Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Hydrogenation of C_6H_5CHOHCOOH over Rh-Al_2O_3 catalyst in methanol gives :

Answer»

`C_6H_5CH_2COOH
`C_6H_11CHOHCOOH`
`C_6H_5CHOHCH_2OH`
`C_6H_11CH_2COOH`

ANSWER :B
Discussion
2.

Hydrogenation of benzoyl chloride in the presence of Pd on BaSO_(4) gives

Answer»

benzyl ALCOHOL
BENZOIC acid
benzaldehyde
PHENOL

ANSWER :C
Discussion
3.

Hydrogenation of benzene is done by

Answer»

`NaBH_4`
`H_2/Ni`
HCL
`NaNH_2`

ANSWER :B
Discussion
4.

Hydrogenation of alkenes and alkynes takes place in presence of certain catalysts. In Sabatier Senderen's reaction, the addition of hydrogen takes place I presence of Raney nickel catalyst. Platininm and palladium can also be used as catalyst and used in finley divided state. Experimentally, it is observed that less crowded alkenes adsorb H_(2) with faster rate. Controlled hydrogenation of alkyne H_(2) with faster rate. Controlled hydrognetation of alkyne in presence of Lindlar's catalyst yields cis producti.e.,'cis' alkene. Thus, in presence of Lindlar's catalyst 'syn' addition takes place. The relative rate of hydrogenation follows the order: Non-terminal alkynes are reduced in presence of Na or Li metal dissolved in liquid ammonia. In this reaction, anti-addition of hydrogen results into the trans-product. Powdered nickel is more effective than grannular nickel because:

Answer»

Surface AREA of powdered nickel is maximum
Free VALENCIES are LARGE in number
Powdered nickel increases the activation energy of the reaction.
Powdered nickel increases the intermolecular collision of REACTANT molecules.

Solution :N//A
Discussion
5.

Hydrogenation of alkenes and alkynes takes place in presence of certain catalysts. In Sabatier Senderen's reaction, the addition of hydrogen takes place I presence of Raney nickel catalyst. Platininm and palladium can also be used as catalyst and used in finley divided state. Experimentally, it is observed that less crowded alkenes adsorb H_(2) with faster rate. Controlled hydrogenation of alkyne H_(2) with faster rate. Controlled hydrogenation of alkyne in presence of Lindlar's catalyst yields cis producti.e.,'cis' alkene. Thus, in presence of Lindlar's catalyst 'syn' addition takes place. The relative rate of hydrogenation follows the order: Non-terminal alkynes are reduced in presence of Na or Li metal dissolved in liquid ammonia. In this reaction, anti-addition of hydrogen results into the trans-product. In which of the following cases, the reaction is most exothermic?

Answer»




SOLUTION :N//A
Discussion
6.

Hydrogenwill not reduce

Answer»

heatedcupric OXID
heated ferric oxide
heated STANNIC oxide
heated ALUMINIUM oxide

Answer :D
Discussion
7.

Hydrogen will not reduce :-

Answer»

HEATED CUPRIC oxide
Heated FERRIC oxide
Heated STANNIC oxide
Heated ALUMINIUM oxide

Answer :D
Discussion
8.

Hydrogen sulphide reacts with lead acetate forming a black compound which reacts with H_2O_2 to form another compound. The colour of the compound is:

Answer»

Black
Yellow
White
Pink

Answer :C
Discussion
9.

Hydrogen sulphide is not a group reagent for (basic radical)

Answer»

`2^(ND)` group RADICALS
`3^(RD)` group radicals
`4^(th)` group radicals
`5^(th)` group radicals

Answer :B::D
Discussion
10.

Hydrogen sulphide is important in cation analysis. Explain.

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Solution :Group 2 and Group 4cations are PRECIPITATED as their respective SULPHIDES.
Group 2 cations have lower solubility. Hence they are precipitated in dilute acid medium.
Discussion
11.

Hydrogen sulfide is a monobasic acid in solution.

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SOLUTION :HYDROGEN SULFIDE is a DIBASIC ACID.
Discussion
12.

Hydrogen shows :

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+1 OXIDATION STATE only
-1 oxidation state only
ZERO oxidation state only
+1, -1 and zero oxidation states

Answer :D
Discussion
13.

Hydrogen resembles in many of its properties with:

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ALKALI metals
Halogens
BOTH (A) AND (B)
NONE of these

Answer :C
Discussion
14.

Hydrogen produced in contact with substance which is to be reduced is :

Answer»

ORTHO `H_2`
PARA `H_2`
ACTIVE H
NascentH

Answer :D
Discussion
15.

Hydrogen peroxide solution (20 ml) reacts quantitavely with a solution of KMnO_(4) (20 ml) acidified with dilute H_(2)SO_(4). The same volume of the KMnO_(4) solution is just decolourized by 10 ml of MnSO_(4) in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO_(2). The brown precipitate is dissolved in 10 of 0.2 M sodium oxalate under boiling condition in the presence of dilute H_(2)SO_(4). Write the balanced equations involved in the reactions and calculate the volume strength of H_(2)O_(2).

Answer»

SOLUTION :Meq. Of `MnO_(2) = " Meq. of " Na_(2)C_(2)O_(4)`
` = 10 xx 0.2 xx 2 = 4`
` (C_(2)O_(4)^(2-) to 2CO_(2), ` n- factor = 2)
Millimoles of `MnO_(2) = 4/2 = 2`

or `2MnO_(4)^(-) + 3Mn^(2+) to 5MnO_(2) `(MOLE ratio is reciprocal of n-factor ratio )
`("Millimoles of " MnO_(4)^(-))/("Millimoles of " MnO_(2))= 2/5`
Millimoles of `MnO_(4)^(-) = 2/5 xx ` Millimoles of `MnO_(2)`
` = 2/5 xx 2 = 4/5`
`2KMnO_(4) + 5H_(2)O_(2) + 3H_(2)SO_(4) to 2MnSO_(4) + K_(2)SO_(4) + 8H_(2)O + 5O_(2)`
` ("Millimoles of " H_(2)O_(2))/("Millimoles of " MnO_(4)^(-)) = 5/4 `
` :. ` Millimoles of `H_(2)O_(2) = 5/2 xx 4/5 = 2`
`N_(H_(2)O_(2)) xx 20 = 2 xx 2("n- factor for " H_(2)O_(2) = 2) `
`:. N_(H_(2)O_(2)) = 0.2`
Volume STRENGTH of `H_(2)O_(2) = 5.6 xx N_(H_(2)O_(2))`
` = 5.6 xx 0.2 = 1.12`
Discussion
16.

Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO_(4) (20 mL) acidfied with dilute H_(2)SO_(4).The same volue of theKMnO_(4) solution is just decoulurisedby 10 mL of MnSO_(4) in neutral medium forming a dark brown precipitate of hydrated MnO_(2) . the brownprecipitate isdissolved in 10 mL of 0.2M solution oxalate under boilingcondition in the presence of dilute H_(2)SO_(4).Writethe balancedequationsinvolved in teh reactions and calculate the molarity of H_(2)O_(2) .

Answer»

Solution :m.e of `H_(2)O_(2)` in 20 mL solution
= m.e of 20 mL of `KMnO_(4)`
= m.e of 20 mL of `MnsO_(4)`
= m.e of `MnO_(2)`
= m.e of `Na_(2)C_(2)O_(4)`
` = 0.4 xx 10 ` (NORMALITY = MOLARITY `xx` change in ON)
= 4
EQ. of `H_(2)O_(2)` in 20 mL solution
` = 4/1000`
Moleof `H_(2)O_(2)//20 mL = 4/1000 xx1/2 `
Molarity (mole.L) = `4/2000 xx 1000/20`
`0.1 M `
The reactions involved are ,
`5H_(2)O_(2)+2KMnO_(4)+3H_(2)SO_(4)=2MnSO_(4)+K_(2)SO_(4)+8H_(2)O+5O_(2)`
`2KMnO_(4) +3MnSO_(4) +2H_(2)O = 5MNO_(2) +2H_(2)O `
`MnO_(2) +Na_(2)C_(2)O_(4) +2H_(2)SO_(4) = MnSO_(4) +Na_(2)SO_(4) +CO_(2)+2H_(2)O `
Discussion
17.

Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO_(4) (20mL) acidified with dilute of H_(2)SO_(4). The same volume of the KMnO_(4) solution is just decolourised by 10 mL of MnSO_(4) in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO_2. The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H_(2)SO_(4). Write the balanced equations involved in the reactions and calculate the molarity of H_(2)O_(2).

Answer»


ANSWER :0.1
Discussion
18.

Hydrogen peroxide works as :

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An OXIDANT only
A REDUCTANT only
An ACID only
An oxidant , a reductant and an acid

ANSWER :D
Discussion
19.

Hydrogen peroxide on heating gives

Answer»

HYDROGEN and OXYGEN.
water and oxygen.
`H_4O_2`
HO

ANSWER :B
Discussion
20.

Hydrogen peroxide, povidone - iodine, chlorine compounds, benzalkonium chloride.

Answer»

SOLUTION :CHLORINE COMPOUNDS. It is a DISINFECTANT whereas others are antiseptic.
Discussion
21.

Hydrogen peroxide molecules are

Answer»

diatomic and form `HO_(2)^(-)`ions
diatomic and form `HO^(-)`ions
monoatomic and form `H_(2)^(2-)`ions
monoatomic and form `H^(-)`ions.

Solution :`H_(2)O_(2)` is diatomic and FORMS `H^(+)+HO_(2)^(-)` (HYDRO PEROXIDE ion).
Discussion
22.

Hydrogen peroxide reacts with ethylene to give

Answer»

ETHANE
ethanal
ethylene GLYCOL
ETHANOL

ANSWER :C
Discussion
23.

Hydrogen peroxide linerates iodine from acidified potassium iodide. Identify the oxidation reaction.

Answer»

SOLUTION :`H_(2)O_(2) + H_(2)SO_(4) + 2KI RARR I_(2) + K_(2)SO_(4) + 2H_(2)O`
In this reaction, iodide is oxidised to iodine.
Hydrogen peroxide acts as oxidant in this reaction.
Oxidation reaction is : `2I^(-) rarr 2E^(-) + I_(2)`
Discussion
24.

Hydrogen peroxide is usedas an antiseptic under the name

Answer»

BLEACHING powder
perhydrol
nessler REAGENT
CATECHOL

Answer :B
Discussion
25.

Hydrogen peroxide is now generally prepared on industrial scale by the:

Answer»

Action of `H_2SO_4` on BARIUM peroxide
Action of `H_2SO_4` on sodium peroxide
Electrolysis of `50% H_2SO_4`
BURNING HYDROGEN in excess of oxygen

Answer :C
Discussion
26.

Hydrogen peroxide has a non-planar structure. In gas phase of H_(2)O_(2), dihedral angle is v^(@) and in solid phase at 110 K, its dihedral angle is y^(@). What are x and y ?

Answer»

`90^(@), 90^(@)`
`90.2^(@), 101.5^(@)`
`111.5^(@), 90.2^(@)`
`90.2^(@), 119.5^(@)`

SOLUTION :Structure of `H_(2)O_(2)` is slightly different is GAS phase and CRYSTALLINE phase.
Discussion
27.

Hydrogen peroxide has ----- structure .

Answer»

LINEAR structure
Closed CHAIN structure
Closed book structure
Half open book structure or BENT structure

Answer :D
Discussion
28.

Hydrogen peroxide for the first, time was prepared by:

Answer»

Priestley
Thenard
Gay-Lussac
Bernard

Answer :B
Discussion
29.

Hydrogen peroxide does not

Answer»

liberate iodide from KI
turn TITANIUM SALT yellow
gives silver peroxide with MOIST silver oxide
trun mixture of aniline, `KClO_(3)` and dil. `H_(2)SO_(4)` vciolet

Solution :No such reaction is KNOWN (c).
Discussion
30.

Hydrogen peroxide does not:

Answer»

Liberate iodine from KI
Turn the TITANIUM salt yellow
Give SILVER peroxide with moist silver oxide
Turn the mixture of aniline, `KClO_3` and Dil. `H_2SO_4`

Answer :C
Discussion
31.

Hydrogen peroxide cannot be concentrated easily because

Answer»

it decomposes at its boiling point
it is not miscible with water
it is HIGHLY VOLATILE in nature
it has a very HIGH boiling point.

Solution :N.A
Discussion
32.

Hydrogenperoxide can be prepared from

Answer»

NAOH
`BaO_(2).8H_(2)O`
`CA(OH)_(2)`
`Na_(2)O`

ANSWER :B
Discussion
33.

Hydrogen - oxygen fuel cells are used in space prograins to supply

Answer»

Power
drinking water
oxygen
Both (1) and (2)

ANSWER :D
Discussion
34.

Hydrogen-oxygen fuel cell are used in space-craft to supply

Answer»

POWER for heat and light
Power for pressure
Oxygen
Water

SOLUTION :Fuel-cell CONVERTS the chemical ENERGY into electrical energy.
Discussion
35.

Hydrogen of vegetable ghee at 250^(@)C reduces pressure of H_(2) from 2 atm to 1.2 atm in 30 minutes. The rate of reaction in terms of molarity per secondis

Answer»

`1.03xx10^(-6)`
`1.03xx10^(-5)`
`1.03xx10^(-7)`
`1.03xx10^(-9)`

Solution :`r=(2-1.2)/30=(0.8/(60x30))XX1/(0.0821xx523)=1.03xx10^(-5)`M/sec
Discussion
36.

Hydrogenof acetadehydeare nothighlyacidicmediumexplain .

Answer»

Solution :In weakly ACIDIC MEDIUM, the carbonyl group is protonated. As a result of PRESENCE of +ve charge on carbon, the protonated carbonyl group UNDERGOES nucleophilic attack of hydrazine easily
`[-underset(| )C= O +H^+ hArr - underset(| ) C=overset(+)OH harr- underset(|) C^(+)- underset(* *) overset(**) H]`
In strongly acidic medium, the hydrazine being basic in nature, forms its salt by protonating the unshared pair of electrons on N atom.
`H_2overset(* *) n overset(**)NH_2 overset(H^+ ) hArrH_3^(-) N NH_2`
This salt cannot act as nucleophile. Thus, to CARRY out these reactions, we have to control the pH between 3 and 4.
Discussion
37.

Hydrogen may be prepared by heating a solution of caustic soda with:

Answer»

Mg
Zn
Fe
Ag

Answer :B
Discussion
38.

Hydrogen is not used for ?

Answer»

Manufacture of vegetable ghee
Production of HIGH temperature
As ROCKET FUEL with kerosene
As a reducing agent

Answer :C
Discussion
39.

Hydrogen is not obtained when zinc reacts with:

Answer»

Cold water
Dil.`H_2SO_4`
Dil.`HCL`
HOT `20% NaOH`

Answer :A
Discussion
40.

Hydrogen is evolved by the action of cold dilute HNO_(3) on

Answer»

Al
Mg
Cu
Fe

Answer :B
Discussion
41.

Hydrogen is evolved by the action of cold dilute HNO_3 on :

Answer»

FE
Mn
Cu
Al

Answer :B
Discussion
42.

Hydrogen is a common reductantof organic chemicals, but it isnotwidelyused in metallurgy. Substantiate.

Answer»

Solution :METALLURGICAL reductions are usually high temperature processes. At elivated temperatures, hydrogen may form hydrides withmetals. Metallurgical OPERATIONS in open VESSELS lead to be combination of hydrogen with oxygento form water.
Further, Ellinghamcurve has a positive slopeand is parallel to the CURVES of several metal OXIDES.
Discussion
43.

Hydrogen ion concentration of an aqueous solution is 1 xx 10^(-8) M. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol dm^(-3) is

Answer»

`1 xx 10^(-8)`
`1 xx 10^(-6)`
`2 xx 10^(-10)`
`0.5 xx 10^(-10)`

SOLUTION :`[OH^(-)]` of aqueous solution `=(1xx10^(-14))/(1 xx 10^(-4))`
`=1 xx 10^(-10)M`
When the solution is diluted by adding EQUAL volume of water, then concentration of the `OH^(-)` ions is HALVED.
`i.e., (1)/(2) xx 1xx 10^(-10) =0.5 xx 10^(-10) M`
Discussion
44.

Hydrogen ion concentration of an aqueous solution is 1 xx 10^(-4)M. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol dm^(-3) is

Answer»

`2 XX 10^(-10)`
`0.5 xx 10^(-10)`
`1 xx 10^(-8)`
`1 xx 10^(-6)`

Solution :After dilution with EQUAL volume of water `[H^(@)] = 0.5 xx 10^(-4)M`
`rArr [OH^(-)] = (10^(-14))/(0.5 xx 10^(-4)) = 2 xx 10^(-10)` MOLES `dm^(3)`
Discussion
45.

Hydrogen ion concentration of an aqueous solution is 1 xx 10^(-4) M. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol dm^(-3) is

Answer»

`1 xx 10^(-8)`
`1 xx 10^(-6)`
`2 xx 10^(-10)`
`0.5 xx 10^(-10)`

Answer :C
Discussion
46.

Hydrogen ion concentration of a solution whose pH is zero, would be

Answer»

`2.0g ION L^(-1)`
`1.0g ionL^(-1)`
`10^(-7) g ion L^(-1)`
`10^(-14)G ionL^(-1)`

Solution :`PH =- log[H^(+)] or ph =- log 1=0`
Discussion
47.

Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be

Answer»

`3.98 XX 10^(8)`
`3.88 xx 10^(6)`
`3.68 xx 10^(-6)`
`3.98 xx 10^(-6)`

SOLUTION :`ph = -LOG[H^(+)]`
`5.4 = -log[H^(+)], [H^(+)] = 3.98 xx 10^(-6)`.
Discussion
48.

Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be :

Answer»

`3.98xx10^(-8)`
`3.88xx10^(6)`
`3.68xx10^(-6)`
`3.98xx10^(-6)`

SOLUTION :`PH=-log[H^(+)]`
or`[H^(+)]=`ANTILOG `(-pH)`
= Antilog `(-5.4)=3.98xx10^(-6)M`
Discussion
49.

Hydrogen iodide cannot be prepared by the action of conc. H_(2)SO_(4) on KI because :

Answer»

`HI` is stronger than `H_(2)SO_(4)`
`HI` is more volatile than `H_(2)SO_(4)`
`H_(2)SO_(4)` is an oxidising agent
`H_(2)SO_(4)` FORMS complex.

Answer :C
Discussion
50.

Hydrogen iodide is injected into a container at 458^(@)C. Certain amount of HI dissociates to H_(2) and I_(2) At equilibrium, concentration of HI is found to be 0.421M while [H_(2)] and [I_(2)] each equal to 6.04xx10^(-2)M, at 458^(@)C. Calculate the value of the equilibrium constant of the dissociation of HI at the same temperature.

Answer»


ANSWER :`K_(C)=0.06xx10^(-2)`
Discussion