangleG^o=nFE_(cell)^o=0.

E^(c-) of Cu is + 0.34 V while that of Zn is - 0.76V . Explain.

Ebullioscopy is concerned with

Ebonite is a

Ebullioscopy is concerned with:

Ebonite is

Easily ___________ gases are readily adsorbed by the adsorbent.

Ease of nucleophilic addition in the given compounds is

Easiest to perform in the following is

Ease of sulphonation of alkanes is:

Ease of abstraction of hydrogen is greater when attached to:

EAN rule is followed by the complexes. (i)[Fe(pi-C_5H_5)_2](ii)[Mn_2(CO)_10](iii)[V(CO)_6]^(-)

EAN of the elements (*) are equla in :

EAN of Fe in K_3[Fe(CN)_6] is.

EAN of cobalt is 36 in [Co(NH_(3))_(2)O_(2)(en)Cl]. Knowing that atomic number of Co is 27, O_(2) is present is

EAN of a metal carbonyl M(CO)_(x) is 36. If atomic number of metal M is 26, what is the value of x?

EAN of Cr in [Cr(NH_3)_6]Cl_3 is: