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.

Main source of lead is PbS. It is converted to Pb by (I) PbS underset(Delta)overset("air")rarr PbO + SO_(2) (II) Self reduction process is

Answer»

I
II
Both A & B
none of these

Answer :B
Discussion
2.

Main productP of the givenreaction isCH_(3)COOH + HCOOH overset("MnO")underset(570K)to P

Answer»

`CH_(3)CHO`
`CH_(3)COCH_(3)`
HCHO
`(CH_(3)CO)_(2) O`

Solution :` CH_(3)COOH + HCOOH underset(570K) overset(MnO) toCH_(3)CHO + CO_(2) + H_(2)O`
Discussion
3.

Main product of the reaction , CH_3CONH_2+HNO_2rarr…….is

Answer»

`CH_3COOH`
`CH_3CH_2NH_2`
`CH_3NH_2`
`CH_3COONH_4`

ANSWER :A
Discussion
4.

The product of reaction,

Answer»

`CH_3COOH+H_2`
`CH_3CH_2COOH`
`CH_3CH(OH)CH_3`
`CH_2CH_2OH`

ANSWER :C
Discussion
5.

Main product of the reaction is/are:CH_3COCH_3overset(Pd)underset(H_2)rarr ?

Answer»

`CH_3COOH+H_2`
`CH_3CH_2COOH`
`CH_3CH(OH)CH_3`
`CH_3CH_2OH`

ANSWER :C
Discussion
6.

Main ore of lead is gaalena.This is mined and separated from other minerals by froth flotation.There are two methods of extracting the lead. (i)First method involves the roasting of ore followed by reduction with coke or CO. (ii)Second method involves the partial roasting of ore followed by self reduction. Which of the following statement is incorrect about the extraction of lead from galena ?

Answer»

Galena is a sulphide ore and therefore , it is concentrated by froth flotation process
Self reduction takes place in absence of air
COMPLETE ROASTING of galena gives PbO which is then reduced by coke or CO to GIVE metallic lead
`FeSiO_3` is obtained as slag.

Solution :`FeSiO_3` is obtained as slag in the extraction of copper from copper pyrites.
Discussion
7.

Main product of the following reactions is:

Answer»




NONE of these

Solution :
Discussion
8.

Main pollutant released from iron and steel industry are :

Answer»

`CO, CO_2` and `SO_2`
`NO`,`SO_3` and `H_2S`
`CO_2,H_2S`and `NO_2`
`CO_2,NO_2` and `SO_3`

ANSWER :A
Discussion
9.

Most hazardous metal pollutant of automobile exhausts is :

Answer»

CO
`CO_2`
`NO`
Hydrocarbons

Answer :A
Discussion
10.

Main ore of lead is gaalena.This is mined and separated from other minerals by froth flotation.There are two methods of extracting the lead. (i)First method involves the roasting of ore followed by reduction with coke or CO. (ii)Second method involves the partial roasting of ore followed by self reduction. In self reduction process, the species which bring about the reduction is :

Answer»

`O_2`
`S^(2-)`
`O^(2-)`
`H_2S`

Solution :`Pboverset(2-)S + 2PbO UNDERSET("absence of AIR")overset(DELTA)to3Pb(L)+overset(+4)(SO_2)uarr`
Discussion
11.

Main ore of aluminium is :

Answer»

CRYOLITE
Kaolin
Bauxite
Felspar

Answer :C
Discussion
12.

Main elements present in lipids are

Answer»

C
H
O
C,H,O

Answer :D
Discussion
13.

Main function of roasting is

Answer»

to remove volatile SUBSTANCES
oxidation
reduction
slag formation

Solution :To remove moisture and non-metallic impurities like S, P and As are oxidised and are removed as volatile substances.
`S_(8)+8O_(2) to 8SO_(2) uarr : P_(4)+5O_(2) rarr P_(4)O_(10) uarr`
`4As+3O_(2) to 2As_(2)O_(3) uarr`
Discussion
14.

Main constituents of the cell walls of plants is :

Answer»

Cellulose
Glycogen
Lactose
Chlorophyll

Answer :A
Discussion
15.

Mainconstituent(S)LPGis / are :

Answer»

METHANE
`H_(2),CH_(4)` iso - butane
iso-butane, produce
none of these

ANSWER :C
Discussion
16.

Main constituent of marsh gas is

Answer»

`C_2H_2`
`CH_4`
`H_2S`
CO

Solution :MARSH GAS MAINLY CONSISTS of METHANE
Discussion
17.

Main constituent of dynamite is

Answer»

Nitrobenzene
Nitroglycerine
Picric acid
TNT

Answer :B
Discussion
18.

Main constituent of dynamite is ………………….. .

Answer»

NITRO BENZENE
nitro GLYCERINE
PICRIC ACID
TNT

Solution :nitro glycerine
Discussion
19.

Main cause of allergy is

Answer»

SECRETION of demetne in our body
secretion of SELDANE in our body
secretion of HISTAMINE in our body
secretion of DIMETAPP in our body

Answer :C
Discussion
20.

Main axis of diatomic molecule is Z. The orbitals P_x and P_y overlap to form

Answer»

`PI` - MOLECULAR ORBITAL
`sigma` - molecualr orbital
`delta`- molecular orbital
no nond is formed

Solution :The `P_(x)` and `P_(y)` ORBITALS do not overlap thus no BOND is formed .
Discussion
21.

Main air pollutant is

Answer»

`N_2`
`CO`
`CO_2`
Sulphur

Answer :B
Discussion
22.

Magnitude of dipole moment generated in chloro benzene will be same as that of dipole moment generated in :

Answer»

benzene
para-dichloro benzene
meta-dichloro benzene
cannot be SIMILAR with any MOLECULE?

ANSWER :C
Discussion
23.

Magnetite is concentrated by

Answer»

gravity METHOD
FROTH floatation PROCESS
electromagnetic method
All of the above

Solution :MAGNETITE `(Fe_(3)O_(4))` being a magneticore of IRON can be concentrated by electromagneticseparation.
Discussion
24.

Magnetite is a metal of....

Answer»

FE 
ZN 
Al
Cu 

ANSWER :A
Discussion
25.

Magnetite is an ore of :

Answer»

IRON
CALCIUM
COPPER
Zinc

Answer :A
Discussion
26.

Magnetic separation it is based on the difference in the _______ of the ore and the impurities.

Answer»

MAGNETIC properties
chemical properties
physical properties
melting point

Answer :A
Discussion
27.

Magnetic separation is used in the concentration of:Copper pyrites,Chromite,Bauxite,Cinnabar.

Answer»

COPPER pyrits
Chromite
Bauxite
Cinnabar

Answer :B
Discussion
28.

Magnetic oxide (Fe_(3)O_(4)) when when heated with hydrogen is reduced to iron and water is also produced. Write balanced equation for the reaction.

Answer»

Solution :The reaction is : `"Magnetic oxide of iron"+"Hydrogen"rarr"Iron "+"Water"`
`"The skeleton EQUATION is :"Fe_(3)O_(2)rarrFe+H_(2)O`
The equation contains hydrogen as an elementary gas. Writing it in the atomic state, we have
`Fe_(3)O_(4)+HrarrFe+H_(2)O`
The biggest formula is `Fe_(3)O_(4)`. Hence, order of selection of atoms for balancing as FE, O and H.
(i) To equalise the number of atoms of Fe and O on both sides, MULTIPLY Fe by 3 and `H_(2)O` by 4. We have
`Fe_(3)O_(4)+Hrarr 3Fe+4H_(2)O`
(ii) The above equation has 8 atoms of H on the R.H.S. as against 1 on L.H.S. To equalise, multiply H atoms by 8 and convert it into molecular state as `4H_(2)`. We have
`Fe_(3)O_(4)+4H_(2)rarr 3Fe+4H_(2)O`
This is the required balanced molecular chemical equaiton.
Discussion
29.

Magnetic separation is used in the concentration of…

Answer»

COPPER PYRITES 
BAUXITE 
CASSITERITE 
Cinnabar 

Answer :C
Discussion
30.

Magnetic nature of C2 molecule will be:-

Answer»

PARAMAGNETIC (1.7 BM)
DIAMAGNETIC
Paramagnetic (3.8 BM)
Paramagnetic (2.8 BM)

ANSWER :D
Discussion
31.

Magnetic moment (spin only) of octahedron complex having SFSE=-0.8Delta_(0) and surrounded by weak field ligands can be:

Answer»

`sqrt(15)` BM
`sqrt(8)`BM
(a) and (B) both
none of these

Answer :C
Discussion
32.

Magnetic moments arise due to the presence of unpaired electrons’ Calculated magnetic moments of two transition metal ions are given below: Justify these observations on the basis of spin only formula.

Answer»

SOLUTION :Magnetic moment, `M=sqrt(n(n+2))BMSC^(+3)is[Ar]3d^0,` no unpaired ELECTRON `TI^(+3)is[Ar]3d^1` ONE unpaired-electron `therefore` Magnetic moment `=sqrt(1(1+2))=sqrt3=1.73` BM
Discussion
33.

Magnetic moment of transition metal ion is 5.92 BM (spin-only formula). The number of unpaired electrons in the metal ion is

Answer»

4
5
2
3

Answer :B
Discussion
34.

Magnetic moment of the complex [Fe(CN)_(6)^(3+) is approximately

Answer»

5.91 BM
4.89 BM
2.84 BM
1.73BM

Answer :D
Discussion
35.

Magnetic moment of (NH_(4))_(2)[MnBr_(4)] is __________B.M.

Answer»

5.91
4.91
3.91
2.46

Solution :`[MnBr_(4)]^(2-)` has Mn in +2 state (`d^(5)` ION) and in `sp^(3)` hybridisedform

`mu_("EFF")=sqrt(5(5+2))=5.91B.M.`
Discussion
36.

Magnetic moment of O_(2) is nearly

Answer»

1.8 BM
2.8 BM
3.8 BM
ZERO

ANSWER :B
Discussion
37.

Magnetic moment of [MnCl_(4)]^(2-) is 5.92 BM. Explain giving reason.

Answer»

Solution :The MAGNETIC moment of 5.92 BM corresponds to the presence of five unpaired electrons in the dorbitals of `MN^(2+)` ion. As a result the hybridisation involved is `sp^(3)` rather than `dsp^(2)`. THUS tetrahedral structure of `[MnCl_(4)]^(2-)` complex will show 5.92 BM magnetic moment value.
Discussion
38.

Magnetic moment of [MnCl_(4)]^(2-) is 5.92 BM. Explain giving reason

Answer»

Solution :`mu=sqrt(N(n+2))BM, mu=5.92` BM means n=5, i.e., 5 unpaired ELECTRONS. `Mn^(2+)=3d^(5)4S^(0)4p^(0)`. To form `[MnCl_(4)]^(2-)`, HYBRIDISATION will be `sp^(3)`. Hence, the structure will be tetrahedral.
Discussion
39.

Magnetic moment of Cr^(2+) is nearest to …..

Answer»

`Fe^(2+)`
`Ni^(2+)`
`MN^(2+)`
`CO^(2+)`

Solution :`Cr^(2+) and Fe^(2+)` have same number of unpaired ELECTRONS i.e
Discussion
40.

Magnetic moment of diamagnetic substance in Bohr magnetons is

Answer»

1.73
2.83
2.83
0

Answer :D
Discussion
41.

Magnetic moment of 2.84 B.M. is given by : (At. Nos. Ni=28, Ti = 22, Cr= 24, Co = 27)

Answer»

`Cr^(2+)`
`Co^(2+)`
`Ni^(2+)`
`Ti^(3+)`

SOLUTION :Magneticmoment, `mu = sqrt(N(n+2))`
`mu = 2.84 BM` whn `n=2`,i.e., there are 2 UNPAIRED electrons.
`Cr^(2+) = 3d^(4)` ( unpaired electrons= 4 )
`Co^(2+ = 3d^(7)`( unpaired electrons `= 2 )`
`Ni^(2+) = 3d^(8) ( `unpaired electrons =2)
`Ti^(2+) = 3d^(1) `( unpaired electrons `=1)`
Discussion
42.

Magnetic moment of [Ag(CN)_2]^_ is zero. How many unpaired electrons are there:

Answer»

Zero
4
3
1

Answer :A
Discussion
43.

Magnetic moment, ionic conductance and colligative properties are useful in deciding structure/constitution of a given unknown complex compound. Q. A metal M having elecftronic configuration (n-1)d^(8)ns^(2) forms complexes with co-ordination No.=4 and 6, if it forms diamagnetic complexes then permissible oxidation states of metal cation and geometry is:

Answer»

`+2`, OCTAHEDRAL
`+4`, octahedral
`+2`, square planar
(B) and (C) both

Answer :D
Discussion
44.

Magnetic moment, ionic conductance and colligative properties are useful in deciding structure/constitution of a given unknown complex compound. Q. The cyano complex that exhibit highest value of paramagnetism is:

Answer»

`[MN(CN)_(6)]^(4-)`
`[Co(CN)_(6)]^(3-)`
`[FE(CN)_(6)]^(3-)`
`[Cr(CN)_(6)]^(3-)`

ANSWER :D
Discussion
45.

Magnetic moment, ionic conductance and colligative properties are useful in deciding structure/constitution of a given unknown complex compound. Q. if molar conductivity of complex is almost equal to that of NaCl and it does not exhibits stereioisomerism then the complex will be:

Answer»

`[Co(CO_(3))(EN)_(2)]Br`
`[Co(CO_(3))(H_(2)O)_(2)(NH_(3))_(2)]Br`
`[Co(CN)(NH_(3))_(5)]Br_(2)`
`[Co(CO_(3))(NH_(3))_(4)]Br`

Answer :D
Discussion
46.

Magnetic moment 2.84 BM is given by

Answer»

`Cr^(2+)`
`Co^(2+)`
`Ni^(2+)`
`Ti^(3+)`

SOLUTION :`Cr^(2+)-3d^(4):4 "unpaired ELECTRON"`
`Co^(2+)-3d^(7),3:4 "unpaired electron"`
`Ni^(2+)-3d^(8):2"unpaired electron"`
`Ti^(3+)-3d^(1):1"unpaired electron"`
Magnetic moment`(MU)=SQRT(n(n+2))=sqrt(2(2+2))=2.86BM`
It shows that `Ni^(2+)` with 2 unpaired electrons MATCHES the value of magnetic moment.
Discussion
47.

Magnetic moment 2.84BM is given by: (Atoms No: Ni= 28, Ti= 22,Cr= 24, Co= 27)

Answer»

`Ni^(2+)`
`Ti^(3+)`
`Cr^(2+)`
`Co^(2+)`

SOLUTION :
NUMBER of UNPAIRED electrons (n)=2
`:. MU =sqrt(n(n+2))`
Hence `mu` 2.8BM, paramagnetic
Discussion
48.

Magnetic moment 2.84B.M. is given by (Atomic number, Ni = 28, Ti = 22, Cr = 24, Co = 27)

Answer»

`CR^(2+)`
`CO^(2+)`
`Ni^(2+)`
`Ti^(3+)`

Answer :C
Discussion
49.

Magnetic moment 2.83 BM is given by which of the following ions ?

Answer»

`TI^(3+)`
`Ni^(2+)`
`CR^(3+)`
`Mn^(2+)`

ANSWER :B
Discussion
50.

Magneta is

Answer»

ALKALINE PHENOLPHTHALEIN
METHYL red
p- rosaniline hydrochloride
red litmus

Answer :C
Discussion