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.

Lines of Balmer series appear in ________ region.

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SOLUTION :VISIBLE
2.

Linear form of carbon dioxide molecule has two polar bonds. Yet the molecule has zero dipole moment. Why?

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Solution :(i) The linear form of carbon DIOXIDE has zero DIPOLE moment, even though it has two polar bonds.
(ii) In `CO_2` there are two polar bonds [C =O], which have dipole moments that are equal in magnitude but have opposite direction
(III) Hence the NET dipole moment of the `CO_(2)` is `MU= mu_1 + mu_2 = mu_1+(-mu_1)=0`
(iv) `overset(O=)underset(mu_1)rarr overset(C)("") overset(=O)underset(mu_1)larr`
(v)In this case, `mu=vecmu_1 + vec mu_2 = vecmu_1 + (-vecmu_1)=0`
3.

Linear polyenes on ozonoysis give two moles of acetaldehyde and one mole of propandial.Linear polyene will be

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Alkadiene
Allkatriene
Alkatetraene
Alkapentaene

Solution :Polyene is alkadiene
`R- CH = CH_2 - CH = CH - R to R - CHO + OHC - CH_2 - CHO + RCHO`
4.

Linear form of carbondioxide molecule has two polar bonds. Yet the molecule has zero dipole momentwhy?

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Solution :(i) The linear form of carbon dioxide has zero dinole moment, even thougn it has two polar bonds.
(ii) In `CO_(2),`there are two polar bonds [C = O] which have dipole momens that are equal in magnitude but have OPPOSITE direction.
(iii) HENCE the net dipole moment of the `CO_(2) " " is " "mu = mu_(1)+ (-mu_(1))=0`
(iv)
(v) In this case,
(iv) `underset(mu_(1))overset(O=)RARROVERSET(C=O)underset(mu_(1))LARR`
(v) In this case. `mu=vecmu_(1)+vecmu_(2)=vecmu_(1)+(-VEC(mu_(1)))=0`
5.

Line spectrum is characteristic of

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Atoms
Molecules
Any SUBSTANCE in SOLID STATE
Any substance in LIQUID state

Answer :A
6.

Lindlar's catalyst cannot be used to carry out the following process A) CH_(3)-C-=C-CH_(3)overset(H_2)rarrtrans Butene-2 B) CH_(3) -C-=C-CH_(3) overset(H_2)rarr Butane C) C_3H_2overset(H_2)rarrC_2H_6 D) CH_3-C-= CH_(3) overset(H_2)rarrcis Butene - 2

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A only
A, B and D only
A, B and C only
A, C and D only

ANSWER :C
7.

Lindlar's catalyst is

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PT in ethanol
`PD + BaSO_4 `
Ni in ethanol
NA in LIQ. `NH_3 `

Solution :Pd + `BaSO_4`
8.

Limiting line of any spectral series in the hydrogen spectrum is the line when n_(2) in the Rydberg's formula is .........

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SOLUTION :INFINITY
9.

Limited amount of Grignard's reagent and ethyl acetate reacts to give

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Acid
Ketone
Aldehyde
Alcohol

Solution :`H_(3)C - overset(O)overset(||)(C) - OC_(2)H_(5) + H_(3) overset(-)(C) overset(+)(M) gI overset("dry ether")RARR H_(3) C - UNDERSET(CH_(3))underset(|)overset(OMgI)overset(|)(C) - OC_(2)H_(5) underset(underset(-Mg(OH)I)(-C_(2)H_(5)OH))overset(H_(2)O // H^(+))rarr`
`H_(3)C - underset(CH_(3))underset(|)overset(O - H) overset(|)(C) - O - H overset(-H_(2)O)rarr underset("Acetone")(H_(3)C - overset(O)overset(||)(C) - CH_(3))`
10.

Limiting compositions of f-block hydrides are

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`MH_(2)&MH_(3)`
`MH_(3)&MH_(5)`
`MH_(2)&MH_(8)`
`MH_(2)&MH_(6)`

ANSWER :A
11.

Lime water turning milky is due to the formation of ______________

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CALCIUM carbonate
calcium hydroxide
Calcium oxide
Calcium chloride

Answer :A
12.

Lime water in aqueous solution gives white precipitate by passing carbondioxide, but the precipatate dissolves in excess carbondioxide. Why?

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Solution :Lime water gives insoluble calcium carbonate with CARBONDIOXIDE
`Ca(OH)_(2)+CO_(2)rarrCaCO_(3)darr+H_(2)O`
Calcium carbonate is soluble in water in the PRESENCE of carbondioxide, GIVING calcium bicarbonate
`CaCO_(3)+H_(2)O+CO_(2)rarrCa(HCO_(3))_(2)`
13.

Lime stone is heated at elevated temperature ........ furnace to convert it in quick lime.

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Rotary
Reverberatory
Blast
All

Solution :`CaCO_(3) OVERSET(Delta)hArr CaO+CO_(2(G))`
14.

Mortar is a mixture of

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I part of SLAKED lime, 1 part of sand and WATER
1 part of slaked lime, 3 part of sand and water
3 parts of slaked lime, 3 parts of sand and water
3 parts of slaked lime, 1 part of sane and water

Solution :Lime MORTAR is prepared by mixing of 1 part of slaked lime and 3 parts of sand and water.
15.

Like COwhy its analog of SiOis notstable.

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Solution :CO isresonance hybridof the following two structure :

Thus, CO contains`ppi-ppi`multiplebonds. This is DUE to the reason that carbonhas a STRONG tendency to FORM `ppi-ppi`multiplebonds due toits smallsize and highelectronegativity. Silicon, on theother hand , because of itsbigger size andlowerelectronetivityhas not tendency to form `ppi- ppi` multiple bonds andhence Si does not form`SIO` molecule analogousto `CO`molecule.
16.

Like aluminum Beryllium is inert towards ...... acid.

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HCl
`H_(2)SO_(4)`
`HNO_(3)`
`CH_(3)COOH`

Answer :C
17.

Like alkali metals, alkaline earth metals are also soluble in .......

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Ether
Ammonia
Alcohol
Water

Answer :B
18.

Like Al_(4)C_(3), Be_(2)C is produce ….. gas.

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Oxygen
Hydrogen
Ethane
Methane

Answer :D
19.

LiHCO_(3) and Ca(HCO_(3))_(2) are not found in solid state .

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SOLUTION :They are HIGHLY SOLUBLE in WATER
20.

LiH is an example of _______________ hydride.

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ionic
saline
covalent
both a and b

Answer :D
21.

Light waves travel with the speed of 3 xx 10^(8) "m s"^(-1). Express the speed in miles are per hour.

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Solution :`1.6 xx 10^(3)m=1" MILE" :. "conversion factor"=(("1 mol"))/((1.6xx10^(3)m))`
`3.6 xx 10^(3)s=1HR:. "conversion factor"=((1hr))/((3.6xx10^(3)s))`
`(3xx10^(8)m)/(1s)=((3xx10^(8)m)xx("conversion factor"))/(1sxx("conversion factor"))`
`=(3xx10^(8)m)xx(("1 mile"))/((1.6xx10^(3)m))xx(1)/((1s))xx((3.6xx10^(3)s))/((1hr))=6.75xx10^(8) "miles/hr"`.
22.

Light of wavelength 4000 Å falls on the surface of cesium. Calculate the energy of the photoelectron emitted. The critical wavelength for photoelectric effect in cesium is 6600 Å.

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SOLUTION :`1.95 XX 10^(-19)` JOULE
23.

Light scattering takes place in

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SOLUTIONS of electrolyte
Colloidal solution
Electrodialysis
Electroplating

Answer :B
24.

Lifetimesof themoleculesin theexcitedstatesare oftenmeasuredby usingpulsedradiationsourceof durationnearlyin the nanosecondrange. If theradiationsourcehas theemittedduringthe pulsesourceis 2.5 xx 10^(15).Calculatethe energyof the source .

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Solution :radiationduration`=2 ns = 2 xx 10^(9) s`
Frequency= `(1) /(" radiation period") = (1)/( 2xx 10^(9) s)`
total= NOOF photon(N ) `xx ` energyof 1photon(hv )E = Nhv
`=(2.5 xx 10^(15) )(1.626 xx 10^(34) J s) (5.0 xx 10^(8) s)`
`=82 .825 xx 10^(11) J`
`=8.282xx 10^(10)J`
25.

LiF is..........in water due to its high.........energy.

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SOLUTION :INSOLUBLE, LATTICE
26.

LiF is less soluble in water than KF because

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LIF more is covalent than KF
LiF has higher LATTICE ENERGY than KF
LiF has higher enthalpy of hydration than KF
Li+ ions are not extensively HYDRATED than `K^(+)` ions

Solution :Conceptual
27.

LiF is ionic but BeF_2 is covalent. Comment.

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Solution :The electronegativity difference between elements of LiF is 3 and of `BeF_(2)` is 2.5. Both are expected to be ionic. But DUE to high POLARISATION ABILITY of `Be^(2+)` ion, `BeF_(2)` is not pure ionic. It is more covalent.
28.

Liebig test is used ot estimate

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H
C
C and H both
N

Answer :C
29.

LiCl is soluble in water whereas LiBrand Lil are soluble in organic solvent. Give reason

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SOLUTION :LITHIUM chloride (LiCl) is ionic in nature and it is soluble in POLAR solvent water, whereas lithium bromide and lithium IODIDE are covalent and are soluble in non-polar ORGANIC solvents
30.

LICIO_(4) is more soluble than NaCIO_(4)Why?

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SOLUTION :The small size of the Lition MEANS that it has a very high enthalpy of hydration and so lithium salts are MUCH more soluble than the salts of other GROUP 1. E.g. `LiClO_(4)`, is upto 12 TIMES more soluble than `NaClO_(4)`
31.

Liberation of dihydrogen from methane by steam reaction ?

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Solution :METHANE reacts with steam at 1273 K in the PRESENCE of nickel catalyt to form carbon monoxide and dihydrogen. This method is used for industrial preparation of dihydrogen gas.
`UNDERSET("Methane")(CH_(4(g)))+underset("Steam")(H_(2)O_((g))) underset(1273 K)overset(Ni)rarr underset("monoxide")underset("Carbon")(CO_((g))) + underset("Dihydrogen")(3H_(2(g)))`
32.

LiAlH_(4)is used as ______.

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anoxidisingagent
areducingagent
a MODERATOR
a watersoftener

ANSWER :B
33.

LiA//H_4 is used as :

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an OXIDANT
a REDUCTANT
a mordant
water softner

Solution :(B) It is a FACT
34.

Li_2CO_3 decomposes readily whereas other carbonates are not. Why?

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Solution :The carbonates `(M_2 CO_3)` of ALKALI metals are REMARKABLY more stable UPTO 1270 K above which they first melt and then decompose to FORM their oxides, whereas `Li_2 CO_3` is less stable and decomposes readily .
`Li_2 CO_3 + Delta to Li_2 O + CO_2 uarr`
This is DUE to the large size differnece between `Li^+` and `CO_3^(2-)` which makes the crystal lattice unstable.
35.

Li_(2)SO_(4) is not isomorphous with sodium sulphate

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Due to SMALL size of lithium
Due to high COORDINATION number of lithiu
Due to high ionisation energy of lithium
Both (B) and (c)

SOLUTION :Due to small size of lithium `Li_(2)SO_(4)` is ot ISOMORPHOUS with `Na_(2)SO_(4)`
36.

Li_(2)CO_(3) decomposes readily whereas other carbonates are not. Why?

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Solution :The carbonates `(M_(2)CO_(3))` of alkali METALS are remarkably more stable upto 1270K above which they FIRST melt and then decompose to FORM their oxides, whereas `Li_(2)CO_(3)` is less stable and decomposes readily
`Li_(2)CO_(3) + /_\ to Li_(2)O + CO_(2)uarr`
This is due to the large size difference between `Li^(+)` and `Co_(3)^(2-)` which makes the crystal LATTICE unstable
37.

Li on reaction with nitrogen form which type of compound?

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Lithium-nitrite
Lithium-nitrate
Lithium - nitride
None of above

Answer :C
38.

Li, Na, K are Dobereiner triads. If the atomic weight are 7 and 39 respectively then what would be the atomic weight of Na ?

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Solution :Li (Lithium) = 7 atomic number
K (Potassium) = 39 atomic number
MIDDLE element of triads is NA
` :. ` Atomic weight of sodium = `((Li +K))/2= (7 + 39)/2 = 46/2 = 23`
39.

Li , Na and K react withdioxygen giving ……… , ………and …………respectively .

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SOLUTION :`Li_(2)O , Na_(2)O_(2) , KO_(2)`
40.

Li is not used in ......

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AERO PLANE industries.
nuclear reactor.
alloy formation.
as REDUCING agent.

Answer :B
41.

Li - Mg alloy used to prepare ......

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SOLUTION :ARMOUR PLATES
42.

Li+have high hydration fraction and therefore its salts are usually hydrated in nature.

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SOLUTION :TRUE STATEMENT
43.

Li is less vigorously react then Na, because its...

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SMALLER SIZE and HIGH HYDRATION enthalpy.
Larger size and high IONIZATION enthalpy.
High electronegativity.
Larger size and high hydration enthalpy.

Answer :A
44.

Li has the following abnormal behaviour in its grou

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Lithium CARBONATE decomposes into its oxide on heating unlike other element
LiCl is covalent in nature
Li3N is a STABLE compound
LiCl is a poor CONDUCTOR of electricity in molten state

Solution :Due to SMALL SIZE and high charge / radius ratio
45.

PCl_(5) +SO_(2) rarr A+B A overset(3H_(2)O)rarrHCl C overset("Red hot")rarrD +H_(2)O Compound 'A' is

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0
1
-2
`-1

Solution :
46.

Li (520), Be (899) and C (1086) kJ "mol"^(-1) ionisation enthalpy then B contain which are ionisation enthalpy 800 or 900 kJ/mol why ?

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Solution :`800 KJ" mol"^(-1)` because order of IONISATION ENTHALPY in second period is `Li lt B lt Be lt C`.
47.

Lewis symbol for second period

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SOLUTION :
48.

Lewis acids among the following are a) CCl_(4) b) SiCl_(4) c) GeF_(4)

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only a and B
only b and c
only a and c
a, b and c

Answer :B
49.

Let vartheta_(1) be the frequency of the series limit of the Lyman series and vartheta_(2) be the frequency of the first line of the Lyman series and vartheta_(3) be the frequency of the series limit of Balmar series, then

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`v_1 -v_2 = v_3`
`v_2 - v_1 = v_3`
`v_3 -1/2 (v_1 +v_2)`
`v_1 + v_2 = v_3`

Solution :`bar(upsilon)=1/(LAMBDA) = R(1)/(n_1^2) - 1/(n_2^2) , upsilon_1 = CR,`
`upsilon_2 = (3cR)/(4) , upsilon_3 = (cR)/(4) THEREFORE upsilon_1 - upsilon_2 = upsilon_3`
50.

Let u_(av), u_(rms)and u_(mp)are average, root means square and most probable speed the molcules in an ideal monoatomic gas at absolute temperature T. The mass of molecule is m, then :

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none of the moelcules can have a speed greater than `sqrt(2)u_(rms)`
none of the molecules can have a speed less than `sqrt(2)u_(MP)`. 
`u_(av) lt u_(rms) lt u_(mp)`
the AVERAGE kinetic energy of molecule is `3/4m u_(mp)^2`

SOLUTION :`K.E. = 1/2 mU_(rms)^2 = 1/2 m xx 3/2 xx U_(mp)^2 = 3/4 U_(mp)^2`.