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.

Order of estrification of alcohols is :

Answer»

`T gt S gt P`
`S gt T gt P`
`P gt S gt T`
NONE of these

Solution :In esterification of alcoholscleavage of `O-H` BOND in alcohols takes PLACE. The ORDER of CLEAVAGE of this bond in alcohols is `1^(@) gt 2^(@) gt 3^(@)`.
Discussion
2.

Order of arrangement of the following compounds with increasing dipole moment is:

Answer»

`I LT IV lt II lt III`
`IV lt I lt II lt III`
`IV lt I lt III lt II`
`IV lt II lt I lt III`

ANSWER :B
Discussion
3.

Orbital is the region in an atom where the probability of finding the electron around the nucleus. Orbitals do not specify a definite path according to the uncertainty principle. An orbital is described with the help of wave function Phi. Whenever an electron is described by a wave function, we say that an electron occupies that orbital. Since, many wave functions are possible for an electron, there are many atomic orbitals in an atom. Orbitals have different shapes, except s-orbitals, all other orbitals hae different shapes, except s-orbitals, all other orbital have different shapes, except s-orbitals, all other orbitals have directional character. Number of spherical nodes in an orbital is equal to (n-l-1). Orbital angular momentum of an electron is sqrt(l(l+1))h. Q. Which of the following orbitals has/have two nodal planes?

Answer»

`d_(XY)`
`d_(YZ)`
`d_(X^(2)-y^(2))`
All of these

Answer :D
Discussion
4.

Orbits are also called as stationary states. Say whether the above statement is true or false. Justify you answer.

Answer»

Solution :The statement is TURE. According to Bohr, as long as an ELECTRON remains in a particularorbit, it does not lose or gain ENERGY. This means that energy of an electron in a particular path remains constant. Therefore, these orbits are also CALLED STATIONARY states.
Discussion
5.

Orbital frequency of electron in nth orbit of hydrogen is twice that of 2nd orbit. The value of n is:

Answer»


ANSWER :1
Discussion
6.

Orbital which is not possible

Answer»

2p
3d
3S
3F

ANSWER :D
Discussion
7.

Orbital angular momentum of 3s orbitals will be ...........

Answer»


ANSWER :0
Discussion
8.

Orbital angular momentum depends on .....

Answer»

`l`
N and l
n and m
m and s

Solution :Orbital angular MOMENTUM `= SQRT(l(l +1)) (H)/(2pi)`. Hence, it depends only on 'l'
Discussion
9.

Orbital angular momentum depends on ....

Answer»

l
N and l
n and m
m and s

Solution :Orbital ANGULAR momentum MVR=`(h)/(2pi)sqrt(l(l+1))` So, it depends only on ‘Z.. Z can have VALUES ranging from 0 to (n - 1). (A) When I - 0, the subshell is 5 and orbital is spherical in shape. I (B) When Z = 1, the subshell is p and orbital is dumb-bell shaped. (C) When I = 2, the subshell is d and orbital is double dumb-bell shaped. (D) When 1 = 3, the subshell is f and orbital is COMPLICATED in shape.
Discussion
10.

Orange solution 'B' is due to formation of:

Answer»

`BiI_(3)`
`[BiI_(4)]^(-)`
`BI(OH)_(2)I`
`BIOI`

Answer :B
Discussion
11.

Orange red coloured monoxide is

Answer»

`K_(2)O`
`Na_(2)O`
`Cs_(2)O`
`Li_(2)O`

ANSWER :D
Discussion
12.

Optically active molecules among the following are

Answer»




ANSWER :A::C::D
Discussion
13.

Optically active among the following is

Answer»

MESO tartaric ACID
DL - tartaric acid
Meso - 2, 3- butanediol
Erythro - 2, 3 - dihydroxy butanoic acid

Solution :Erythro-2,3-dihydroxy butanoic acid.
Discussion
14.

Optically active compounds among the following is

Answer»

`CH_(3)-CH_(2)-CHCl-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CDCl-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CHD-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CHD-CH_(3)`

SOLUTION :
Due to the PRESENCE of chiral centre it is optically active
Discussion
15.

Optical activity is shown by a molecule which

Answer»

Contains at least THREE asymmetric centres
Is asymmetric as a WHOLE
Contains a double bond
Has a centre of symmetry

Solution :Optical activity is SHOWN by a MOLECULE which is asymmetric as a whole. (i.e. plane of symmetry or CENTER of symmetry are absent.)
Discussion
16.

Optical activity is observed for alkynes with least no.of carbons with a alkyne having minimum(no.of carbon atoms)

Answer»


SOLUTION :`H-UNDERSET(CH_2CH_3)underset(|)OVERSET(CH_3)overset(|)C-C -= C-H`
Discussion
17.

Oppenauer oxidation is the reverse process of

Answer»

Wolff Kishner REDUCTION
Resenmund reaction
clemmenson's reduction
Meerwein-Ponndorf VERLEY reduction

Solution :Meerwein-Ponndorf-Verley Reduction
Discussion
18.

Only valence electrons of atoms take part in chemical combination. Draw the Lewis representation of NF_3

Answer»

SOLUTION :
Discussion
19.

Only two isomeric monochloro derivatives are possible for…….

Answer»

n-butane
2, 4-dimethyl pentane
benzene
2-methyl propane

Answer :A::D
Discussion
20.

Only two isomeric monochloro derivatives are possible for : -

Answer»

n-PENTANE
2,4 -DIMETHYL pentane
Tollune
2,3-Dimethyl butane

Answer :D
Discussion
21.

Only three pentanoic acids have alpha- hydrogen and hence undergo HVZ reaction. These are CH_(3)CH_(2)CH_(2)CH_(2)COOH CH_(3)CH_(2)CH (CH_(3))COOH (CH_(3))_(2) CHCH_(2) COOH

Answer»
Discussion
22.

Only two isomeric mono chloro derivatives are possible for

Answer»

n-butane
2,3-dimethylbutane
benzene
2-methyl PROPANE

SOLUTION :
Discussion
23.

Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using only covalent bonds, e.g., diamond structure. The latter is based on an fcc lattice where four out of the eight tetrahedral holes are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four others. Germanium, silicon and grey tin also crystallize in the same way as diamond ("Given " sin 54^(@) 44'=0.8164) Total number of unit cells in one gram of such sample is

Answer»

`4.95times10^(21)`
`2.01times10^(22)`
`6.27times10^(21)`
`4.95times10^(22)`

Solution :`(8times12)/(6.02times10^(23)) G " Present in one cell"`
`therefore"total NUMBER of unit CELLS" = (6.02times10^(23))/(8times12)`.
Discussion
24.

Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using only covalent bonds, e.g., diamond structure. The latter is based on an fcc lattice where four out of the eight tetrahedral holes are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four others. Germanium, silicon and grey tin also crystallize in the same way as diamond ("Given " sin 54^(@) 44'=0.8164) If the edge length is same,density of carbon atom is

Answer»

`3.42 g//C c`
`2.40 g//c c`
`1.63 g//c c`
`2.58 g//c c`

SOLUTION :`d=(8times12)/(6.02times10^(23)TIMESA^(3))`
Discussion
25.

Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using only covalent bonds, e.g., diamond structure. The latter is based on an fcc lattice where four out of the eight tetrahedral holes are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four others. Germanium, silicon and grey tin also crystallize in the same way as diamond ("Given " sin 54^(@) 44'=0.8164) Total number of effective atoms in a cube is

Answer»

1
2
4
8

Solution :Total atoms at CCP = 4
Total atoms at TETRAHEDRAL VOIDS = 4.
Discussion
26.

Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using only covalent bonds, e.g., diamond structure. The latter is based on an fcc lattice where four out of the eight tetrahedral holes are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four others. Germanium, silicon and grey tin also crystallize in the same way as diamond ("Given " sin 54^(@) 44'=0.8164) If edge length of the cube is 3.60 Å, then radius of carbon atom is

Answer»

0.78 Å
0.92 Å
0.64 Å
0.35 Å

Solution :`r=a/4.62`
Discussion
27.

Only those atoms which form four covalent bonds produce a repetitive three dimensional structure using only covalent bonds, e.g., diamond structure. The latter is based on an fcc lattice where four out of the eight tetrahedral holes are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four others. Germanium, silicon and grey tin also crystallize in the same way as diamond ("Given " sin 54^(@) 44'=0.8164) Fraction of volume unoccupied is

Answer»

0.34
0.66
0.26
0.48

Solution :`sqrt2a=8r sin54^(@)44''thereforea=4.62 R`
`"PACKING FRACTION"=(8times4/3pir^(3))/a^(3)=0.34`
Discussion
28.

Only the surface atoms in an adsorbent ,play an active role in adsorption.These atoms posses some residual forces likes van der waal's forces and chemical forces. ltbRgt In the process of adsorption ,weak adsorbate is constituted by strong adsorbate. Activated charcoal used in gas mask is already exposed to the atmospheric air, so the gases and water vapours in the air are adsorbed on its surface. When the mask is exposed to chlorine atmosphric, the gases are displaced by chlorine. Porous and finely powered solids ,e.g, charcoal and Fuller's earth adsorb more as compared to the hard non-porous material.It is due to this property that the powered charcoal is used in gas masks. In general, easily liquefiable gases likesCO_(2) ,NH_(3) ,Cl_(2) and SO_(2) ,etc. are adsorbed to a greater extent than the elemental gases , e.g, H_(2), N_(2) ,O_(2), He, etc. Which of the following gas molecules has maximum value of enthalpy of physisorption in a gas mask ?

Answer»

`C_(2)H_(6)`
NE
`H_(2)O`
`H_(2)`

ANSWER :C
Discussion
29.

Only one element of .......... forms hydride.

Answer»

GROUP-6
group-7
group-8
group-9

Solution :Only one ELEMENT of group 6 i.e. Cr FORMS hydride. It is CRH.
Discussion
30.

Only one element of …………. Forms hydride

Answer»

group 6
group 7
group 8
group 9

Solution :Only ONE element of group 6 i.e., CR forms HYDRIDE.
Discussion
31.

Only one element of __________ forms hydrode.

Answer»

GROUP 6
group 7
group 8
group 9

Solution :From group 6, only one element i.e., chromium FORMS CRH.
Discussion
32.

Only mono-substitution is possible in benzene, give its structure and name.

Answer»

Solution :Mono-substituted BENZENE compounds : Benzene has all 6 SIMILAR HYDROGEN atoms, and therefoer it form only one mono-substituted product, e.g.,
Discussion
33.

Only….. Model of molecule indicate both bond as well as atoms.

Answer»

SOLUTION :BOLL and STICK MODEL
Discussion
34.

Only HCOOH reduces Tollen's reagent

Answer»
Discussion
35.

One word answers are given for the following, Mark the example which is not correct.

Answer»

Alkali metals with lowest melting point- Cs
Alkaline earth metal with higher HYDRATION ethalpy- `Ba^(2+)`
Alkaline earth metal which imparts brick red colour to the flame-`Ca^(2+)`
OXIDE of alkaline earth metal which is amphoteric in nature -BeO

Solution :Hydration ENTHALPY DECREASES with increases in ionic size down the group. `Be^(2+) gt Mg^(2+) gt Ca^(2+) gt Sr^(2+) gt Ba^(2+)`
Discussion
36.

One way of writing the equation of state for real gases is : PV=RT[1+(B)/(V)] where B is a constant. Derive an approximate expression for B in terms of the van der Waals constant a and b

Answer»

SOLUTION :According to van DER Waals equation,
`(P+(a)/(V_(2)))(V-b)=RT ""rArr "" P=(RT)/(V-b)-(a)/(v_(2)) rArr ""PV=(RTV)/(V-b)-(a)/(V)`
or `"" PV=RT (1-(b)/(V))^(-1)-(a)/(V)=RT(1+(b)/(V))-(a)/(V)""` (Neglecting higher powers of b/V)
`rArr "" PV=RT (1+(b)/(V)-(a)/(VRT))=RT[1+(1)/(V)(b-(a)/(RT))]`
Comparing with the given form of TEH equation, `B=b-(a)/(RT)`
Discussion
37.

One word answers are givenn for the following, Mark the example which is not correct.

Answer»

Alkali metals with lowest melting point- Cs
ALKALINE earth METAL with higher hydration ethalpy- `Ba^(2+)`
Alkaline earth metal which imparts brick red colour to the FLAME-`CA^(2+)`
Oxide of alkaline earth metal which is amphoteric in nature -BeO

Solution :Hydration enthalpy decreases with increases in ionic SIZE down the group. `Be^(2+) gt Mg^(2+) gt Ca^(2+) gt Sr^(2+) gt Ba^(2+)`
Discussion
38.

One unknown compound X on hydrocarbon. Which on reaction with HgSO_(4)(H^(+)) gives compound (B). (B) on oxidation gives compound (C). (C) on reaction with NaOH and soda lime gives methane. So identify A, B and C by giving chemical reaction.

Answer»

SOLUTION :The FOLLOWING GIVEN reaction is of conversion of methane from the ETHYNE.

So, A = Ethyne, B = Ethanal and C = Ethanoic acid
Discussion
39.

One unpaired electron in an atom contributes a magnetic moment of 1.1 B.M. Calculate the magnetic moment for chromium

Answer»

SOLUTION :E.C. of Cr `= [Ar] 3d^(5) 4S^(1)`, i.e., it has 6 unpaired electrons
Hence, its magnetic moment `= 6 XX 1.1= 6.6 B.M`
Discussion
40.

One unknown compound X on chlorination followed by wurtz reaction and dehydro halogenation gives ethene. So give whole reaction with stepwise manner.

Answer»

Solution :In this REACTION final product is ethene, so the WHOLE reaction is OCCURRED in following manner.

In this reaction ethene is formed from methane.
Discussion
41.

One 'U' stands for the mass of

Answer»

An ATOM of CARBON -12
`1//12^th` of the carbon-12
`1//12^th` of HYDROGEN atom
One atom of any of the element

Answer :A::B::C
Discussion
42.

One tonne of air contains 2 xx 10^(-3) g of carbon as smoke. Calculate the concentration of carbon in ppm in air.

Answer»


ANSWER :`2 xx 10^(-3)` ppm
Discussion
43.

One 'u' stands for :

Answer»

An ATOM of CARBON (C-12)
1/12 of the carbon atom (C-12)
1/12th of HYDROGEN atom
one atom of all the elements

Answer :B
Discussion
44.

One substance exist in three states - Give its example. With physical properties, chemical composition and characteristics.

Answer»

Solution :One SUBSTANCE exist different physical state. e.g., water, can exist as ice which is a solid, it can exist as liquid and it exist in the GASEOUS state as water VAPOUR or steam.
Physical properties of ice, water and steam are very different. Each three states it is `H_(2)O` and their chemical composition are same in three states.
Characteristics of three states is DEPENDS upon energy of molecules and how can molecules COMBINED in `H_(2)O`.
Discussion
45.

One saturated hydrocarbon possess two carbon and it has same type of Hydrogen atoms, so give the structure and type of hydrogen.

Answer»

Solution :Molecule is `C_(2)H_(6)` and STRUCTURE is ethane which CONTAIN all hydrogen of `1^(@)`.
`H-UNDERSET(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-H`
Discussion
46.

One reactant X on ozonolysis gives 2-product. In which molecular weight =72 mu, in structure 16 C-H sigma bond, 7C-C sigma bond and 1pi bond is there.

Answer»

Solution :(i) Molecular weight `= 72 mu`, molecular formula is `C_(4)H_(8)O.(48+8+16 = 72)`
Ozonolysis product has C=O (28) and other is alkyl group C 44 molecular weight `C_(3)H_(8)`.
(ii) Isomers are formed, having molecular formula `C_(4)H_(8)O`, X product is formed on ozonolysis.
`underset((A) C_(4)H_(8)O)(CH_(3)CH_(2)CH_(2)CHO)` and `underset((B)C_(4)H_(8)O)(CH_(3)COCH_(2)CH_(3))`
(iii) Therefire, UNSATURATED hydrocarbon structure and name is as follows :
Structure of X :
`overset(7)(CH_(3))overset(6)(CH_(2))overset(5)(CH_(2))overset(4)(CH)=underset(CH_(3))underset(|)overset(3)(C)-overset(2)(CH_(2))overset(1)(CH_(3))overset(O_(3))rarr A+B`
Name (X) : 3-methyl-hept-3-ene
(iv) Structure : `C_(8)H_(16)`
There are 16 C-H `SIGMA` bond
`7 C -C sigma`-bond
`1C-C pi`-bond
Discussion
47.

One reason on heating in excess supply of air K, Rb and Cs from superoxide on preference to oxide and peroxides?

Answer»

Solution :K, Rb and Cs are more reactive therefore, they form superoxide in preference to oxides and PEROXIDES `K^(+), Rb^(+)` and `Cs^(+)` ions are large cations and SUPEROXIDES ion `O_(2)^(-)` is also large. LARGER cations STABILIZE larger anions, therefore, they form superoxide.
Discussion
48.

One poise in kg ms^(-1) is numerically equal to

Answer»

`10^2`
`10`
`10^(-1)`
`10^(-2)`

ANSWER :C
Discussion
49.

One organic compound has empirical formula C_(2)H_(4)O.If its molecular mass is 88 "gmole"^(-1) then what will be molecular formula ?

Answer»

`C_(4)H_(8)O`
`C_(2)H_(4)O`
`C_(4)H_(8)O_(2)`
`C_(4)H_(2)O_(8)`

Answer :B::C::D
Discussion
50.

One organic compound has C=40%, O=53.34% andH=6.60% then its empirical formula is....

Answer»

`CH_(2)O`
CHO
`CH_(4)O_(2)`
`C_(2)H_(2)O`

ANSWER :A::B::C
Discussion