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.

Limestone is present in the blast furnance production of iron in order to: (I) provide a source of CaO (II) remove some impurities (III) supply CO_(2)

Answer»

I, II, III
I, II
II, III
I only

Answer :B
Discussion
2.

Like O_(2), S_(2) vapours are.............due to the presence of two unpaired electrons in the ..................orbitals.

Answer»

SOLUTION :PARAMAGNETIC, antibonding MOS
Discussion
3.

Like heterogeneous catalyst ........... can be recovered and recycled.

Answer»

SOLUTION :NANO CATALYST
Discussion
4.

Like ammonia, amines combine with metal ions to form coordination compounds. Discuss.

Answer»

Solution :Like AMMONIA, AMINES combine with metal ions like `AG^(+)`and `Cu^(2+)` to form coordination compounds. It is due to the presence of a lone pair of electrons on the nitrogen atom of the amines which can form a coordinate BOND with the metal ion. For EXAMPLE,
Discussion
5.

Which is the lightest transition metal ?

Answer»

SOLUTION :SCANDIUM.
Discussion
6.

Lightest transtion metal is titanium.

Answer»

SOLUTION :SCANDIUM.
Discussion
7.

Lightest transition element is :

Answer»

Fe
Sc
Os
Co

Answer :B
Discussion
8.

Lightest construction metal in industry is

Answer»

Lithium
Calcium
Aluminium oxide
Magnesium

Solution :Magnesium is the lightest CONSTRUCTION METAL in industry. Its density is LESS than 2/3 of aluminium.
Discussion
9.

Light with a wavelength 310 nm fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is 1.5 eV. Then [Given : lambda_(e) = sqrt((150)/(DeltaV)) Å where Delta V= Voltage difference of battery]

Answer»

de-Broglie wavelength of electron is `10 Å`
Work fuction of strontium is `2.5 eV`
Threshold wavelength for strontium metal will `496 nm`
All EJECTED phot ELECTRONS will have kinetic energy `= 1.5 eV`

Solution :`Delta E = (1240)/(310 nm) eV = 4.0 eV`
`(KE_("max"))_(e) = 1.5 eV = q Delta V`
`Delta V = 1.5 V`
(A) `lambda_(e) = sqrt((150)/(Delta V))Å = sqrt((150)/(1.5)) Å = 10 Å`
(B) `Delta E = KE_(e) + w.f`
`3 = 1.5 + w.f.`
WF `= 2.5 eV`
(C) `lambda = (1240)/(w.f.) nm = (1240)/(2.5) nm = 496 nm`
Discussion
10.

Light of wavelength lamda shines on a metal surface with intensity x and the metal emits y electrons per second of average energy, z. What will happen to y and z if x is doubled ?

Answer»

y will be DOUBLED and Z will become half
y will remain same and z will be doubled
bothy and z will be doubled
y will be doubled but z will remain same.

Solution :When intensity is doubled, number of ELECTRONS emitted per second is also doubled but average ENERGY of photoelectrons emitted REMAINS the same.
Discussion
11.

Light scattering method is used

Answer»

to find CONCENTRATION
to find MOLAR MASS of POLYMER
to test elements
to find number of molecules.

Solution :to find molar mass of polymer
Discussion
12.

Lightof wavelenght lamda shines oinlametal surface with intensity x and the metal emits y electrons per second of average energy, z. What will happen to y and z if x is doubled?

Answer»

y will be DOUBLED and Z will BECOME half
y will REMAIN same and z will be doubled
both y and z will be doubled
y will be doubled but z will remain same.

Answer :D
Discussion
13.

Light is an electromagnetic phenomenon. A beam of light consists of_______mutuall perpendicular oscillating fields

Answer»

three
two
four
infinite

Solution :an OSCILLATING ELECTRIC FIELD and an oscillating MAGNETIC field.
Discussion
14.

Light green (Compund A) overset(Delta)rarr White Reside (B) overset("High")underset("Temperature")rarrC+D+E i) D & E are two acidic gases ii) D is pased through HgCl_2 solution to give yellow ppt iii) E is passed through water first and then H_2S is passed, whith turbidity is obtained (iv) A is wter soluble and addition of HgCl_2 in it, white pp it obtained but white ppt does no turn into grey on addition of excess solution of A 'C' is soluble. In

Answer»

DIL HCl
dil . `H_2SO_4`
`CONC, CH_3COOH`
Boile . Conc. HCl

Solution :
Discussion
15.

Light green solution of (A)does not give bluecoloured ppt . With K_(4)[Fe(CN)_(4)] but on adding a drop of HNO_(3) blue ppt .(B) appears .However (A) gives blue colour (C ) with K_(4)[Fe(CN)_(6)] Example the formation of (B) and (C ) identify (A) if (A) also gives while ppt , with AgNO_(3) solution

Answer»

Solution :(A) `FeCI_(2)`
(B) `Kfe^(III)[Fe^(II)(CN)_(6)]` prassian blue
(C ) `KFe^(II) [Fe^(II)(CN)_(6)]` Turnull's blue
Discussion
16.

Liquid hydrocarbon can be converted to a mixture of gaseous hydrocarbon by

Answer»

Hydrolysis
Oxidation
Cracking
Distillation under REDUCED pressure

Answer :C
Discussion
17.

Light green (compound 'A')overset(Delta)(to)"white Residue(B)"overset("high")underset("Temp.")C+D+E (i)'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl_(2) solution to give yellow ppt. (iii) 'E' is passed through water first and then H_(2)S is passed, white turbidity is obtained. (iv) A is water soluble and addition of HgCl_(2) in it, yellow ppt. is obtained but white ppt does not turn into grey on addition of excess solution of 'A' Q. The no. of water crystallisation in 'A' is :

Answer»

`0`
`2`
`7`
`5`

Answer :C
Discussion
18.

Light green (Compund A) overset(Delta)rarr White Reside (B) overset("High")underset("Temperature")rarrC+D+E i) D & E are two acidic gases ii) D is pased through HgCl_2 solution to give yellow ppt iii) E is passed through water first and then H_2S is passed, whith turbidity is obtained (iv) A is wter soluble and addition of HgCl_2 in it, white pp it obtained but white ppt does no turn into grey on addition of excess solution of A Yellow ppt in the above observation is

Answer»

MERCURIC oxide
BASIC MERUCRY (II) sulphite
Basic MERCURY (II) sulphate
Mercurous oxide

Solution :
Discussion
19.

Light green (compound 'A')overset(Delta)(to)"white Residue(B)"overset("high")underset("Temp.")C+D+E (i)'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl_(2) solution to give yellow ppt. (iii) 'E' is passed through water first and then H_(2)S is passed, white turbidity is obtained. (iv) A is water soluble and addition of HgCl_(2) in it, yellow ppt. is obtained but white ppt does not turn into grey on addition of excess solution of 'A' Q. 'C' is soluble in :

Answer»

dil. HCl
dil.`H_(2)SO_(4)`
Conc.`CH_(3)COOH`
Boiled conc. HCl

Answer :D
Discussion
20.

Light green (Compund A) overset(Delta)rarr White Reside (B) overset("High")underset("Temperature")rarrC+D+E i) D & E are two acidic gases ii) D is pased through HgCl_2 solution to give yellow ppt iii) E is passed through water first and then H_2S is passed, whith turbidity is obtained (iv) A is wter soluble and addition of HgCl_2 in it, white pp it obtained but white ppt does no turn into grey on addition of excess solution of A 'D' & 'E' are

Answer»

`SO_2 & SO_3`
`SO_3 & SO_2`
`SO_2 & CO_2`
`CO_2 & CO`

SOLUTION :
Discussion
21.

Light green (compound 'A')overset(Delta)(to)"white Residue(B)"overset("high")underset("Temp.")C+D+E (i)'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl_(2) solution to give yellow ppt. (iii) 'E' is passed through water first and then H_(2)S is passed, white turbidity is obtained. (iv) A is water soluble and addition of HgCl_(2) in it, yellow ppt. is obtained but white ppt does not turn into grey on addition of excess solution of 'A' Q.Yellow ppt in the above observation is :

Answer»

MERCURIC oxide
Basic MERCURY(II) sulphite
Basic mercury (II) sulphate
Mercuric iodine

Answer :C
Discussion
22.

Light green (compound 'A')overset(Delta)(to)"white Residue(B)"overset("high")underset("Temp.")C+D+E (i)'D' and 'E' are two acidic gas. (ii) 'D' is passed through HgCl_(2) solution to give yellow ppt. (iii) 'E' is passed through water first and then H_(2)S is passed, white turbidity is obtained. (iv) A is water soluble and addition of HgCl_(2) in it, yellow ppt. is obtained but white ppt does not turn into grey on addition of excess solution of 'A' Q.'D' and 'E' are respectively.

Answer»

`SO_(2) and SO_(3)`
`SO_(3) and SO_(2)`
`SO_(2) and CO_(2)`
`CO_(2) and CO`

ANSWER :B
Discussion
23.

Light blue colour of nitrous acid is due to dissolved:

Answer»

`O_2`
`N_2`
`N_2O`
`N_2O_3`

ANSWER :D
Discussion
24.

Ligands with which linkage isomerism is possible A) NO_(2) B) CN^(-) C) SCN^(-)

Answer»

A only
A & B
B &C
A, B &C

ANSWER :D
Discussion
25.

Ligands in a complex salt are :

Answer»

Anions linked by COORDINATE bonds to a CENTRAL METAL atom or ion
Cations linked by coordinate bonds to a central metal or ion
Any species linked by coordinate bonds to a central metal or ion
Ions or molecules linked by coordinate bonds to a central atom or ion

Answer :D
Discussion
26.

Ligands for which Delta_(0) < P are known as weak field ligands and form high spin complexes.

Answer»


ANSWER :1
Discussion
27.

Ligands contains :

Answer»

Ione PAIR of electrons
incomplete OCTET
UNPAIRED electrons
SHARED pair of electrons.

Answer :A
Discussion
28.

Ligands can be classified by various ways, based upon charges, denticity and interaction between ligand and central atom. Which of the following ligand is unsymmetrical bidentate ligand as well as having chiral centre which cannot be made symmetrical at all giving rotation around any single bond ?

Answer»

pn
bn
`GLY^(-)`
`DMG^(-)`

ANSWER :A
Discussion
29.

Ligands can be classified by various ways, based upon charges, denticity and interaction between ligand and central atom. Which of the following ligand is of ambidentate type ?

Answer»

`NO_(3)^(-)`
`C_(2)O_(4)^(2-)`
`S_(2)O_(3)^(2-)`
NONE of these

Answer :C
Discussion
30.

Ligands can be arranged in a series called ______ in order of increasing field strength.

Answer»


ANSWER :SPECTROCHEMICAL SERIES
Discussion
31.

Ligands can be arranged in a series in order of increasing field strength. What is that series called ?

Answer»

SOLUTION :SPECTROCHEMICAL
Discussion
32.

Ligands are electron pair donor groups. Ambidentate ligands are those ligands where two different atoms can donate electron pair. The complex where ambidentate ligands are present shows linkage isomerism. Total possible linkage isomers of K_(4)[Fe(CN)_(5)(NO_(2))] is :

Answer»

12
10
6
4

Solution :`K_(4)[FE(CN)_(5)(NO)_(2)]`
`6xx2=12`
Discussion
33.

Ligands are electron pair donor groups. Ambidentate ligands are those ligands where two different atoms can donate electron pair. The complex where ambidentate ligands are present shows linkage isomerism. The correct name of linkage isomer of [Co(NH_(3))_(5)(SCN)]SO_(4)

Answer»

Pentaamminethiocyanato-S-cobalt (III) sulphate
Pentaamminethiocyanato-N-Cobalt (II) sulphate
Pentaamminethiocyanato-N-Cobalt (III) sulphate
Pentaamminecyanato-N-Cobalt (III) sulphate

Solution :`[CO(NH_(3))_(5)(NCS)]SO_(4)`
Discussion
34.

Ligands are electron pair donor groups. Ambidentate ligands are those ligands where two different atoms can donate electron pair. The complex where ambidentate ligands are present shows linkage isomerism. Total number of linkage isomers that are present in Na_(4)[Fe(CN)_(5)NOS] is :

Answer»

18
6
8
16

Solution :
Discussion
35.

Ligands are broadly classified into two classes classical and non-classical ligands, depending on their donor annd acceptor ability. Classical ligands form classical complexes while non-classical ligands form non-classical complex. Bonding mechanism in non-classical is called synergic bonding. Q. In compound [M(CO)_(n)]^(x), the correct match for highest 'M-C' bond length for given M, n and z respectively:

Answer»

M-Cr, n-6, z-0
M-V, n-6, z- `-1`
M-Ti, n-6, z- `-2`
M-Mn, n-6, z- `+1`

Solution :Higher the positive oxidation STATE of metal, poor will be `M overset(pi) to l`. HENCE `overset(+1)(Mn) -C` bond will longer AMONG given complexes.
Discussion
36.

Ligands are broadly classified into two classes classical and non-classical ligands, depending on their donor annd acceptor ability. Classical ligands form classical complexes while non-classical ligands form non-classical complex. Bonding mechanism in non-classical is called synergic bonding. Q. Which is not pi-acceptor ligand?

Answer»


`sigma-C_(5)H_(5)^(-)`
`PH_(3)`
B_(3)N_(3)H_(6)`

SOLUTION :No vacant .d. or `pi^(**)` orbitals on
Discussion
37.

Ligands are broadly classified into two classes classical and non-classical ligands, depending on their donor annd acceptor ability. Classical ligands form classical complexes while non-classical ligands form non-classical complex. Bonding mechanism in non-classical is called synergic bonding. Q. Synergic bonding is absent in:

Answer»

`[Mo(CO)_(6)]`
`[CR(CO)_(3)(B_(3)N_(3)H_(6)]`
`[Sc(CO)_(6)]^(3+)`
`[NI(CN)_(4)]^(4-)`

Solution :`overset(+III)(Sc):3d^(@)4S^(@)`, as d-orbitals at `overset(+III)(Sc)` are empty, hence, there is no synergic bonding in `[Sc(CO)_(6)]^(3+)`.
Discussion
38.

Ligand Which is used to treat hard water

Answer»

`NH_3`
` O X `
` EN `
` "EDTA"`

ANSWER :D
Discussion
39.

Ligand to form a complex which is used to treat hard water

Answer»

`NH_(3)`
ox
en
EDTA

Answer :D
Discussion
40.

Ligand NO is named as :

Answer»

Nitrosonium
Nitronium
Nitrosyl
Nitro

Solution :nitrosyl
Discussion
41.

Ligand in the complex [FeCN)_6]^(3-) ion is：

Answer»

`CN^-`
N
Fe
C

Answer :A
Discussion
42.

Ligand in a metal carbonyl complex is

Answer»

`CO_2`
CO
`CoCl_2`
`C_2O_4^(2-)`

ANSWER :B
Discussion
43.

Life Saving drug(s) used in cancer therapy are- ( a )Cisplatin ,( b ) AZT , ( c ) Taxol

Answer»

( a ) & ( C )
( a ) & ( B )
( a ) , ( b ) & ( c )
( b ) & ( c )

ANSWER :A
Discussion
44.

LIFis a/ an______ Crystal

Answer»

IONIC
METALLIC
Covalent
molecular

ANSWER :A
Discussion
45.

LiF is a/an:

Answer»

IONIC CRYSTAL
Metallic crystal
Covalent crystal
MOLECULAR crystal

Answer :A
Discussion
46.

Liebig test is used to estimate:

Answer»

H
C
C and H both
BEILSTEIN's TEST

SOLUTION :N//A
Discussion
47.

liebermann nitroso reaction is used for the detetion of _____amines.

Answer»

SOLUTION :SECONDARY.
Discussion
48.

Liebermann's test is answered by

Answer»

Aniline
Methylamine
Ethyl benzoate
Phenol

Answer :D
Discussion
49.

Lichens do not like to grow in cities

Answer»

because of absence of the right type of algae and ftmgi
because of LACK of moisture
Because of `SO_(2)` pollution
because NATURAL habitat is missing

Solution :Because they are very sensitive to sulphur dioxide and 2 in cities the AMOUNT of `SO_(2)` is high so LICHEN do not grown in cities.
Discussion
50.

Liberman[s nitroso test is used to detect……………………..

Answer»


ANSWER :SECONDARY AMINE
Discussion