52671 + Interview Questions in Chemistry in Class 11

1.	Neither q nor w is a state function but q + w is a state function.Explain why?
Answer» Solution :Both Q and w are not state FUNCTION but `q + w` is EQUAL to `DELTAU` which is state function.

Discussion

2.	Neils Bohr was the first to explain quantitatively the general features of hydrogen atom structure and its spectrum. Calculate the wave number of radiation due to transition of an electron from fourth orbit to second orbit (R_h=109677cm^-1)
Answer» SOLUTION :`_V^-=R_H[1/(n_1^2)-1/n_2^1]=109677[1/2^2-1/4^2]=109677xx3/16=2.05xx10^6m^(-1)`

Discussion

3.	Negatively charged carbon species having eight electrons in its outermost orbit is called as
Answer» carbonium ion carbanion carbocation both(a) and (C ) SOLUTION :CARBON atom has ONE electron more.

Discussion

4.	Negative soil pollution is
Answer» Reduction is soil productivity DUE to erosion and over use Reduction is soil productivity due to addition of PESTICIDES and industrial wastes Converting FERTILE land into barren land by DUMPING ash sludge and garbage None of the above Solution :Reduction in soilproductivity due to erosion and over use

Discussion

5.	Negative catalyst is that which
Answer» takes the REACTION in BACKWARD direction retards the rate of the reaction promotes the SIDE reactions None of these. Answer :B

Discussion

6.	Negative catalyst for the decomposition of H_2O_2 is
Answer» SILICA `MnO_2` ALUMINA ACETANILIDE ANSWER :D

Discussion

7.	Necessary conditions for spontanity of a reaction ________
Answer» `(G)_("products") lt (G)_("reactants")` `(S)_("products") lt (S)_("reactants")` `[(Delta S)_("reaction") + (Delta S)_("surr")] GT 0` `[- ((Delta H)_("SYS"))/(T ) + (Delta S)_("sys")] gt 0` Solution :Conditions for spotaneity of process are `Delta G lt 0 RARR (Delta H)_(S.T) lt 0 rArr (Delta H)_(S.P) lt 0` `Delta S_("universe") gt 0 rArr` So Ans a, c, d

Discussion

8.	Necessary conditions for Fiels-Alder reactions are :
Answer» `(I),(II),(III)` `(I),(II),(IV)` `(II),(III)` `(II),(IV)` ANSWER :A

Discussion

9.	NCl_(3) is an endothermic compound while NF_(3) is an exothermic compound.
Answer» Solution :DUE to smaller size of F as compared to Cl, the N-F bond is much stronger `(272k kJ MOL^(-1))` than N-Cl bond `(201 kJ mol^(-1))` while bond dissociation energy of `F_(2)(158.8 kJ mol^(-1))` is much lower than that of `Cl_(2)(242.6 kJ mol^(-1))`. Therefore, energy RELEASED during the formation of `NF_(3)` molecule is more than the energy needed to break `N_(2)(941.4 kJ mol^(-1))` and `F_(2)` molecules into individual ATOMS. In other words formation of `NF_(3)` is an exothermic reaction or `NF_(3)` is an exothermic compoud. `{:(N_(2)(g)+3F_(2)(g)to2NF_(3)(g),,DeltaH=-214.2Jmol^(-)),(N_(2)(g)+3Cl_(2)(g)to2NCl_(3)(g),,DeltaH=-463.2kJmol^(-1)):}` In contrast, energy released during the formation of `NCl_(3)` molecule is les than the energy needed to break `N_(2)andCl_(2)` molecules into individual atoms. In other words, formation of `NCl_(3)` is an endothermic reaction or `NCl_(3)` is an endothermic compound.

Discussion

10.	NCl_(3) gets hydrolysed to from NH_(3)andHOCl while PCl_(3) on hydrolysis gives H_(3)PO_(3)andHCl. Explair why ?
Answer» Solution :N does not have d-orbitals to accommodate the ELECTRONS donated by Oof `H_(2)O`, therefore, attack of `H_(2)O` OCCURS on the Cl atom which has d- orbitals to accommodate the extra electrons donated by `H_(2)O` Consequently, Cl-O bond is FORMED LEADING to the FORMATION of `HOClandNH_(3)` as shown below : In contrast, both P and Cl have d- orbitals to accommodate electrons by `H_(2)O`. But P-O bond much stronger than Cl-O bond. As a result, attack of `H_(2)O` molecules occurs preferentially on P of PC to form `H_(3)PO_(3)andHCl`.

Discussion

11.	NCl_(3) gets readily hydrolysed while NF_(3) does not. Why ?
Answer» Solution :In `NCl_(3)`, Cl has vacant d-orbitals to accept the lone PAIR of electrons donated by O-atom of `H_(2)O` molecule but in `NF_(3)`, F does not have d-orbitals. Thus, `NCl_(3)` undergoes hydrolysis but `NF_(3)` does not. `NCl_(3)+3H_(2)OtoNH_(3)+HOCl","NF_(3)+H_(2)Oto` No REACTION

Discussion

12.	(NCERT Examplar Problem) Molecular orbitals are formed by the overlap of atomic orbitals . Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and antibonding molecular orbital (ABMO) . Energy of antibonding orbitals is raised above the parent atomic orbitals that have combined and the energy of the bonding orbitals is lowered than the parent atomic orbitals . Energies of verious molecular orbitals for elements hybrogen to nitrogen increase in the order : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt (pi_(2p_(x)) = pi_(2p_(y))) lt sigma2p_(z) lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) and for oxygen and fluorine , order of energy of molecular orbitals is given below : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt sigma 2p_(x) lt (pi_2p_(x) = pi_2p_(y))lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation . Further , if the overlapping is headon, the molecular orbital is called 'sigma', (sigma) and if the overlap is lateral, the molecular orbitals is called 'pi' , (pi) . The molecular orbitalsare filled with electrons according to the same rules as followed for filling ofatomic orbitals . However, the order for filling is not Whihc of the following statements is correct ?
Answer» In the formation of dioxygen form OXYGEN atoms, 10 molecular orbitals will be formed. All the molecular orbitals in the dioxygen will be completely filled Total number of bonding molecular orbitals will not be same as total number of antibonding orbitals in dioxygen. Number of filled bonding orbitals will be same as number of filled antibonding orbitals Solution :`O_(2)` `sigma_((1s)/(1))^(2) sigma_((1s)/(2))^(2) sigma_((2s)/(3))^(2) sigma_((2s)/(4))^(2) sigma_((2p_(z))/(5))^(2) pi_((2p_(X))/(6))^(2)pi_((2p_(y))/(7))^(2)pi_((2p_(x))/(8))^(1)pi_((2p_(y))/(9))^(1)pi_((2p_(z))/(10))^(**0)`

Discussion

13.	(NCERT Examplar Problem) Molecular orbitals are formed by the overlap of atomic orbitals . Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and antibonding molecular orbital (ABMO) . Energy of antibonding orbitals is raised above the parent atomic orbitals that have combined and the energy of the bonding orbitals is lowered than the parent atomic orbitals . Energies of verious molecular orbitals for elements hybrogen to nitrogen increase in the order : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt (pi_(2p_(x)) = pi_(2p_(y))) lt sigma2p_(z) lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) and for oxygen and fluorine , order of energy of molecular orbitals is given below : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt sigma 2p_(x) lt (pi_2p_(x) = pi_2p_(y))lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation . Further , if the overlapping is headon, the molecular orbital is called 'sigma', (sigma) and if the overlap is lateral, the molecular orbitals is called 'pi' , (pi) . The molecular orbitalsare filled with electrons according to the same rules as followed for filling ofatomic orbitals . However, the order for filling is not In whihc of the following moelcular,sigma 2p_(z), molecular orbitals is filled afterpi 2p_(x) and pi 2p_(y) molecular orbitals ?
Answer» `O_(2)` `Ne_(2)` `N_(2)` `F_(2)` Solution :In` N_(2), sigma_(2p_(z))` is filled after `pi_(2p_(X)) and pi_(2p_(y))`

Discussion

14.	(NCERT Examplar Problem) Molecular orbitals are formed by the overlap of atomic orbitals . Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and antibonding molecular orbital (ABMO) . Energy of antibonding orbitals is raised above the parent atomic orbitals that have combined and the energy of the bonding orbitals is lowered than the parent atomic orbitals . Energies of verious molecular orbitals for elements hybrogen to nitrogen increase in the order : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt (pi_(2p_(x)) = pi_(2p_(y))) lt sigma2p_(z) lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) and for oxygen and fluorine , order of energy of molecular orbitals is given below : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt sigma 2p_(x) lt (pi_2p_(x) = pi_2p_(y))lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation . Further , if the overlapping is headon, the molecular orbital is called 'sigma', (sigma) and if the overlap is lateral, the molecular orbitals is called 'pi' , (pi) . The molecular orbitalsare filled with electrons according to the same rules as followed for filling ofatomic orbitals . However, the order for filling is not Which of the followingpair is expected to have the same bond order ?
Answer» `O_(2), N_(2)` `O_(2)^(+), N_(2)^(-)` `O_(2)^(-),N_(2)^(+)` `O_(2)^(-) , N_(2)^(-)` Solution :BOND order are : `O_(2) = 2, , N_(2) = 3 , O_(2)^(+) = 2.5` , `N_(2)^(+) = 2.5 , O_(2)^(-) = 1.5, N_(2)^(-) = 2.5`

Discussion

15.	(NCERT Examplar Problem) Molecular orbitals are formed by the overlap of atomic orbitals . Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and antibonding molecular orbital (ABMO) . Energy of antibonding orbitals is raised above the parent atomic orbitals that have combined and the energy of the bonding orbitals is lowered than the parent atomic orbitals . Energies of verious molecular orbitals for elements hybrogen to nitrogen increase in the order : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt (pi_(2p_(x)) = pi_(2p_(y))) lt sigma2p_(z) lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) and for oxygen and fluorine , order of energy of molecular orbitals is given below : sigma 1s lt sigma ^()1s lt sigma 2s lt sigma^()2s lt sigma 2p_(x) lt (pi_2p_(x) = pi_2p_(y))lt (pi^()2p_(x) = pi^() 2p_(y)) lt sigma^() 2p_(z) Different atomic orbitals of one atom combine with those atomic orbitals of the second atom which have comparable energies and proper orientation . Further , if the overlapping is headon, the molecular orbital is called 'sigma', (sigma) and if the overlap is lateral, the molecular orbitals is called 'pi' , (pi) . The molecular orbitalsare filled with electrons according to the same rules as followed for filling ofatomic orbitals . However, the order for filling is not Which of the following molecular orbitals has maximum number of nodal planes ?
Answer» `sigma^()1S` `sigma ^()2p_(z)` `pi_(2p_(x))` `PI^()2p_(z)` SOLUTION :Nodal planes are : `sigma_(1s)^()= 1, sigma_(2p_(z))^() = 1 , pi_(2p_(x)) = 1, pi_(2p_(y))^() = 2 `

Discussion

16.	NBS is a specific reagent for
Answer» NUCLEOPHILIC SUBSTITUTION reaction Electrophilic substitution Electrophilic addition ALLYLIC substitution Solution :Allylic substitution

Discussion

17.	Nbenzeen reacts with hydrogen in the presence of pt to yield
Answer» SOLUTION :CYCLOHEXANE

Discussion

18.	Nature of Sb_4O_6 is
Answer» ACIDIC NEUTRAL BASIC AMPHOTERIC ANSWER :D

Discussion

19.	Nature of CO_(2) and SiO_(2) are respectively
Answer» ACIDIC, Basic Basic, Basic Acidic, Acidic Basic, Acidic Answer :C

Discussion

20.	The nature of 0.1 M solution of sodium bisulphate is
Answer» Acidic Alkaline NEUTRAL AMPHOTERIC SOLUTION :`NaHSO_4` is an acid salt and gives ` H^(+) `

Discussion

21.	Natural sources of air pollution are ?
Answer» Volcanic eruptions Forest FIRES VEGETATION decay Automobile exhausts ANSWER :A::B::C

Discussion

22.	Natural abundance of heavy water in water is 1: 6000. How many heavy water molecules are present in one drop of water ? (one mL water is 20 drops)
Answer» Solution :ONE mL water =one gram (d=1 g per mL) weight of one drop of water `=(1)/(20)g` weight of `D_(2)O` in one drop of water `=(1)/(20 xx 6000)g` Number of moles of `D_(2)O =("weight")/(GMW)=(1)/(20 xx 6000 xx 20)` Number of `D_(2)O` molecules =Number of moles `xx N_(A)` `=(6.022 xx 10^(23))/(20 xx 6000 xx 20)=2.51 xx 10^(17)` Number of HEAVY water molecules is `2.51 xx 10^(17)`

Discussion

23.	Native silver metal forms a water soluble complex with a dilute solution of NaCN in the presence of
Answer» nitrogen oxygen carbon dioxide argon Solution :N//A

Discussion

24.	Nassler's reagent is :
Answer» `NaHgCl_(4)` `K_(2)HgI_(4)` `Hg(NH_(3))_(2)Cl` `K_(2)HgI_(4) + KOH` SOLUTION :It is the correct answer :

Discussion

25.	Nascent hydrogen consists of.....
Answer» Hydrogen ATOMS with excess energy Hydrogen molecules with excess energy Hydrogen IONS in the EXCITED state Solvated protons Answer :B

Discussion

26.	Nascent hydrogen consists of
Answer» HYDROGEN ions in the EXCITED state Hydrogen molecules with exceis energy Salvated protoms Hydrogen atoms with EXCESS energy Solution :Part of energy preduced in the reaction is absorbed by newly born hydrogen atoms.

Discussion

27.	Nascent hydrogen consists of:
Answer» Hydrogen atoms with excess energy Hydrogenmolecules with excess energy Hydrogen IONS with in the EXCITED state Solvated protons Solution :(a) Nascent HYDROEN [H] consists of hydrogen atoms with excess energy

Discussion

28.	Napthalene is an aromatic compound justify the statement using Huckle rule.
Answer» <P> Solution :ACCORDING to Hucle rule, if NAPHTHALEIN is an aromic compound it should posses `(4n+2)pi` electrons delocalized in the RING. In naphthalein `n=2` Number of delocalized p electrons `=(4n+2)=4xx2+2` `=10pi` electrons Hence naphthalein is an aromatic compound.

Discussion

29.	Napthalene is a/an
Answer» Ionic SOLID Covalent solid Metallic solid Meolcular solid Solution :Napthalene is a molecular solid. If the crystals CONTAINS only individuals atoms, as in solid ARGON or krypton or if they are composed of non polar MOLECULES as in napthalene, the only attraction between the molecules are the LONDON forces.

Discussion

30.	Naphthalene is an aromatic compound.
Answer» ANSWER :T

Discussion

31.	Naphthalene is a volatile solid. It is best purified by :
Answer» crystallisation distillation steam distillation sublimation Answer :D

Discussion

32.	NaOH(Solid)+Cooverset(200^(@)C)rarrX, product X is :
Answer» `NaHCO_(3)` `Na_(2)CO_(3)` `HCOONa` `H_(2)CO_(3)` Solution :N//A

Discussion

33.	NaOH_((aq)) + HCl_((aq)) rarr NaCl_((aq)) + H_(2)O, Delta H= -13.6 Kcal. H_(2(g)) + (1)/(2) O_(2(g)) rarr H_(2)O , Delta H = - 68 Kcal. What is hat of formation of OH^(-) ?
Answer» `-54.4` KCAL `+54.4` kCal `+13.6` kCal `-13.6` kCal Solution :`NaOH + HCL rarr NACL + H_(2)O` can be written as `OH^(-) + H^(+) rarr H_(2)O, Delta H = -13.6` `Delta H = Delta H_(f_(H_(2)O)) - Delta H_(f_(OH^(-))) rArr - 13.6 = - 68- Delta H_(f_(OH^(-)))` `Deta H_(f_(OH^(-))) = -54.4` Kcal

Discussion

34.	NaOH is least soluble in
Answer» `H_(2)O` ETHANOL `C Cl_(4)` Dil. HCl Answer :C

Discussion

35.	NaOH liberates NH_(3) with
Answer» `NH_(4)Cl` `(NH_(4))_(2)SO_(4)` `(NH_(4))_(2)CO_(3)` All the above Answer :D

Discussion

36.	NaOH exhibits disproportionation reaction with
Answer» `AL` `HCL` `H_(2)SO_(4)` `Cl_(2)` ANSWER :D

Discussion

37.	NaOH+D_(2)OrarrNaOD+HDO is known as
Answer» EXCHANGE reaction Deuterolysis reaction Hydrolysis reaction Softening reaction Answer :A

Discussion

38.	NaOH+D_(2)O rarr NaOD+HDO is known as
Answer» EXCHANGE reaction Deuterolysis reaction Hydrolysis reaction Softening reaction Answer :A

Discussion

39.	NaOH +D_(2) to NaOD+HDO is known as
Answer» EXCHANGE reaction Deuterolysis reaction Hydrolysis reaction Softening of HARD water Solution :Exchange of ..H.. with ..D..

Discussion

40.	NaOH+Counderset(5-10 atm)overset(200^(@)C)rarrA. The product A is
Answer» HCOONa `Na_(2)CO_(3)` `NaHCO_(3)` `H_(2)CO_(3)` Solution :`NaOH+Counderset(5-10 ATM)overset(200^(@)C)rarrHCOONa`

Discussion

41.	NaOH and Na_(2)CO_(3) are dissolved in 200ml aqueous solution. Now methyl orange is added in the same solution titrated and requires 2.5ml of the same HCl. Calculate the normality of NaOH & NaCO_(3).
Answer» `(0.5)/(200),(1.5)/(200)` `(1.5)/(200),(0.5)/(200)` `(0.5)/(200).(0.5)/(200)` `(1.5)/(200),(1.5)/(200)` Solution :Let M moles of NaOH = a, `Na_(2)CO_(3)=B` With phenolphthalein, `eq_(HCl)` = `eq_(NaOH)+(1)/(2)eqNa_(2)CO_(3)` `17.5xx0.1=axx1+bxx1` With methyl organe, `eq_(HCl)=(1)/(2)eqNa_(2)CO_(3)` `2.5xx0.1=bxx1` `implies a=1.5,b=0.25` implies `[NaOH]=(1.5)/(200)M=(1.5)/(200)N` `[Na_(2)CO_(3)]=(0.25)/(200)M=(0.5)/(200)N`

Discussion

42.	NaNO_(3) overset(500^(@)C)rarr (A)+(B) (A) overset(800^(@)C)rarr (C)+(D)+(E) Find the number of correct statements: (a) Compound (B) is paramagnetic in nature (b) Compound (B) when undergoes dimerisation the dimer product is diamagnetic in nature (c) Bond order of compound (B) is two (d) (D) is N_(2) gas (e) Compound (B) and (E) are same gas
Answer» SOLUTION :(A) `rarr NaNO_(2)` (B) `rarr O_(2)` (C) `rarr Na_(2)O` (D) `rarr N_(2)` (E) `O_(2)`

Discussion

43.	NaNO_(3)overset(500^(@)C)rarrA+X_((g)) , A+NH_(2)CONH_(2)+H^(+)rarrY_((g))+z_((g))+H_(2)O. What are the gases X, Y and Z ?
Answer» Solution :`2NaNO_(3)overset(500^(@)C)rarr2NaNO_(2)+O_(2)` `2NaNO_(2)+NH_(2)CONH_(2)+2H^(+)rarr2N_(2)+CO_(2)+3H_(2)O+2Na^(+)` X is `O_(2)` , Y is `N_(2)` and Z is `CO_(2)`

Discussion

44.	Names of some compounds are given. Which one is not in IUPAC system ?
Answer» `{:(""CH_(3)),("\|"),(CH_(3)-CH_(2)-CH_(2)-CH-CH-CH_(2)CH_(3)),("\|"),(""CH_(2)CH_(3)),("3-Methyl-4-ethylheptane"):}` `{:(CH_(3)-CH-CH-CH_(3)),("\|""\|"),(""OH""CH_(3)),("3-Methyl-2-butanol"):}` `{:(CH_(3)-CH_(2)-CH-CH-CH_(3)),("\|\|""\|"),(""CH_(2)" "CH_(3)),(" 2-Ethyl-3-methylbut-1-ene"):}` `underset("4-Methyl-2-pentyne")(CH_(3)-C-=C-CH(CH_(3))_(2))` Solution :The means of the ALKYL subtituents are not in alphabetical order. The CORRECT NAME is 4-ethyl 3-methylheptane.

Discussion

45.	Name unknown group of elements at the time of Mendeleef
Answer» ANSWER :INERT GASES

Discussion

46.	Name twophenomena that can be explained on the basis of surface tension.
Answer» SOLUTION :Two phenomena that can be explained on the basis of surface TENSION are as FOLLOWING. (i) Rise or FALL of the LIQUID in a capillary (capillary action). (ii) Spherical shape of small liquid drops.

Discussion

47.	Name two types of hydrogen bonding.
Answer» SOLUTION :Inter molecular and INTRA molecular HYDROGEN BOND.

Discussion

48.	Name two solvents which are commonly used to dissolve organic solids.
Answer» SOLUTION :ETHYL ALCOHOL and ANHYDROUS ETHER.

Discussion

49.	Name two phenomena that can be explained on the basis of surface tension.
Answer» Solution :Surface tension can explain (i) CAPILLARY ACTION, i.e., RISE or FALL of a liquid in a capillary, (ii) spherical shape of small liquid drops.

Discussion

50.	Name two methods which can be safely used to purify aniline.
Answer» SOLUTION :VACUUM DISTILLATION and STEAM distillation.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Neither q nor w is a state function but q + w is a state function.Explain why?

Neils Bohr was the first to explain quantitatively the general features of hydrogen atom structure and its spectrum. Calculate the wave number of radiation due to transition of an electron from fourth orbit to second orbit (R_h=109677cm^-1)

Negatively charged carbon species having eight electrons in its outermost orbit is called as

Negative soil pollution is

Negative catalyst is that which

Negative catalyst for the decomposition of H_2O_2 is

Necessary conditions for spontanity of a reaction ________

Necessary conditions for Fiels-Alder reactions are :

NCl_(3) is an endothermic compound while NF_(3) is an exothermic compound.

NCl_(3) gets hydrolysed to from NH_(3)andHOCl while PCl_(3) on hydrolysis gives H_(3)PO_(3)andHCl. Explair why ?

NCl_(3) gets readily hydrolysed while NF_(3) does not. Why ?

NBS is a specific reagent for

Nbenzeen reacts with hydrogen in the presence of pt to yield

Nature of Sb_4O_6 is

Nature of CO_(2) and SiO_(2) are respectively

The nature of 0.1 M solution of sodium bisulphate is

Natural sources of air pollution are ?

Natural abundance of heavy water in water is 1: 6000. How many heavy water molecules are present in one drop of water ? (one mL water is 20 drops)

Native silver metal forms a water soluble complex with a dilute solution of NaCN in the presence of

Nassler's reagent is :

Nascent hydrogen consists of.....

Nascent hydrogen consists of

Nascent hydrogen consists of:

Napthalene is an aromatic compound justify the statement using Huckle rule.

Napthalene is a/an

Naphthalene is an aromatic compound.

Naphthalene is a volatile solid. It is best purified by :

NaOH(Solid)+Cooverset(200^(@)C)rarrX, product X is :

NaOH_((aq)) + HCl_((aq)) rarr NaCl_((aq)) + H_(2)O, Delta H= -13.6 Kcal. H_(2(g)) + (1)/(2) O_(2(g)) rarr H_(2)O , Delta H = - 68 Kcal. What is hat of formation of OH^(-) ?

NaOH is least soluble in

NaOH liberates NH_(3) with

NaOH exhibits disproportionation reaction with

NaOH+D_(2)OrarrNaOD+HDO is known as

NaOH+D_(2)O rarr NaOD+HDO is known as

NaOH +D_(2) to NaOD+HDO is known as

NaOH+Counderset(5-10 atm)overset(200^(@)C)rarrA. The product A is

NaOH and Na_(2)CO_(3) are dissolved in 200ml aqueous solution. Now methyl orange is added in the same solution titrated and requires 2.5ml of the same HCl. Calculate the normality of NaOH & NaCO_(3).

NaNO_(3)overset(500^(@)C)rarrA+X_((g)) , A+NH_(2)CONH_(2)+H^(+)rarrY_((g))+z_((g))+H_(2)O. What are the gases X, Y and Z ?

Names of some compounds are given. Which one is not in IUPAC system ?

Name unknown group of elements at the time of Mendeleef

Name twophenomena that can be explained on the basis of surface tension.

Name two types of hydrogen bonding.

Name two solvents which are commonly used to dissolve organic solids.

Name two phenomena that can be explained on the basis of surface tension.

Name two methods which can be safely used to purify aniline.