94420 + Interview Questions in Chemistry in Class 12

1.	In the isolation of fluorine a number of difficulties were encountered which statement is correct
Answer» Fluorine reacts with moist GLASS vessels Fluorine has great AFFINITY for hydrogen Electrolysis of aqueous HF gives ozonised oxygen The POTENTIAL required for the discharge of the fluoride IONS is the lowest Answer :D

Discussion

2.	In the isolation of fluorine, a number of different of difficulties were encountered. Which statement is correct
Answer» The potential required for the discharge of the FLUORIDE ions is the LOWEST Fluorine reacts with most glass vessels fluorine has great affinity for hydrogen Electrolysis of AQUEOUS HF gives ozonised oxygen ANSWER :A

Discussion

3.	In the following groups (1) -OA c (2) -OM e (3) -OSO_(2)Me (4) -OSO_(2)CF_(3) the order of leaving group ability is
Answer» `1gt2gt3gt4` `4gt3gt1gt2` `3gt2gt1gt4` `2gt3gt4gt1` ANSWER :B

Discussion

4.	In the isoelectronic speciesthe ionic radii (Å) of N^(3-), O^(2-) and F^(-) are respectively given by
Answer» 1, 36, 1.40, 1.71 1.36, 1.71, 1.40 1.71, 1.40, 1.36 1.71, 1.36, 1.40 Answer :C

Discussion

5.	In the isoelectronic series of metal carbonyl, the CO bond strength is expected to increase in the order:
Answer» `[MN(CO)_(6)]^(+) lt [Cr(CN)_(6)] lt [V(CO)_(6)]^(-)` `[V(CO)_(6)]^(-) lt [Cr(CO)_(6)] lt [Mn(CO)_(6)]^(+)` `[V(CO)_(6)]^(-) lt [Mn(CO)_(6)]^(+) lt [Cr(CO)_(6)]` `[Cr(CO)_(6)] lt [Mn(CO)_(6)]^(+) lt [V(CO)_(6)]^(-)` SOLUTION :CO bond strength is reciprocal to the extent of back DONATION involved in synergic bonding i.e., `M hArr CO`

Discussion

6.	In the following groups,
Answer» I`gt `II` gt `III `gt`IV IV` gt` III` gt` I` gt` II III` gt `II` gt` I `gt` IV II` gt` III` gt` IV` gt` I Solution :F

Discussion

7.	In the isoelectric species the ionic radii (Å) of N^(3-) O^(2-) and F^(-) are respectively,
Answer» `1.36, 1.40, 1.71` `1.36, 1.71, 1.40` `1.71, 1.40, 1.36` `1.71, 1.36, .40` Solution :IONIC RADII DECREAE in the isoelectronic series as `N^(3-)gtO^(2-)gtF^(-)`

Discussion

8.	In the following gaseous phase first order reaction A_((g))to2B_((g))+C_((g)) Initial pressure was found to be 400mm and it changed to 1000mm after 20 minutes. Calculate the rate.
Answer» Solution :`{:(A_((g)),,to,,2B_((g)),,+,,C_((g))),(P_(i),,,,0,,,,0),((P_(1)-X),,,,2x,,,,x):}` `P_(t)=P_(i)-x+2x+x=(P_(i)+2x)` `:.x=(P_(t)-P_(i))/(2)` `:.(a-x)=P_(i)-((P_(t)-P_(i)))/(2)=(3P_(i)-P_(t))/(2)` `=(3xx400-1000)/(2)=100mm` `K=(2.303)/(t)LOG((a)/(a-x))` `=(2.303)/(20)log.(400)/(100)=0.0693min^(-1)`

Discussion

9.	In the ionic equation for the reaction : IO_(3)^(-)+6H^(+)+ae^(-)toI^(-)+3H_(2)Othe value of a is :
Answer» 2 4 6 10 Answer :C

Discussion

10.	In the following gaseous phase first order reaction A(g) rarr 2B(g)+C(g) initial pressure was found to be 400 mmof Hgand it changed to 1000 mm of Hg after 20min. Then
Answer» half life for A is 10 MIN rate constant is 0.0693 `"min"^(-1)` partial PRESSURE of C at 30 min is 350 mm of Hg total pressure after 30 min is 1100 mm of Hg Answer :A::B::C::D

Discussion

11.	In the iodometric estimation in laboratory, which process is involved
Answer» `Cr_(2)O_(7)^(2-)+H^(+)+I^(-)to2Cr^(3)+I_(2),I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)` `MnO_(4)^(-)+H^(+)+I^(-)toMnO_(2)+I_(2)` `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)` `MnO_(4)^(-)+OH^(-)+I^(-)toMnO_(2)+I_(2)` `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)` `Cr_(2)O_(7)^(2-)+OH^(-)+I^(-)to2Cr^(3+)+I_(2)` `I_(2)+S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)+I^(-)` ANSWER :B

Discussion

12.	In the following first order competing reactions: A + ReagentrarrProduct B + Reagent rarrProduct The ratio of K_1//K_2 if only 50% of B will have been reacted'when 94% of A has been reacted is:
Answer» 4.06 0.246 2.06 0.06 Answer :A

Discussion

13.	In the ionic equation - BiO_(3)^(-) + 6H^(+) + Xe^(-) → Bi^(3+) + 3H_(2)O the values of x is
Answer» 6 2 4 3 Solution :`BiO_(3)^(-) + 6H^(+) + xe^(-) RIGHTARROW Bi^(3+) + 3H_(2)O` By electroneutrality principle, `-1 + 6 + X(-1) = 3 + 0` or `+5 + x(-1) = 3 Rightarrow x = 2`

Discussion

14.	In the following figure, when the two stopcocks are opened, the total pressure inside the flask will be
Answer» 1.41 atm 2.41 atm 3.41 atm 1.12 atm Answer :A

Discussion

15.	In the ideal gas equation, the gas constant R has the units :
Answer» litre `MOL^(-1)` erg `mol^(-1)` litre atm `mol^(-1)` `K^(-1)` ml atm `mol^(-1) K` Answer :C

Discussion

16.	In the following first order competing reactions: A + ReagentrarrProduct B + Reagent rarrProduct The ratio ofK_1/ K_2 if only 50% of B will have been reacted.when 94% of A has been reacted is:
Answer» 4.06 0.246 2.06 0.06 Answer :A

Discussion

17.	In the hydrolytic equilibrium A^- + H_2O hArr HA + OH^-K_a = 1.0 xx 10^(-5) . The degree of hysrolysis of 0.001 M solution of the salt is :
Answer» `10^(-3)` `10^(-4)` `10^(-5)` `10^(-6)` ANSWER :A

Discussion

18.	In the following fields, how adsorption is applied? (i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer» Solution : (i) Medicine: - DRUGS cure diseases by adsorption on body tissues. (ii) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ore particles are adsorbed by PINE oil. (iii) Mordants and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the substances USED for fixing dyes onto the fabric. (iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after GETTING absorbed on PRECIPITATE. It is used to indicate the end point of filtration.

Discussion

19.	In the hydrolytic equilibrium , A^(-) + H_(2) O hArr HA + OH^(-) K_a = 1 .0 xx 10^(-5) . The degree of hydrolysis of 0.001 M solution of the salt is
Answer» `10^(-6)` `10^(-4)` `10^(-5)` `10^(-3)` ANSWER :D

Discussion

20.	In the following equilibrium N_(2)O_(4)(g) rarr 2NO_(2)(g)When 5 moles of each is taken and the temperature is kept at 298 k, the total pressure was found to be 20 barGiven : DeltaG^(@)_(f)(N_(2)O_(4) = 100 KJ)DeltaG^(@)_(f)(NO_(2)) = 50 KJFind DeltaG^(@) of the reaction at 298 K
Answer» <P>`-4.68 KJ` `-6.04 KJ` `-5.705 KJ` `0.4 KJ` Solution :REACTION quotient = `p[NO_(2)]^(2)/p[N_(2)O_(4)] = 100/10 = 10` `DeltaG^(@) = 2DeltaG^(@)_(f)(NO_(2))-DeltaG^(@)_(f)(N_(2)O_(4)) = 2 xx 50 - 100 = 0` `DeltaG = DeltaG^(@) - 2.303RT log_(10)Q_(p) = 0 - 2.303 xx 8.314 xx 298 log_(10)10 = -5705.8 J = -5.705 KJ`

Discussion

21.	In the hydrolysis of an organic chloride in presence of large excess of water,RCl+H_2OrarrROH+HCl
Answer» MOLECULARITY and order of REACTION both are 2 molecularity is 2 but order of reaction is 1 molecularity is 1 but order of reaction is 2 molecularity is 1 and order of reaction is also 1. Solution :SINCE water is present in excess, the reaction is pseudo unimolecular.

Discussion

22.	In the following fields, how adsorption is applied? (i) Medicine(ii) Metallurgy (iii) Mordant & Dyes (iv) indicators
Answer» Solution :(i) Medicine :- Drugs cure diseases by adsorption on BODY tissues. (II) Metallurgy:- Sulphide ores are concentrated by a process called froth floation in which lighter ORE particles are adsorbed by pine oil. (iii) MORDANT and Dyes:- Most of the dyes are adsorbed on the surface of the fabric. Mordants are the sustances used for fixing dyes onto the fabric. (iv) In the precipitation titrations, the end point is indicated by an external indicator which changes its colour after GETTING absorbed on precipitate. It is used to indicate the end point of filtration.

Discussion

23.	In the hydrogenation of the following reaction the state of carbon changes from : C_(2)H_(2)toC_(2)H_(4)toC_(2)H_(6)
Answer» `SP-sp^(2)-sp^(3)` `sp^(3)-sp^(2)-sp` `sp^(2)-sp-sp^(3)` `sp^(3)-sp-sp^(2)` Solution :`C_(2)H_(2)(sp)toC_(2)H_(4)(sp^(2))toC_(2)H_(6)(sp^(3))`

Discussion

24.	In the following equilibrium, N_2O_4(g)to2NO_2(g) when 5 moles of each is taken and the temperature is keptat 298K, the total pressure was found to be 20 bar. Given : /_\G_f^0(N_2O_4)=100kJ,/_\G_f^0(NO_2)=50kJ Find the direction of the reaction.
Answer» SOLUTION :`prop=SQRT((K_a)/(C))=sqrt((1.8xx10^(-5))/(0.001))=0.1341` i.e., 13.41%

Discussion

25.	In the following equilibrium , N_2O_4(g) hArr 2NO_2(g) When 5 moles of each are taken , the temperature is kept at 298 K the total pressure was found to be 20 bar . Given that DeltaG_f^@(N_2O_4)=100 KJ DeltaG_f^@(NO_2)=50 kJ (i) Find DeltaG of the reaction. (ii) in which direction the equilibrium of reactionwill shift ?
Answer» Answer :(i) 56.0304 Lt ATM (ii) therefore, reaction will shift TOWARDS BACKWARD DIRECTION

Discussion

26.	In thehydroboration- oxidationreactionof propanewithdibrane , H_(2)O_(2)and NaOH, theorganiccompoundformedis :
Answer» `CH_(3)CH_(2)OH` `CH_(3)CHOHCH_(3)` `CH_(3)CH_(2)CH_(2)OH` `(CH_(3))_(3)COH` Solution :`CH_(3)CH = CH_(2)overset (B_(2)H_(6)) underset (H_(2)O_(2),NaOH) toCH_(3)CH_(2)CH_(2)OH`

Discussion

27.	In the following equilibrium, N_2O_4(g)to2NO_2(g) when 5 moles of each is taken and the temperature is keptat 298K, the total pressure was found to be 20 bar. Given : /_\G_f^0(N_2O_4)=100kJ,/_\G_f^0(NO_2)=50kJ Find /_\G for the reaction at 298K
Answer» <P> Solution :For the given reaction, `/_\G^0=2/_\G_f(NO_2)^0-/_\G_(f(N_2O_4)^0` `=(2xx50-100)kJ=0` We know, `/_\G^0=-RTinK_p` or `0=-RTInK_p` `:.K_p=1` In the mixture, total NUMBER of MOLES of `N_2O_4` and `NO_2=(5+5)mol=10mol` So, in the mixture, `p_(N_2O_4)=5/(10)xx20"bar"=10"bar"` `p_(NO_2)=5/(10)xx20"bar"=10"bar"` `:.Q_p=(p_(NO_2)^2)/(p_(N_2O_4)^2)=((10^2)/(10))`bar=10bar `Q_pgtK_p`, reaction will occur to a greater extent towards LEFT.

Discussion

28.	In the hydration of an alkene carbocation is formed from :
Answer» carbanion oxonium ION hydroxide ion HYDRIDE ion ANSWER :B

Discussion

29.	In the following elimination reaction, hybridisation of carbon atom to which halogen is attached changes from
Answer» `sp^2` to `sp^3` `sp^3` to `sp^2` `sp^2` to `sp^2` `sp^3` to `sp^3` ANSWER :B

Discussion

30.	In the hydrides of VI A group elements, the acidic strength gradually increases from top to bottom. This is due to
Answer» decreasesin the EN of the CHALCOGENS increase in their Ka values increase in the METALLIC strengthof chalcogen increase in the m.p. of chalcogen ANSWER :B

Discussion

31.	In the hydrides of VIA elements largest bond angle and bond length is observed respectively in
Answer» `H_(2)O,H_(2)O` `H_(2)Po, H_(2)O` `H_(2)O, H_(2)Po` `H_(2)S, H_(2)SE` ANSWER :C

Discussion

32.	In the following dehydration of diol with H_(3)PO_(4) following product is formed such that isotopic .^(18)O goes H_(2)O. Explain.
Answer» SOLUTION :N//A

Discussion

33.	In the following electron dot structure, calculate the formal charge from left ot righ nitrogen atom. underset(..)overset(..)(N)=N=underset(..)overset(..)(N)
Answer» `-1,-1,+1` `-1,+1,-1` `+1,-1,-1` `+1,-1,+1` Solution :FORMAL charge on an atom- total no. of VALENCE electron in FREE atom- total no. of non bonding electrons `-1/2` total no. of bonding electrons `underset(1)underset(..)OVERSET(..)(N)=underset(2)(N)=underset(3)underset(..)overset(..)(N)` `N_(1)=5-4-1/2xx4=5-6=-1` `N_(2)=5-0-1/xx8=5-4=+1` `N_(3)=5-4-1/2xx4=-1`

Discussion

34.	In the hydrides of group 16th elements the bond angles decrease in the order, H_2O, H_2S, H_2Se, H_2Te. How would you account for this
Answer» SOLUTION :`80.19%`

Discussion

35.	In the following electrochemical cell : Zn(s)abs(zn^(+2)) abs(H^(+))Pt(H_(2)),E^(o)""_(cell) = E_(cell) This will be true when :
Answer» <P>`[Zn^(2+)]=[H^(+)]=1 M and P_(H_(2))= 1 atm` `[Zn^(2+)]=0.01 M,[H^(+)]=0.1 M and P_(H_(2))= 1 atm` `[Zn^(2+)]=1 M,[H^(+)]=0.1 M and P_(H_(2))= 0.01 atm` all of the above are TRUE ANSWER :D

Discussion

36.	In the hydrated choride of magnesium MgCl_(2).(H_(2)O)_(x)then 'x' is
Answer» ANSWER :6

Discussion

37.	In the following electrochemical cell Zn underset(Ө)(\|)Zn^(2+)\|\|H^(+)underset(o+)(\|)Pt(H_(2)) E_(cell)^(o)=E_(cell), this will be when:
Answer» `[Zn^(2+)]=[H^(+)]=1M and pH_(2)=1atm` `[Zn^(2+)]=0.01M,[H^(+)]=0.1M and pH_(2)=1atm` `[Zn^(2+)]=1M,[H^(+)]=0.1M and pH_(2)=0.1atm` `[Zn^(2+)]=[H^(+)]=0.1M and pH_(2)=0.1atm` Answer :A::B::C::D

Discussion

38.	In the hydrated chloride (MgCl_(2)xH_(2)O) of Me, the value of x'is
Answer» Solution :Carnllite `-KCI MG Cl_(2).6H_(2)O`

Discussion

39.	In the following conversion Which of the following reagents is suitable ?
Answer» `NH_(2)NH_(2),KOH,DMSO` `NaBH_(4)` `Zn,Hg` concentrated `H_(2)SO_(4)` `LiAlH_(4)` Solution : ACETAL is hydrolysed in acidic MEDIUM so clemmensen REDUCTION is not used.

Discussion

40.	In the Hofmann 's method for separation of 1^(@), 2^(@) and 3^(@)amines, the reagent used is
Answer» ACETYL chloride benzene SULPHONYL chloride diethyl oxalate nitrous acid. Answer :C

Discussion

41.	In the following concentration cell Ag(s) // AgCl "(saturated)"// // AgNO_3(aq) (0.1M)//Ag_((s)), K_(SP) " of AgCl"=1 xx 10^(-10)The cell potential will be
Answer» `E_("CELL") = 0.295V ` `E_("cell")=0.236 V ` `E_("cell")=(0.059)/(1)log""([Ag^(+)]_("cathode"))/(sqrt(K_(SP) " of " AgCl))` `E_("cell")=E_("cell")^(@) + (0.059)/(1)log ""(sqrt(K_(SP) " of AgCl"))/([Ag^(+)]_("cathode"))` Solution :`E=0 - (0.0591)/(1) log""([Ag^(oplus)]_a)/([Ag^(oplus)]_c), E = (-0.0591)/(1) log""(10^(-5))/(10^(-1)) = - (0.0591)/(1) xx (-4) = 0.2364V `

Discussion

42.	In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br_(2) used per mole of amine produced are
Answer» one mole of NAOH and one mole of `Br_(2)` FOUR MOLES of NaOH and two moles `Br_(2)` two moles of NaOH and two moles of `Br_(2)` four moles of NaOH and one mole of `Br_(2)` Solution :`RCONH_(2)+Br_(2)+ 4NaOH to R-NH_(2)+Na_(2)CO_(3)+2NaBr+2H_(2)O`

Discussion

43.	In the following compounds the order of basicly is
Answer» `IVgtIgtIIIgtII` `IIIgtIgtIVgtII` `IIgtIgtIIIgtIV` `IgtIIIgtIIgtIV` ANSWER :D

Discussion

44.	In the Hansging method Magnesia is reduced using.
Answer» SI `CaC_(2)` `FESI` Cocke Answer :D

Discussion

45.	In the following compounds : The order of basicity is
Answer» `IV gt I gt III gt II` `III gt I gt IV gt II` `II gt I gt III gt IV` `I gt III gt II gt IV` Solution :See Ultimate PREPARATORY PACKAGE of this unit.

Discussion

46.	In the following compounds the order of basicity is
Answer» `IV gt I gt II gt II` `III gt I gt IV gt II` `II gt I gt III gt IV` `I gt III gt II gt IV` SOLUTION :Delocalization of 1.p on .N.-atom `UARR b.c darr`

Discussion

47.	In the Hanging method of reduction of Magnesia with carbon, the backward reaction is prevented by quenching the mixture of magnesium vapour and cartoon monoxide with
Answer» HCl GAS CO gas `CAF_(2)` `H_(2)` ANSWER :D

Discussion

48.	In the following compounds, the order of acidity is:
Answer» IIIgtIVgtIgtII IgtIVgtIIIgtII IIgtIgtIIIgtIV IVgtIIIgtIgtII Answer :D

Discussion

49.	In the following compounds The order of basicity is
Answer» `IV gt I gt III gt II` `III gt I gt IV gt II` `II gt I gt III gt IV` `I gt III gt II gt IV` Solution :Piperidine (I) is a `2^(@)` alphatic amine. As such it is most basic with `K_(b)=2xx10^(-3)`. Pyrrole (IV) with `K_(b)=2.5xx10^(-14)` is least basic of all the four. In pyrrole, the N is `sp^(3)` hybridised and the lone pair on N is involved in `pi`-electron cloud. Thus is due to non availability of lone pair on N for sharing with acids, it is least basic of the four. In pyridine (II) N is `sp^(2)` hybridised and the UNPAIRED electron in the unhybridised p-orbital is involved in `pi`-electron cloud. As such the lone pair on N PRESENT in `sp^(2)`-hybridised orbital is available for sharing with acids. Thus it is more basic than pyrrole. The value of `K_(b)` for pyridine is `2.3 xx 10^(-9)`. In (III) the lone pair on N is LESS readily available because of electronegative O atom present in the skeleton RING. Thus (III) is less basic than (I).

Discussion

50.	In the following compounds piperidine (I), pyridine (II), morpholine (III) and pyerrole (IV), the order of basicity is :
Answer» `IVgtIgtIIIgtII` `IIIgtIgtIVgtII` `IIgtIgtIIIgtIV` `IgtIIIgtIIgtIV` ANSWER :D

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In the isolation of fluorine a number of difficulties were encountered which statement is correct

In the isolation of fluorine, a number of different of difficulties were encountered. Which statement is correct

In the following groups (1) -OA c (2) -OM e (3) -OSO_(2)Me (4) -OSO_(2)CF_(3) the order of leaving group ability is

In the isoelectronic speciesthe ionic radii (Å) of N^(3-), O^(2-) and F^(-) are respectively given by

In the isoelectronic series of metal carbonyl, the CO bond strength is expected to increase in the order:

In the following groups,

In the isoelectric species the ionic radii (Å) of N^(3-) O^(2-) and F^(-) are respectively,

In the following gaseous phase first order reaction A_((g))to2B_((g))+C_((g)) Initial pressure was found to be 400mm and it changed to 1000mm after 20 minutes. Calculate the rate.

In the ionic equation for the reaction : IO_(3)^(-)+6H^(+)+ae^(-)toI^(-)+3H_(2)Othe value of a is :

In the following gaseous phase first order reaction A(g) rarr 2B(g)+C(g) initial pressure was found to be 400 mmof Hgand it changed to 1000 mm of Hg after 20min. Then

In the iodometric estimation in laboratory, which process is involved

In the following first order competing reactions: A + ReagentrarrProduct B + Reagent rarrProduct The ratio of K_1//K_2 if only 50% of B will have been reacted'when 94% of A has been reacted is:

In the ionic equation - BiO_(3)^(-) + 6H^(+) + Xe^(-) → Bi^(3+) + 3H_(2)O the values of x is

In the following figure, when the two stopcocks are opened, the total pressure inside the flask will be

In the ideal gas equation, the gas constant R has the units :

In the following first order competing reactions: A + ReagentrarrProduct B + Reagent rarrProduct The ratio ofK_1/ K_2 if only 50% of B will have been reacted.when 94% of A has been reacted is:

In the hydrolytic equilibrium A^- + H_2O hArr HA + OH^-K_a = 1.0 xx 10^(-5) . The degree of hysrolysis of 0.001 M solution of the salt is :

In the following fields, how adsorption is applied? (i) Medicine (ii) Metallurgy (iii) Mordant & Dyes (iv) indicators

In the hydrolytic equilibrium , A^(-) + H_(2) O hArr HA + OH^(-) K_a = 1 .0 xx 10^(-5) . The degree of hydrolysis of 0.001 M solution of the salt is

In the following equilibrium N_(2)O_(4)(g) rarr 2NO_(2)(g)When 5 moles of each is taken and the temperature is kept at 298 k, the total pressure was found to be 20 barGiven : DeltaG^(@)_(f)(N_(2)O_(4) = 100 KJ)DeltaG^(@)_(f)(NO_(2)) = 50 KJFind DeltaG^(@) of the reaction at 298 K

In the hydrolysis of an organic chloride in presence of large excess of water,RCl+H_2OrarrROH+HCl

In the following fields, how adsorption is applied? (i) Medicine(ii) Metallurgy (iii) Mordant & Dyes (iv) indicators

In the hydrogenation of the following reaction the state of carbon changes from : C_(2)H_(2)toC_(2)H_(4)toC_(2)H_(6)

In the following equilibrium, N_2O_4(g)to2NO_2(g) when 5 moles of each is taken and the temperature is keptat 298K, the total pressure was found to be 20 bar. Given : /_\G_f^0(N_2O_4)=100kJ,/_\G_f^0(NO_2)=50kJ Find the direction of the reaction.

In thehydroboration- oxidationreactionof propanewithdibrane , H_(2)O_(2)and NaOH, theorganiccompoundformedis :

In the following equilibrium, N_2O_4(g)to2NO_2(g) when 5 moles of each is taken and the temperature is keptat 298K, the total pressure was found to be 20 bar. Given : /_\G_f^0(N_2O_4)=100kJ,/_\G_f^0(NO_2)=50kJ Find /_\G for the reaction at 298K

In the hydration of an alkene carbocation is formed from :

In the following elimination reaction, hybridisation of carbon atom to which halogen is attached changes from

In the hydrides of VI A group elements, the acidic strength gradually increases from top to bottom. This is due to

In the hydrides of VIA elements largest bond angle and bond length is observed respectively in

In the following dehydration of diol with H_(3)PO_(4) following product is formed such that isotopic .^(18)O goes H_(2)O. Explain.

In the following electron dot structure, calculate the formal charge from left ot righ nitrogen atom. underset(..)overset(..)(N)=N=underset(..)overset(..)(N)

In the hydrides of group 16th elements the bond angles decrease in the order, H_2O, H_2S, H_2Se, H_2Te. How would you account for this

In the following electrochemical cell : Zn(s)abs(zn^(+2)) abs(H^(+))Pt(H_(2)),E^(o)""_(cell) = E_(cell) This will be true when :

In the hydrated choride of magnesium MgCl_(2).(H_(2)O)_(x)then 'x' is

In the following electrochemical cell Zn underset(Ө)(|)Zn^(2+)||H^(+)underset(o+)(|)Pt(H_(2)) E_(cell)^(o)=E_(cell), this will be when:

In the hydrated chloride (MgCl_(2)xH_(2)O) of Me, the value of x'is

In the following conversion Which of the following reagents is suitable ?

In the Hofmann 's method for separation of 1^(@), 2^(@) and 3^(@)amines, the reagent used is

In the following concentration cell Ag(s) // AgCl "(saturated)"// // AgNO_3(aq) (0.1M)//Ag_((s)), K_(SP) " of AgCl"=1 xx 10^(-10)The cell potential will be

In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br_(2) used per mole of amine produced are

In the following compounds the order of basicly is

In the Hansging method Magnesia is reduced using.

In the following compounds : The order of basicity is

In the following compounds the order of basicity is

In the Hanging method of reduction of Magnesia with carbon, the backward reaction is prevented by quenching the mixture of magnesium vapour and cartoon monoxide with

In the following compounds, the order of acidity is:

In the following compounds The order of basicity is

In the following compounds piperidine (I), pyridine (II), morpholine (III) and pyerrole (IV), the order of basicity is :