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.

The 4^(th) higher homologer of ethane is

Answer»

BUTANE
PENTANE
HEXANE
heptane.

Answer :C
Discussion
2.

The 4n series starts from thorium-232 and ends at :

Answer»

Lead-208
Bismuth-209
Lead-206
Lead-207.

Answer :A
Discussion
3.

The 4n series starts from Th - 232 and ends at

Answer»

Pb - 208
Bi-209
Pb- 206
Pb- 207

Solution :`Th^(232)` & `Pb^(208) IMPLIES 4n` - series
Discussion
4.

The +4 ion of which one of the following has half filled 4f subshell.

Answer»

Gd
Tb
Tm
Sm

Answer :B
Discussion
5.

The +3 oxidation states of langthanum ( Z= 57), gadolinium (Z= 64) and lutetium (Z= 71) are especially stable. Why ?

Answer»

SOLUTION :This is because in the `+3` oxidation STATE, THEYHAVE empty, half-filled and completely filled4f sub-shell respectively.
Discussion
6.

The 3d-transition series contains elements from atomic number:

Answer»

22 to 30
21 to 30
21 to 31
21 to 29

Answer :B
Discussion
7.

The +3 ion of which one of the following has filled 4f sub shell ?

Answer»

La
Nd
Gd
Ac

Solution :`Gd = [XE] 4f^(7) 5D^(1) 6S^(2)`
Discussion
8.

The +3 ion of which one of the following has half filled 4f subshell ?

Answer»

La
Lu
Gd
Ac

Answer :C
Discussion
9.

The 2nd most basic among the following in aqueous medium is

Answer»

`C_(2)H_(5)NH_(2)`
`(C_(2)H_(5))NH`
`(C_(2)H_(5))_(3)N`
`NH_(3)`

Solution :The basic strength of amines in aqueous medium depends upon the STABILITY of the conjugate acid.

The stability of conjugate acid is `2 gt 3 gt 1 gt 4`.
(2) is most stable because of less steric hinderance and greater +I effect.
When base (c) CONVERTS to conjugate acid (3), the relief in strain is more than from (a) to (1) because in (c) there is severe lone pair-bond pair repulsion due to three bulky ethyl groups than in (a), thus (3) is more stable than (1). (4) is least stable because in all others, i.e., (1), (2), (3) there is +I effect which DECREASES the +ve charge density.
Thus if conjugate acid stability is `2 gt 3 gt 1 gt 4`. HENCE basic strength order is `b gt c gt a gt d`.
So `(C_(2)H_(5))_(3)N` is second most acidic basic. Choice (c) is correct and (a), (b) and (d) are incorrect.
Discussion
10.

The 2 R - X + 2 Na overset("ether")(to) R - R + 2 Na X is an example of

Answer»

KOLBE's REACTION
FRIEDEL- CRAFTS reaction
WURTZ reaction
Cannizzaro reaction .

Answer :C
Discussion
11.

The 19th electron of chromium has which of the following sets of quantum numbers ?

Answer»

N = 3, l = 0, m = 0 , s = 1/2
n = 3, l = 2, m = -2 , s = 1/2
n = 4, l = 0, m = 0 , s = 1/2
n = 4, l = 1, m = -1, s = 1/2

Answer :C
Discussion
12.

The ""^(14)C content of a piece of wood after it is cut from a tree :

Answer»

DECREASES
increases
remains same
piece of WOOD does not have

Answer :A
Discussion
13.

The 14 possible three dimenstional crystal lattices are called͟͟͟͟͟͟͟͟

Answer»


ANSWER :BRAVAIS LATTICES
Discussion
14.

The 11^("th") most abundant element is ……………… .

Answer»

SOLUTION :PHOSPHOROUS
Discussion
15.

The 0.001 M Solution of Mg (NO_(3))_(2) is adjusted to pH9, K_(sp) of Mg (OH)_(2) is 8.9xx10^(-12). At this pH

Answer»

`Mg(OH)_(2)` will be precipitated.
`Mg(OH)_(2)` is not precipitated.
`Mg(OH)_(3)` will be precipitated.
`Mg(OH)_(3)` is not precipitated.

Solution :`pH=9,[H^(+)]=10^(-9),[OH^(-)]=10^(-5),`
`[HG^(2+)]=1xx10^(-3),[Mg^(2+)][OH^(-)]^(2)=1xx10^(-13)`
GIVEN `K_(sp)` of `Mg(OH)_(2)=8.9xx10^(-12)` which is more than `1xx10^(-13)`. HENCE `Mg(OH)_(2)` will not precipitate
Discussion
16.

The 0.1 molal aqueous solution of glucose boils at 100. 16^(@)C. The boiling point of 0.5 molal aqueous solution of sucrose will be

Answer»

`500.80 ^(@)C`
`100.80^(@) C`
`20.16 ^(@)C`
`20.8^(@)C`

Solution :As the SOLVENT is same in both solution `K_(B)` is constant we have, `Delta T = K_(b) XX m`
For 0.1 MOLE solution, `Delta T = 100.6 - 100 = 0.16^(@)C`
:. For 0.5 molal solution `delta T =0.5 xx 0.16 = 0.8^(@) C`
:. B. pt. of sucrose solution `= 100.8^(@) C`
Discussion
17.

That which is not used as gaseous fuel

Answer»

Gasoline
Acetylene
CARBON monoxide
Methane

Answer :A
Discussion
18.

Thallium (I) is oxidized Ce(IV) as follows: Tl^(+)+2Ce^(+4)toTl^(+3)+2Ce^(+3) Following are the elementary steps involved in the above reactionn If rate =k[Ce^(+4)][Mn^(+2)] then catalyst intermediate and rate determining step are respectively

Answer»

`MN^(+2), Mn^(+3),1`
`Mn^(+2),Mn^(+3),2`
`Mn^(+2),Mn^(+4),1`
`TI^(+1),Mn^(+2),1`

SOLUTION :Constant`=Mn^(+2)`
Intermediate `=Mn^(+3),Mn^(+4),r=K[Ce^(+4)[Mn^(+2)]to` 1ST step
Discussion
19.

Thallium shows different oxidation states because :

Answer»

Of inert pair effect
It is a TRANSITION metal
Of its amphoteric character
Of its HIGH reactivity

Answer :A
Discussion
20.

Th octahedral complex of a metal ion M^(3+) with four monodentate ligands L_(1), L_(2), L_(3) and L_(4) absorbs wavelengthsin the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is

Answer»

`L_(1)ltL_(2)ltL_(4)ltL_(3)`
`L_(4)ltL_(3)ltL_(2)ltL_(1)`
`L_(1)ltL_(3)ltL_(2)ltL_(4)`
`L_(3)ltL_(2)ltL_(4)ltL_(1)`

Solution :`overset("V I B G Y O R")underset("Wavelength"(LAMBDA)"increases")rarr`
According to spectrochemical series, greater the adsorption frequency, stronger is the ligand
or CFSE `(Delta_(0)) prop` Strength of ligand `prop` Frequency absorbed (v) or `prop (1)/(lambda)`
Hence, increasing order of ligand strength will be
`underset(("RED"))(L_(1))ltunderset(("Yellow"))(L_(3))ltunderset(("Green"))(L_(2))ltunderset(("BLUE"))(L_(4))`
Discussion
21.

Th esodium carbobate beadtesttest is which Na_(2)CO_(3) is alonginstead of barax it is solutionto chromiam and ____.

Answer»


ANSWER :MN
Discussion
22.

Tetrapyrrole structure has

Answer»

PYRROLE RINGS
METHYLENE linkage
Both 1 and 2
None of the above

Answer :C
Discussion
23.

Tetrahedral voids are surrounded by

Answer»

2 SPHERES.
4 spheres.
6 spheres.
8 spheres.

Answer :B
Discussion
24.

Tetrahedral nature of a carbon atom was first shown by

Answer»

Lewis
Pasteur
Biot
Vant HOFF's and Le-Bel

Answer :D
Discussion
25.

Tetrahedralgeometry of a coordination compound involves the following hybridisation.

Answer»

`sp^(3)`
`DSP^(2)`
`dsp^(3)`
`d^(2)sp^(3)`

ANSWER :A::C
Discussion
26.

Tetrammine diaqua copper (II) hydroxide is given by the formula

Answer»

`[CU(NH_3)_4] (OH)_2 . 2H_2O`
`[Cu(NH_3)_4] (OH)_2 . 2H_2O`
`[Cu(NH_3)_4] (H_2O)_2(OH)_2`
`[Cu(NH_3)_4(H_2O) (OH)_2]`

ANSWER :C
Discussion
27.

Tetrahedral complexes are high spin complexes, because

Answer»

<P>`Delta_t LT P`
`Delta_t GT P`
`Delta_t=P`
`Delta_o gt P`

ANSWER :A
Discussion
28.

Tetrahedral complexes are generally favoured:

Answer»

where the ligands are bulky
when the ligands are stronger
where the ELECTRONIC CONFIGURATION of the central metal is `d^(0) d^(5) or d^(10)` (with weak field ligands ) as there is no CFSE.
when the central metal ion has pseudo NOBLE gas electron configuration ,i.e,(n-1)`d^(10)ns^(0)NP^(0)`

Solution :Tetrahedral complexes as favoured by steric requirements, either simple electrostatic repulsion of charge ligands or vander wall's repulsion of large one.
Tetrahedral complexes are thus generally favoured by large ligands like `Cl^(-),Br^(-),I^(-) and PPh_(3)` and metal ions of three types,
(1) those with a noble gas configuration such as `Be^(2+)(ns^(º))`,
(2) those with pseudo noble gas configuration `(n-1)d^(10)ns^(0)np^(0)`, such as `Zn^(2+),Cu^(+) and Ga^(3+)`, and
(3) those transition metal ions which do not strongly favour other structure by virtue of the CFSE, such as `Co^(2+),d^(7)`.
Discussion
29.

Tetrahedral complexes are generally favoured :

Answer»

where the LIGANDS are bulky
when the ligands are stronger
where the electronic configuration of the central metal is `d^(0)d^(5)" or "d^(10)` (with WEAK field ligands) as there is no CSFE.
when the central metal ion has pseudo noble GAS electron configuration, i.e.`(n-1)d^(10)ns^(0)np^(0).

Answer :A::C
Discussion
30.

Tetrahalides of IVB group 14 elements are:

Answer»

Ionic
Covalent
Polar
Coordinate covalent

Answer :B
Discussion
31.

Tetragonal crystal system has the followingunit cell dimensions :

Answer»

a=b=c and `alpha=beta=gamma=90^(@)`
`a=bnec` and `alpha=beta=gamma=90^(@)`
`anebnec` and `alpha=beta=gamma=90^(@)`
`a=bnec` and `alpha=beta=90^(@),gamma=120^(@)`

Solution :TETRAGONAL SYSTEM has the unit CELL DIMENSION
`a=bnec,alpha=beta=gamma=90^(@)`
Discussion
32.

Tetraethyl lead is used as:

Answer»

FIRE extinguisher
Antiknock compound
Pain killer
Mosquito killer

Answer :B
Discussion
33.

Tetraehtyl lead (TEL)in petrol wasusedas antinockingagent,whichcreateslead pollution in order to avoidlead pollutionthesubstitueof TEL is used in unleaded -petrolthe . Subsitiute is denoted as :

Answer»

AK-33-X
BK-33-K
CK-33-X
DK-33-X

Solution :AK-33-x , ischemicalname iscyclopentadienyl MANGANESE tricarbonyl.
Discussion
34.

Tetracylin, erythromycin and chloramphy are

Answer»

bactericidal
bactericidal antibiortic
antiviral drugs
antimalerial drugs

Answer :B
Discussion
35.

Tetracycline and chloroamphenicols are _____.

Answer»

SOLUTION :BROAD SPECTRUM ANTIBIOTICS
Discussion
36.

Tetrachloromethane or carbon tetrachloride (C""Cl_(4)).

Answer»

Solution :(1) CARBON tetrachloride is USED in the MANUFACTURE of refrierants.
(2) It is used as a dry cleaning agent and as a pesticide for STORED grains.
Discussion
37.

Tetracyanoethylene has a formal C=C double bond length of X in the free ligand but in the complex [Pt{C_(2)(CN)_(4)}Cl_(3)]^(-) the C-C bond length will be :

Answer»

GREATER than X
Less than X
Equal to X
None of these

Answer :A
Discussion
38.

Tetrabromoethane on treatment with Zn gives

Answer»

ETHYL bromide
Ethane
Ethene
Ethyne.

Solution :
Discussion
39.

Tetraaminecopper (II) ion is square planar complex with one unpaired electron. According to valence bond theory the hybrid state of copper should be

Answer»

`sp^3`
`sp^2`
`dsp^2`
`sp^3d`

ANSWER :C
Discussion
40.

Tetra-alkyl ammonium salts are called ____salts.

Answer»

SOLUTION :QUATERNARY AMMONIUM SALTS
Discussion
41.

Tetiary aminedoes not reactwith CH_(3)COCl becauseof

Answer»

leastbasic
no hydrogenatom on nitrogen
three alkyl GROUP
all OFTHESE

ANSWER :B
Discussion
42.

Tests on an aqueous solution of a sodium salt having an anion X^(n-) gave the following results: anion X^(n-) overset(K_(2)Cr_(2)O_(7) // H^(+))to "green solution + gas" overset(Pb(NO_(3))_(2))to "black precipitate". Which one of the following could be X^(n-):

Answer»

`I^(-)`
`NO_(2)^(-)`
`S^(2-)`
`SO_(4)^(2-)`

ANSWER :C
Discussion
43.

Tetahedral complex bis(benzylacetonato) beryllium (II) depicts

Answer»

GEOMETRICAL isomerism
OPTICAL isomerism
Ionisation isomerism
None of the above.

Solution :It shows optical isomerism because it has NON SUPER imposable MIRROR images.
Discussion
44.

Testostrom, progesteron , norethindrone , ethynylestradial. Among them how many compounds contains steroid ring .

Answer»


SOLUTION :All CONTAINS STEROID STRUCTURES
Discussion
45.

Test to differentiatebetween underset(("methanol"))((CH_(3)OH)) and underset(("Phenol"))((Ph-OH)) is/are :

Answer»

Litmus test
`FeCl_(3)`
`Br_(2)//H_(2)O`
All of these

Solution :
`FeCl_(3)` is test for enoland phenol is ENOL.
Discussion
46.

Test used to identify peptide linkage in protein is :

Answer»

BIURET
NINHYDRIN tes
Molisch
2,4-DNP test

Solution :N//A
Discussion
47.

Testosterone is

Answer»

ANIMAL steroid
plant steroid
ester of LONG CHAIN FATTY acid
triolein

Answer :A
Discussion
48.

Test tube (A)contains aqnous zine acetate solutionwhiletest tube (B) contains aq, zine chloride solution .What happens if H_(2)S gas is passed into each solution ?

Answer»

SOLUTION :(A) `Zn(CH_(2)COO)_(2) + H_(2) S RARR ZnS + CH_(3) COOH`
(B) `ZnCI_(2) + H_(2)S rarr ZnS + 2HCI`
`ZnS` is percipitated in (A)
In (B) `HCI` is formed which dissolves `ZnS`
Discussion
49.

Test (Q). A compound 'X' was fused with Na metal and the extract gave a white precipitate with AgNO_(3). The Lassaigne's extract gave red colouration with neutral FeCl_(3). Test R while compound 'Y' when fused with Na metal and subsequent analysis on its Lassaigne's extract did not give any characteristic test. Test S while compound Y on fusion with fusion mixture (sodium carbonate+potassium nitrate) or Na_(2)O_(2), followed by extraction, followed by addition of conc. HNO_(3) and ammonium molybdate gave a yellow precipitate Q. The formula of yellow precipitate is

Answer»

`(NH_(4))_(3)PO_(4)`
`Mo(PO_(4))`
`(NH_(4))_(3)PO_(4).12Mo_(3)O_(3)`
`Mo(PO_(4))_(2)`

ANSWER :C
Discussion
50.

Test (Q). A compound 'X' was fused with Na metal and the extract gave a white precipitate with AgNO_(3). The Lassaigne's extract gave red colouration with neutral FeCl_(3). Test R while compound 'Y' when fused with Na metal and subsequent analysis on its Lassaigne's extract did not give any characteristic test. Test S while compound Y on fusion with fusion mixture (sodium carbonate+potassium nitrate) or Na_(2)O_(2), followed by extraction, followed by addition of conc. HNO_(3) and ammonium molybdate gave a yellow precipitate Q. The chemical reaction taking place in Y, when it is fused with fusioni mixture is

Answer»

`3Na+P+4O to Na_(3)PO_(4)`
`3Na_(2)CO_(3)+2P+[O] to 2Na_(3)PO_(4)+3CO_(2)`
`3KNO_(2)+P+3Oto K_(3)PO_(4)+3NO_(2)+O_(2)`
none of these

Answer :B
Discussion