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.

In the titration between oxalic acid ad acidified potassium permanganate, the manganous salt formed catalyses the reaction. The manganous salt is:-

Answer»

A promoter
A POSITIVE catalyst
An autocatalyst
None of these

Answer :C
Discussion
2.

In the third period the first ionization potential is of the order.

Answer»

`Na GT Al gt MG gt SI gt P`
`Na LT Al lt Mg lt Si lt P `
`Mg gt Na gt Si gt P gt Al`
`Na lt Al lt Mg lt P lt Si`

ANSWER :b
Discussion
3.

In the test of halogen, why the nitric acid is added before the silver nitrate sodium fusion extract?

Answer»

SOLUTION :In sodium fusion extract, the sodium HYDROXIDE is in by the reaction between sodium and water, in which directly `AgNO_(3)` added and form precipitate AgOH. By ADDING of `HNO_(3)`, neutrallisation of NAOH is occurs and the solution become acidic and there is no interference in test of halogen.
Discussion
4.

In the synthesis of sodium carbonate , the recovery of ammonia is done by treating NH_(4)Cl with Ca(OH)_(2) . The by-product obtained in this process is

Answer»

`CaCl_(2)`
`NaCl`
`NaOH`
`NaHCO_(3)`

Solution :`2NH_(4)CL + CA(OH)_(2) to 2 NH_(3) + CaCl_(2) + 2H_(2)O` . THEREFORE `, CaCl_(2)` is the by product .
Discussion
5.

In the system A_((s)) hArr2B_((g)) + 3C_((g)) . If the concentration of C at equilibrium is increased by a factor of 2. It will casuse the equilibrium concentration of B to change to:

Answer»

TWO times the original value
one HALF of its original value
`2sqrt2` times its original value
`(1)/(2sqrt2)` times its original value

Solution :`A_((s)) HARR 2B_((g))+3C_((g))`
`K_(c)=(B)^(2)(C)^(2) IMPLIES (B)^(2)=((B)_(1)sqrt(K_(c)))/(2sqrt2 (C)^(3//2))`
Discussion
6.

In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH_(4)Clwith Ca(OH)_(2). The by-product obtained in this process is

Answer»

`CaCl_(2)`
NaCl
NaOH
`NaHCO_(3)`

SOLUTION :In solvay ammonia soda process, `Na_(2)CO_(3)` is manufactured. The reactions involved in the synthesis are :
`NH_(3)+H_(2)O+underset("Ammonium bicarbonate")(CO_(2) to NH_(4)HCO_(3))`
`NaCl +NH_(4)HCO_(3) underset("Sodium bicarbone")(toNaHCO_(2)darr)+NH_(4)Cl`
`2NaHCO_(3)overset(Delta)to underset("Sodium carbone")(Na_(2)CO_(3))+H_(2)O+CO_(2)`
`NH_(3)` is recovered from `NH_(4)HCO_(3)` and `NH_(4)Cl` and in this process `CaCl_(2)` is FORMED as a by product ACCORDING to the following reactions:
`NH_(4)HCO_(3)overset("heat")to NH_(3)+H_(2)O+CO_(2)`
`underset("chloride")underset("Ammonium")(2NH_(4)Cl)+Ca(OH)_(2)tounderset("Ammonia")(2NH_(3))+underset("chloride")underset("Calcium")(CaCl_(2))+2H_(2)O`
Discussion
7.

In the synthesis of sodium carbonate, the recovery of ammonia is done by treating NH CI with Ca(OH)_(2). The by-product obtained in this process is

Answer»

`CaCl_(2)`
NaCl
NaOH
`NaHCO_(3)`

Solution :`CaCI_(2)` is the by PRODUCTIN SOLVAY PROCESS
Discussion
8.

In the synthesis of HI, the amounts of H_(2(g)), I_(2(g)), and HI_((g)) , at equilibrium were found to be 0.8, 0.8 and 2.4 mole respectively in 10 liter vessel then calculate equilibrium constant of given reaction at constant temp and also calculate equilibrium constant of reverse reaction. H_(2(g)) +I_(2(g)) hArr 2HI_((g)).

Answer»

SOLUTION :`K_c=9` and for REVERSE REACTION `K_c=1/9`
Discussion
9.

In the synthesis of glycerol from propene, the steps involved are

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Glycerol `beta`-chlorohydrin and ALLYL chloride
Glycerol TRICHLORIDE and glycreol `ALPHA`-chlorohydrin
Allyl alcohol and glycerol `beta`-chlorohydrin
Allyl alcohol and monosodium glycerolate

Solution :REFER of synthesis of glycerol from propene.
Discussion
10.

In the sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find the percentage of sulphur in the given compound.

Answer»

Solution :Here, the MASS of the SUBSTANCE taken = 0.468 g
Mass of `BaSO_(4)` FORMED = 0.668 g
Now 1 mole of `BaSO_(4) -= 1g` atom of S or `(137 + 32 + 4 xx 16) = 233 g` of `BaSO_(4) -= 32 g` of S
Applying the relation, Percentage of SULPHUR `= (32)/(233) xx ("Mass of " BaSO_(4) " formed")/("Mass of substance taken") xx 100`
`= (32)/(233) xx (0.668)/(0.468) xx 100 = 19.60`.
Discussion
11.

In the sulhonation, acetylation and formylation of benzene the group of effective electrophiles would be

Answer»

`SO_(3)^(+),CH_(3)overset(o+)(CH_(2)),HC^(+)=O`
`SO_(3),CH_(3)-C-=O^(+),Hoverset(+)(C)O`
`SO_(3)^(+),CH_(3)CHO,CO+HCl`
`HSO_(3),CH_(3)CO,HCO`

Solution :Electrophile in sulphonation is `SO_(3)`.
Discussion
12.

In the study of chlorination of propane, four product (A,B & C,D) of the formula C_(3)H_(6)Cl_(2) were isolated. Each was further chlorinated to provide trichloro products (C_(3)H_(5)Cl_(3)). It was found that A provide one trichloro product, B gave two and C&D each gave three. It is found that D is optically active. Q. Correct formula of the compound D is

Answer»

`CH_(3)C Cl_(2)CH_(3)`
`CLCH_(2)CH_(2)CH_(2)Cl`
`CH_(3)CH_(2)CHCl_(2)`
`ClCH_(2)CHClCH_(3)`

ANSWER :C
Discussion
13.

In the study of chlorination of propane, four product (A,B & C,D) of the formula C_(3)H_(6)Cl_(2) were isolated. Each was further chlorinated to provide trichloro products (C_(3)H_(5)Cl_(3)). It was found that A provide one trichloro product, B gave two and C&D each gave three. It is found that D is optically active. Q. Correct formula of the product of chlorination of B is-

Answer»

`Cl_(2)CHCH_(2)CH_(2)Cl`
`ClCH_(2)CHClCH_(2)Cl`
Both A and B

Answer :C
Discussion
14.

In the study of chlorination of propane, four product (A,B & C,D) of the formula C_(3)H_(6)Cl_(2) were isolated. Each was further chlorinated to provide trichloro products (C_(3)H_(5)Cl_(3)). It was found that A provide one trichloro product, B gave two and C&D each gave three. It is found that D is optically active. Q. Formula of the compound A is

Answer»



`CH_(3)CH_(2)CHCl_(2)`
`CH_(3)CHClCH_(2)CL`

ANSWER :B
Discussion
15.

In the structure of hydrogen peroxide which of the following is correct?

Answer»

`O-H gt O-O` (bond length)
Dihedral angle is 111.5 in GAS phase
Oxygen are BONDED by OXIDE bond
All of these

Answer :B
Discussion
16.

In the structure of silica, each silicon atom is bonded to how many oxygen atoms ?

Answer»


SOLUTION :Each Si atom in SILICA is SURROUNDED by FOUR oxygen atoms
Discussion
17.

In the structure of HNO_(3) molecule, N-O bond (121 pm) is shorter than N-OH bond (140 pm).

Answer»

Solution :`HNO_(3)` is a resonance HYBRID of the following TWO canonical structures :

Due to resonance N-O BOND has some DOUBLE bond character. In contrast, N-OH has only pure single bond character. Since single bonds are longer than double bonds, therefore, N-OH bond is longer (140 pm) than N-O bond (121 pm) in `HNO_(3)`.
Discussion
18.

…….in the structure of heterocyclic compound

Answer»

SOLUTION :HETERO ATOM in CYCLIC CORNER
Discussion
19.

In the structureof diborane

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All hydrogen atoms LIE in ONE PLANE and boronatoms lie in a plane perpendicularto thisplane
2 boron atoms and 4 terminalhydrogenatomslie in the same plane and 2 bridging hydrogen atoms lie in the perpendicularplane
4 bridging hydrogenatoms andboron atomslife inone PLNE and TWO terminalhydrogenatoms lie in a planeperpendicularto this plane
All the atoms are in the same plane.

Answer :B
Discussion
20.

In the structure of diborane....

Answer»

All hydrogen ATOMS LIE in one PLANE and boron atoms lie in a plane perpendicular to this plane.
2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in the perpendicular plane.
4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane.
All the atoms are in the same plane.

Solution :The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms. The four terminal B-H bonds are regular two-centre-two-electron bonds while the two bridge (B-H-B) bonds are different and can be described in terms of three-centre-two-electron bonds as shown in FIGURE :
Discussion
21.

In the structure of diborane

Answer»

All hydrogen atoms lie in ONE plane and boron atoms lie in a plane perpenduicular to this plane
2 boron atoms and 4 TERMINAL hydrogen atoms lie in the same plane and 2 BRIDGING hydrogen atoms lie in the perpendicular plane
4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a plane perpendicular to this plane
All the atoms are in the same plane

ANSWER :2
Discussion
22.

In the structure of CIF_3 , the number of lone pairs of electrons on central atom 'Cl' is

Answer»

three
one
four
TWO

Solution :
T-shape(NUMBER of LONE PAIR ELECTRONS) two.
Discussion
23.

In the stoichiometry of nautral faujasite-a zeolite with formula Na_(x)[(AlO_(2))_(56)(SiO_(2))_(136)] 250H_(2)O the value of x is

Answer»

56
136
250
None of these.

Solution :Here `SiO_(2)` and `H_(2)O` do not carry any charge.The charge present in `AlO_(2)` UNIT is -1(+3 on Al and -4 on two O-atoms). THUS negative charge present on 56 `AlO_(2)` unit is -56. Therefore, 56 `Na^(+)` ions are present in ONE molecule of faujasite.
Discussion
24.

In the standardization of Na_2S_2O_3 using K_2Cr_2O_7 by iodometry, the equivalent weight of K_(2)Cr_(2)O_(7) is

Answer»

(MOLECULAR WEIGHT)/2
(molecular weight)/6
(molecular weight)/3
same as molecular weight

Answer :B
Discussion
25.

In the standardization of Na_(2)aS_(2)O using K_(2)Cr_(2)O_(7) by iodmetry the equivalent weight of K_(2)Cr_(2)O_(7) is

Answer»

`("MOLECULAR weight")//2`
`("molecular weight")//6`
`("molecular weight")//3`
same as molecular weight

Solution :`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O`
Since `K_(2)Cr_(2)O_(7)` accepts 6 electrons for its REDUCTION to `Cr^(3+)` ions
`therefore` Eq. wt. `=` MOL wt/ 6
Discussion
26.

In the standardization of Na_(2)S_(2)O_(3) using K_(2)Cr_(2)O_(7) by eudiometry, the equivalent weight of K_(2)Cr_(2)O_(7) is _____

Answer»

`("MOLECULAR weight")/2`
`("molecular weight")/6`
`("molecular weight")/3`
same as molecular weight.

Answer :B
Discussion
27.

In the standardisation of Na_(2)S_(2)O_(3) using K_(2)Cr_(2)O_(7) by iodometry theequivalent mass of K_(2)Cr_(2)O_(7) is:

Answer»

Molecular `MASS//2`
Molecular `mass//6`
Molecular `mass//3`
Same as the molecular mass.

Solution :The REDOX reaction is:
`(Cr_(2)O_(7))^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O`
Equivalent mass of `K_(2)Cr_(2)O_(7) = "(Molecular )/(6) mass"`
Discussion
28.

In the sphalerite (ZnS) structure S^(2-) ions form a face-centred cubic lattice. Then Zn^(2+) ions are present on the body diagonals at

Answer»

`(1)/(3)` RD of the DISTANCE
`(1)/(4) th` of the distance
`(1)/(6)th` of the distance
`(1)/(8)th` of the distance

Answer :B
Discussion
29.

In the Solvay process , can we obtain sodium carbonate directly by treating the solution containing (NH_(4))_(2) CO_(3) with sodium chloride ? Explain .

Answer»

Solution :In Solvay process , we cannot obtain directly SODIUM carbonate by treating the solution CONTAINING `(NH_(4))_(2)CO_(3)` with sodium chloride because both `Na_(2)CO_(3)` and `NH_(4)Cl` are highly soluble and the equilibrium will not shift in the FORWARD DIRECTION
`(NH_(4))_(2) CO_(3) + 2 NaCl hArr Na_(2)CO_(3) + 2 NH_(4)Cl`
Discussion
30.

In thesolvayprocess,NH_(4)Cl is treatedwithmalkof limeto recover_____.

Answer»

`NACL`
`NH_(4)HCO_(3)`
`NH_(4)`
`CO_(2)`

ANSWER :C
Discussion
31.

In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing(NH_(4))_(2) , CO_(3) with sodium chloride? Explain.

Answer»

SOLUTION :`(NH_(4))_(2), CO_(3)` reacts with NaCl and the products are `Na_(2)CO_(3) and NH_(4)Cl`. Both the products are HIGHLY soluble in water and equilibrium can not shift in forward direction. That.s why `Na_(2)CO_(3)`cannot by prepared directly.
Discussion
32.

In the sodium chloride structure, formula per unit cell is equal to

Answer»

2
8
3
4

Answer :D
Discussion
33.

In the sixth period, the orbitals being filled with electrons are

Answer»

5ss, 5p, 5D
6s, 6p , 6d, 6F
6s, 5F , 6d, 6p
6s, 4F, 5d, 6p

ANSWER :D
Discussion
34.

In the second group of qualitative analysis, H_(2)S is pass through a solution acidified with HCl in order to :

Answer»

LIMIT the concentration of `S^(2-)` ions
increase the solubility of `H_(2)S`
increase the concentration of `S^(2-)` ions
provide EXTRA `Cl^(-)` ions

Solution :HCl suppresses the concentration of `S^(2-)` ions in solution due to common ion EFFECT.
`H_(2)S overset((AQ))hArr 2H^(+) (aq) + S^(2-) (aq)`
`HCl overset((aq))rarr H^(+) (aq) + Cl^(-) (aq)`
Discussion
35.

In the salt bridge KCI is used because

Answer»

it is an electrolyte
it is an IONIC compound
it is a good conductor of electricity
the TRANSPORT NUMBER of `K^(+)` and `CL^(-)` are the same

Answer :D
Discussion
36.

In the ring test of NO_(3)^(-)ion,Fe^(2+) ion reduces nitrate ion to nitric oxide, which combines with Fe^(2+) (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.

Answer»

Solution :`underset("Nitrate ion")NO_(3)^(-)(aq)+3Fe^(2+)(aq)+4H^(+)(aq)tounderset("NITRIC axide")(NO(g))+3Fe^(3+)(aq)+3H_(2)O(l)`
`[Fe(H_(2)O)_(6)]^(2+)+NO(g)tounderset("Pentaaquanitrosnium IORN (I) (Brown complex)")([Fe(H_(2)O)_(5)NO]^(2+)+H_(2)O`
Discussion
37.

If the three interaxial angles defining the unit cell are all equal in magnitude, the crystal cannot belong to the

Answer»

100mg
300 MG
400mg
200mg

Solution :1 CARAT = 0.2 G = 200 mg
Discussion
38.

In the ring test for identification of nitrate ion, what is the formula of the compound responsible for the brown ring formed at the interface of the two liquids ?

Answer»

SOLUTION :The formula of the compound RESPONSIBLE for brown ring in the test for IDENTIFICATION of NITRATE ION is :
`[Fe(H_(2)O)_(5)NO]^(2+)SO_(4)^(-)` Pentaaquanitrosoniumiron (I) sulphate
Discussion
39.

In the representation of specific rotation ([alpha]_(D)^(250C)), .D. indicates

Answer»

DEXTRO rotation
Configuration of -OH GROUP
Configuration of -NH, group
Wave LENGTH of light

SOLUTION :Wave length of light
Discussion
40.

In the Reimer Teimann reaction the intermediate involved is

Answer»




SOLUTION :In REIMER Teimann REACTION, the INTERMEDIATE INVOLVED is
Discussion
41.

In the redox reaction : xKMnO_(4) to +yNH_(3) to KNO_(3)+MnO_(2)+KOH+H_(2)O

Answer»

x=4, y=6
x=3, y=8
x=8, y=6
x=8, y=3

Answer :D
Discussion
42.

In the redox reaction MnO_(4)^(-) +8H^(+)+Br^(-)rarrMn^(2+)+4H_(2)O+5//2br_(2) which one is the reducing agent ?

Answer»

`H^(+)`
`MnO_(4)^(-)`
`Br^(-)`
`MN^(2+)`

Solution :`Br^(-)` is the reducing agent SINCE it reduces `MnO_(4)^(-)` to `Mn^(+)` and itself GETS oxidised to `Br^(2)`
Discussion
43.

In the redox reaction. 2MnO_(4)^(-)+5H_(2)O_(4)^(2-)+16H^(+)hArr2Mn^(2+)+10CO_(2)+8H_(2)O The ion acting as autocatalyst is:-

Answer»

`MnO_(4)^(-)`
`C_(2)O_(4)^(2-)`
`H^(+)`
`Mn^(2+)`

Solution :`Mn^(++)` is a PRODUCT in reaction so it is autocatalyst (ACCORDING to definition).
Discussion
44.

In the redox reactin xKMnO_(4)+NH_(3)rarryKNO_(3)+MnO_(2)+KOH+H_(2)O

Answer»

x=4,y=6
x=3,y=8
x=8,y=6
x=8,y=3

Solution :`MnO_(4)^(-)+2H_(2)O+3e^(-)rarrMnO_(2)+4OH^(-)XX8`
`NH_(3)+9OH^(-)rarrNO_(3)^(-)+6H_(2)O+8e^(-)[XX`
`8MnO_(4)^(-)+3NH_(3)rarr8MnO_(2)+3NO_(3)^(-)+5OH^(-)+2H_(2)O`
or `8KMnO_(4)+3 NH_(3)rarr8MnO_(2)+3KNO_(3)+5KOH+2H_(2)O`
THUS coefficient of `KmnO_(4)=8` and that of `KNO_(3)=3`
Discussion
45.

In the rection 2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)

Answer»

`S_(2)O_(3)^(2-)` gets oxidised to `S_(4)O_(6)^(2-)`
`S_(2)O_(3)^(2-)` gets oxidised to `S_(4)O_(6)^(2-)`
`I_(2) "gets reduced to" l^(-)`
`l_(2) "gets oxidised to"l^(-)`

SOLUTION :`(2overset(+2)S_(2)O_(3))^(2-)+(overset(0)I_(2))rarr(overset(+5//2)S_(4)O_(6))^(2-)+(overset(-1)2I^(-))`
`S_(2)O_(3)^(2-) "is oxidised to" S_(4)O_(6)^(2-)`
`I_(2) "is reduced to" I^(-)`ion
Discussion
46.

In the rection NH_(4)(s) hArr (g)+HCl (g) the value of Deltan_(g) is .......

Answer»

<P>

SOLUTION :`DELTA n_(G) = n_(p)(g) -n_r(g)=2 -0=2`
Discussion
47.

In the redox ionic equation 2MnO_(4)^(-)+Br^(-)+H_(2)O -> 2MnO_(2)+BrO_(3)^(-) +2OH^(-) the equivalent weight of potassium permangante of molar mass M g mol""^(-1) is

Answer»

`M/5`
`M/3`
`M/6`
`M/2`

ANSWER :B
Discussion
48.

In the reactions given below, identify the species undergoing oxidation and reduction : (i) H_(2)S_((g))+Cl_(2(g))to2HCl_((g))+S_((s)) (ii) 3Fe_(3)O_(4(s))+8Al_((s))to9Fe_((s))+4Al_(2)O_(3(s)) (iii) 2Na_((s))+H_(2(g))to2NaH_((s))

Answer»

Solution :(i) `H_(2)S` is oxidised because a more electronegative element, chlorine is added to hydrogen (or a more electropositive element, hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.
(ii) ALUMINIUM is oxidised because oxygen is added to it. Ferrous ferric oxide `(Fe_(3)O_(4))` is reduced because oxygen has been removed from it.
(iii) With the CAREFUL application of the concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced.
Reaction (iii) chosen here prompts us to think in terms of ANOTHER way to define redox REACTIONS.
Discussion
49.

In the reactions Sn Cl_2+HgCl_2toSnCl_4+Hg_2Cl_2 which is the oxidant and which is the reductant?

Answer»

SOLUTION :`SnCl_2` is a REDUCTANT and `HgCl_2` is an OXIDANT.
Discussion
50.

In the reactions given below identify the species undergoing oxidation and recuction (i) H_(2)S(g)+CI_(2)(g)rarr2HCI(g)+S(s)(ii)3Fe_(3)O_(4)(s)+8AI(s)rarr9Fe(s)+4AI_(2)O_(3)(s)(iii)Na(s)+H_(2)(g)rarr2NaH(s)

Answer»

Solution :(i) here `H_(2)S` is oxidised to S because hydorgen has been removed form it whil `CI_(2)` has been reduced to (II) here aluminium is oxidised to aluminium oxid because oxygen has been added to it while ferrous ferric oxide (or magnetic oxide) has been reduced to IRON because oxygen has been removed from it (ii) as SHOWN in SODIUM is oxidised to sodium io while hydrogen is reduced to hydride ion
Discussion