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 Balmer series of hydrogen atom spectrum which electronic transition causes third line ?

Answer»

Fifth Bohr orbit to SECOND one
Fifth Bohr orbit to first one
Fourth Bohr orbit to second one
Fourth Bohr orbit to first one

Solution :`1^(ST)` line is for `n_(3) - n_(2), 2^("ND")` for `n_(4) to n_(2)` and `3^("rd")` for `n_(5)ton_(2)`.
2.

In balancing the half reaction S_(2)O_(3)^(2-)toS_((s)) the number of electrons that must be added is:-

Answer»

4 on the left
3 on the right
2 on the left
2 on the right

Solution :In this REACTION 4 electrons are needed for the reaction VOLUME.
3.

In azo-couplingthe substitutionusuallyoccurs at ____positionto theactivatinggroup

Answer»

ORTHO
META
PARA
ortho andmeta

SOLUTION :ElectrondoN/Ating GROUP
4.

In bakelite , the rings are joined to each other through

Answer»

`-CH_2-`
`-O-`
`-undersetHunderset(|)OVERSET(OH)overset(|)C-H`
`-undersetOunderset(||)C-`

ANSWER :A
5.

In (B), there is increase in conductance before (though slightly) and after the end point. This

Answer»

FORMATION of NaCl type SALT from STRONG acid which is HYDROLYSED releasing NaOH `NaCl+H_2O= NaOH+HCI`
formation of `CH_3COONA` type salt from weak acid which is hydrolysed releasing NaOH
ionization of water
autoprotolysis of water

Answer :B
6.

In AX_(5) type sof moleculeif 'A' undergoes sp^(3)d^(2) hybridisation, then the shape of the molecule is

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T- shape
Octahedral
SQUARE PYRAMIDAL
Tetrahedal

ANSWER :C
7.

In B_2H_6 :

Answer»

There is a DIRECT boron-boron bond
The STRUCTURE is similar to that of `C_2H_6`
The born ATOMS are linked through HYDROGEN bridges
All the atoms are in one plane

Answer :C
8.

In Avogadro's number be 3.01xx10^(23) then the atomic mass of carbon will be:

Answer»


ANSWER :6
9.

In auto catalysis_____acts as catalyst.

Answer»

reactant
product
both
none

Answer :B
10.

In athe extraction of copper, the copper matte is a mixture is

Answer»

COPPER (II) Sulphide and IRON (II) Sulphide
Copper (II) Sulphide and iron (II) Sulphide
Copper (I) Sulphide and iron (II) Sulphide
Copper (I) Sulphide and iron (II) Sulphide

ANSWER :C
11.

In aryl halides, what is the hybridisation of carbon atom to which halogen is attached?

Answer»

Solution :In aryl HALIDES, the hhybridisation of carbon atom to which HALOGEN is attached is `SP^(2)`
12.

In assigning R-S configuration which among the following groups has highest priority?

Answer»

`-SO_(3)H`
`-COOH`
`-CHO`
`-C_(6)H_(5)`

ANSWER :A
13.

In Arrhenius equation , k=Ae^(E_a/(RT)) , A may not be termed as rate constant .

Answer»

When 100% REACTANT will CONVERT into the product
When the temperature BECOMES infinite.
When the fraction of molecule crossing over the energy barrier becomes unity
At very LOW temperature

Answer :D
14.

In Arrhenius equation for a certain reaction, thevalue of A and E_a (activation energy) are 4 x 10^(13) s^(-1) and 98.6 kJ mol^(-1) respectively If the reaction is of first order, then at what temperature will its half life period be 10 minutes? [log 2 88 = 0.4594]

Answer»

350.5 K
270.2 K
453.6 K
311.5 K

Answer :D
15.

In Arrhenius equation, If EA is positive and T_(2)gt T_(1), then ___________

Answer»

`k_(1)=k_(2)`
`k_(2)LT k_(1)`
`k_(1)=(1)/(k_(2))`
`k_(2)GT k_(1)`

ANSWER :D
16.

In Arrhenius equation for a a certain reaction , the values of A and E_a (activation energy ) are 4xx10^(13)"sec"^(-1) and 98.6"kJ mol"^(-1) respectively. If the reaction is of order , at what temperature will its half life period be 10 minutes ?

Answer»

Solution :ACCORDING to the Arrhenius equation,
`k=Ae^(-E_a/(RT))`
or `log_ek=log_eA-(E_a)/(RT)`
or `2.303log_(10)k=2.303log_(10)A -E_a/(RT)`
`t_(1//2)=(0.693)/k(0.693)/600`
`k=(0.693)/(600)sec^(-1)(t_(1//2)=10min=600sec)`
`=1.1xx10^(-3)sec^(-1)`
Hence , LOG `(1.1xx10^(-3)`
`=log(4xx10^(13))-(98.6xx10^3)/(2.303xx8.314xxT)`
`T=310.95K`
17.

In Arrhenius equation K=Ae^-Ea//RT ,the quantity-E_a//RT is referred as :

Answer»

BOLTZMANN factor
Frequency factor
Activation factor
None of the above

Answer :A
18.

In aqueous solutions, the basic strength of amines decreases in the order

Answer»

`CH_(3)NH_(2) GT (CH_(3))_(2)NH_(2) gt (CH_(3))_(3)N`
`(CH_(3))_(2)NHgt(CH_(3))_(3)Ngt CH_(3)NH_(2)`
`(CH_(3))_(3)Ngt(CH_(3))_(2)NHgt CH_(3)NH_(2)`
`(CH_(3))_(2)NH gtCH_(3)NH_(2)gt(CH_(3))_(3)N`

Solution :Basic strength of mcthylamines DECREASES in the order:
`2^(@) gt 1^(@) gt 3^(@) i.e., (CH_(3))_(2)NH gt CH_(3)NH_(2)gt (CH_(3))_(3)N`
19.

In Arrhenius equation

Answer»

The Arrhenius constant has UNITS of RATE of reaction
The Arrhenius constant has the units of rate constant of reaction
If `E_(a)` is high, rate of reaction is slow
If `E_(a)` is high, rate of reaction is high

Solution :rate `prop(1/(E_(a)))` and `K=A.E^(-E_(a)//RT)`
20.

In aqueous solutions H_(2) SO _(4) ionises as: H _(2) SO_(4) + H _(2) O hArr H _(2) SO _(24 ) ^(-) + H _(3) O ^(+) , K a _(1) H_(2) SO _(4) ^(-) + H_(2) O hArr SO _(4) ^(2-) + H_(3) O ^(+) , Ka _(2). The relation between Ka _(1) and Ka _(2)is

Answer»

`Ka_1ltKa_2`
`Ka_1gtKa_2`
`Ka_1=Ka_2`
`2Ka_1=3Ka_2`

ANSWER :B
21.

In aqueous solutions Eu^(2+) acts as

Answer»

an OXIDISING AGENT
a reducing agent
neither OXIDANT nor reductant
can act as redox agent

ANSWER :B
22.

In aqueous solution which of the following colour is exhibited by Nici,

Answer»

PINK
GREEN
green
yellow

Answer :B
23.

In aqueous solutions Cu^+ is less stable than Cu^(2+)

Answer»


ANSWER :A
24.

In aqueous solution the ionization constants for carbonic acid are : K_(1)=4.2xx10^(-7)and K_(2)=4.8xx10^(-11) Select the correct statement for a saturated 0.034 M solution of carbonic acid.

Answer»

The concentration of `H^(+)` and `HCO_(3)^(-)` are approzimately EQUAL.
The concentration of `H^(+)` is double that of `CO_(3)^(2-)`
The concentration of `CO_(3)^(2-)` is 0.034 M.
The concentration of `CO_(3)^(2-)` is GREATER than that of `HCO_(3)^(-)`

Solution :`H_(2)CO_(3)hArr H^(+)+CO_(3)^(2-), K_(2)=4.8xx10^(-11)`
Since `K_(1)gt gt K_(2)` Conc. of `H^(+)` and `HCO_(3)^(-)`are approximately equal.
25.

In aqueous solution, weak electrolytes dissociates:

Answer»

Completely
To a SLIGHT extent
Almost completely
To more the `80 %`

ANSWER :B
26.

In aqueous solution the ionization constant for carbonic acid are K_(1) = 4.2 xx 10^(-7) and K_(2) = 4.8 xx 10^(-11) Select the correct statement for a saturated 0.034 M solution of the carbonic acid

Answer»

The concentration of `H^(+)` is double that of `CO_(3)^(2-)`
The concentration of `CO_(3)^(2-)` is 0.034 M
The concentration of `CO_(3)^(2-)` is GREATER than that of `HCO_(3)^(-)`
The concentration of `H^(+)` and `HCO_(3)^(-)` are approximately equal

Answer :D
27.

In aqueous solution the following undergoes disproportination reaction

Answer»

`Cr^(5+)`
`MN^(6+)`
`CU^+`
All

Answer :D
28.

In aqueoussolutionof anamino acid, the carboxylgroup can gain a proton and amino acidcan losea protons hydrogen bonds.

Answer»


ANSWER :FALSE
29.

In aqueoussolutionof amino aicds mostly existsin ,

Answer»

`NH_(2) - CH (R) - CO OH`
`NH_(2) - CH(R) - CO O`
`H_(3)N^(+) - CH(R) - CO OH`
`H_(3) N^(+) - CH(R) - CO O^(-)`

Answer :D
30.

In Aqueous solution, metal ion M_(1) reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes respectively. An aqueous solution of another metal ion M_(2) always froms tetrahedral complexes with these reagents. Aqueous solution of M_(2) on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given ahead Reagent S is

Answer»

`K_(4)[Fe(CN)_(6)]`
`Na_(2)HPO_(4)`
`K_(2)CrO_(4)`
`KOH

Solution :REAGENT (S) is KOH
31.

In Aqueous solution, metal ion M_(1) reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes respectively. An aqueous solution of another metal ion M_(2) always froms tetrahedral complexes with these reagents. Aqueous solution of M_(2) on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given ahead M_(1), Q and R respectively are

Answer»

`Zn^(2+)` , KCN and HCl
`NI^(2+)` , HCl and KCN
`CD^(2+)` , KCN and HCl
`Co^(2+)` , HCl and KCN

Solution :Among the four available options , `Ni^(2+)` as well as `Zn^(2+)` ions have the ability to form TETRAHEDRAL as well as square planar complexes. This depends upon the availability of the reagents used `Ni^(2+)` ion on reaction with KCN (R) FORMS a square planar complex since `CN^(-)` ion is a strong field ligand. It also reacts with HCl (Q) to form a tetrahedral complex which is quite stable

`Zn^(2+)` ion on the other hand forms tetrahedral complexes with both `CN^(-)` and `CL^(-)` ions because the ion has `d^(10)` configuration which is quite stable.

The sequence of the reaction involved is given.
32.

In aqueous solution of amino acids mostly exists in,

Answer»

`NH_(2)CH( R)-COOH`
`NH_(2)-CH( R)-COO^-`
`H_(3)N^+CH(R )-COOH`
`H_(3)N^(+)-CH( R)-COO^-`

Solution :`H_(3)N^+-CH( R)-COO^-`
33.

In Aqueous solution meta aluminate ion exists as

Answer»

NEUTRAL complex
Cationic complex
Cationic DOUBLE SALT
ANIONIC complex

Answer :D
34.

In aqueous solution H,_2SO_4, ionise in two steps. In first steps it produces ?

Answer»

`SO_(4)^(2-)`
`SO_(3)^(2-)`
`HSO_(4)^(-)`
`S^(2-)`

Answer :C
35.

In aqueous solution, glucose remains as-

Answer»

only in OPEN chain form
only in PYRANOSE form
only in furanose form
in all three forms in equilibrium

Solution :
The six-membered CYCLIC form is generally reffered to as the 'pyranose' form, and the five-membered cyclic form is called the 'furanose' form.
Closure of the ring creates a chiral center at `C-1` resulting in TWO diastereomers (sometime called ''anomers'') the alpha `(alpha)` and beta `(beta)` forms
36.

In aqueous solution glucose remains as

Answer»

only in open CHAIN FORM
only in PYRANOSE form
only in furanose form
in all THREE forms in equilibrium.

Solution :In aqueous solution, glucose exists in all the three forms that is open chain form, pyranose form and furanose form in equilibrium with each other.
37.

In aqueous solution glucose exist in how many isomeric forms.

Answer»


ANSWER :`3.00`
38.

In aqueous solution, H_2 does not reduce:

Answer»

`Pd^(2+)`
`CU^(2+)`
`ZN^(2+)`
`Ag^+`

Answer :C
39.

In aqueous solution, Eu shows stable O.S.

Answer»

`+3`
`+4`
`+3`
`+5`

ANSWER :B
40.

In aqueous solution Eu^(2+) acts as :

Answer»

An OXIDISING AGENT
Reducing agent
Can act as either of these
Can act as REDOX agent

Answer :B
41.

In aqueous solution, cerium shows stable O.S.

Answer»

`+2`
`+4`
`+3`
`+5`

ANSWER :B
42.

In aqueous solution Cd(II) exists as 6-coordinates aqua-complex. The solution on treatment with Br^(-) ion forms finally 4-coordinate Cd(II) complex with Br^(-) ions successively in four step with stepwise formation constants, logK_(1) = 1.56, log K_(2) = 0.54, log K_(3) = 0.06 and log K_(4) = 0.37. The higher value of K_(4) as compound to K_(3) is due to

Answer»

transformation of trigonal COMPLEX into tetrahedral complex.
change from octahedral to tetrahedral geometry
increase in `DeltaH` of the REACTION.
Large increase in the number of particles when fourth `Br^(-)` enters the COORDINATION ZONE.

Answer :D
43.

In aqueous phase basicity of amines is in order NH_(3) lt (CH_(3))_(3)N lt CH_(3)NH_(2) lt (CH_(3))_(2)NH It is due to the following facts

Answer»

ammonium ions in solution are stabilised not only by alkyl GROUPS but also by HYDROGEN bond DONATION to the solvent.
Addition of the proton increases crowding and thus STRAIN set up which being highest in `3^(@)` amines decreases its basic character
both of the above
none of the above

Answer :C
44.

In aqueous phase, most basic among the following is ________.

Answer»

ammonia
trimethylamine
methylamine
dimethylamine

Answer :D
45.

In aqueous alkaline solution, two electron reduction of HO_(2)^(-) gives

Answer»

`HO^-`
`H_2 O`
`O_2`
`O_(2)^(-)`

SOLUTION :
46.

In aqueous medium behavior of SO_2 is similar to that of

Answer»

`SO_(3)`
`CO_(2)`
`NO_(2)`
`CS_(2)`

ANSWER :B
47.

In aquaregia____is responsible for dissolving gold and platinum.

Answer»

SOLUTION :NASCENT CHLORINE
48.

In any unimolecular reaction……

Answer»

only one reacting SPECIES is involved in the rate determining step
the order and the molecularity of slowest step are equal to one.
the molecularity of the reaction is one and order is zero.
Both molecularity of the reaction is one and order is zero.

Solution :Molecularity of reaction =1 so in its rate determine step only one species, So.Order-molecularity=1
49.

In any unimolecular reaction ………………… .

Answer»

only ONE reacting SPECIES is involved in the rate determining step
the ORDER and the molecularity of slowest step are equal to one
the molecularity of the reaction is one and order is zero
both molecularity and order of the reaction are one.

Answer :A::B
50.

In any unimolecular reaction ___________

Answer»

Only two REACTING species is involved in the rate determining STEP.
The ORDER and the molecularity od slowest step are equal to ONE.
The molecularity of the reaction is one and order is zero.
Both molecularity and order of the reaction are one.

Answer :B