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.

Which of the following compounds are readily soluble in water ?

Answer»

`BeSO_(4)`
`MgSO_(4)`
`BaSO_(4)`
`SrCO_(4)`

SOLUTION :Hydration ENERGY of `Be^(+2)`and `Mg^(+2)`is high, it decreases down the GROUP as the size increases.
Therefore, ` BeSO_(4)` and `MgSO_(4)`are easily SOLUBLE.
2.

What is oxidation number of Pt in [Pt(C_(2)H_(4))Cl_(4)]^(-) ?

Answer»

SOLUTION :`Pt+0+4(-1)=-1`
`thereforePt-4=-1`
`thereforePt=+3`
3.

StateKelvin-Planck statementof secondlawof thermodynamics.

Answer»

SOLUTION :Kelvin PLANK statement : It is impossible to CONSTRUCT a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process WITHOUT transferring a part of heat to a cold sink.
4.

Write the IUPAC names of the following compounds (i) {:(CH_(3)-CH-CH-CH_(3)),("||"),(""C_(2)H_(5) " C"_(2)H_(5)):} (ii) (CH_(3))_(3)"CC"_(2)H_(5)

Answer»

SOLUTION :
5.

whatchangesoccursin energyof Bohr'sorbit ?

Answer»

INCREASE awayfromnucleus
does not changeaway fromnucleus
decreaseaway fromnucleus
NONE

ANSWER :A
6.

What will happen if the equilibrium system is kept in ice bath ? underset"brown"(2NO_(2(g))) hArr underset"colourless"(N_2O_(4(g))), DeltaH=-ve

Answer»

COLOUR INTENSITY INCREASES.
Colour intensity does not change.
Colour intensity increases and than REMAINS constant.
Colour intensity decreases.

Answer :D
7.

Which of the following has the highest molarity ? (i) 4 g of NaOH per 100 cm^3 of water (ii) 4 g of HCI per 100 cm^3 of water (iii) 4 g of H_2SO_4 per 100 cm^3 water

Answer»


Solution :`therefore` MOLARITY of a solution is GIVEN by:
`w=(MM.V)/1000`
(w = weight of solute in grams, M. = gram molecular MASS, V = VOLUME of solution in cm ). `therefore` The molarity of NaOH solution is given by
`4=(M xx 39.998 xx 100)/1000`
or M=1.000
Similarly for HCI solution,
`4=(M xx 36.458 xx 100)/1000`
or `M=1.0971`
and for `H_(2)SO_(4)` solution,
`4=(M xx 98.076 xx 100)/1000`
or M = 0.40785
Hence, the HCI solution has the highest molarity.
8.

Which of the reaction does not give quantitative yield of the expected product ?

Answer»




SOLUTION :NITROBENZENE

Maximum YIELD of `93%` is obtained in this reaction.
9.

What are the hybridization and shapes of the following molecules? (i) CH_(3)F (ii) HC-=N

Answer»

(i) `sp^(2)`, TRIGONAL PLANAR, (ii) `sp^(3)`, TETRAHEDRAL
(i) `sp^(3)`, tetrahedral, (ii) sp, linear
(i) sp, linear, (ii) `sp^(2)`, trigonal planar
(i) `sp^(2)`, trigonal planar, (ii) `sp^(2)`, trigonal planar

Answer :B
10.

Which of the following aromatic compounds undergoes Friedel-Crafts alkylation with methyl chloride and aluminum chloride?

Answer»

BENZOIC acid
nitro benzene
aniline
TOLUENE

SOLUTION :Toluene
11.

Which of thefollowingstatements are correct?

Answer»

Helium hashighest firstionisationenthalpyin the periodictable.
Chlorinehas lessnegativeelectrongas enthalpy thanfluorine .
Mercury andbromineare liquidsat room temperature .
In anyperiodatomicradiusof alkali metal the highest .

Solution :Statement(a) iscorrectsinceheliumbeing thesmallernoblegas hasthehighest ionization enthalpy a Dut toelectron- electronrepulsions in thesmallF atom. Theelectrongainenthalpyof F islowerthan THATOF C1 , i.e.,option( b) iswrong . Statement( c) iscorrect .
Statement( d) isbecausein any periodnoblegas has thelargestatomicradius . Please notethat inthe NCERT exemplar the answerto thisstatementis given tobe correctwhichis WRONG .
12.

What is average speed (barc or u_(av)) ?

Answer»

Solution :It is the arithmatic mean of the SPEED of different MOLECULES of the gas at a given TEMPERATURE if `n_(1)` molecules have speed `v_(1),n_(2)` molecules have speed `v_(2)n_(2)` molecules have speed `V_(3)` and so on. Average speed is related to the molar mass (M) of the gas the temperature (T) as follows.
`barC (or U_(av))= sqrt((8RT)/(pi M))`
13.

Which of the following correctly depicts the fact that identical chemical equilibrium can be attained through reversible reaction H_2+I_2=2HI from either direction?

Answer»




Solution :If `1 mol` of `H_2` and `1 mol` of `I_2` are taken in bulb `I` at `500^@C` and `2 mol` of `HI` are taken in an identical bulb `II` at `500^@C`, the intensity of COLOUR in I decreases due to the consumption of `I_2` while the intensity of colour in II INCREASES due to the formation of `I_2`. Ultimately, both have the same intensity of COLOR. In all these graphs, `t_c` is the time at which equilibrium is attained.
14.

Which is correct about tailing of Hg?

Answer»

It is DUE to `Hg_2O`
It is due to HGO
It is REMOVED by `H_2O_2`
It is removed by `O_3`

Solution : He gets oxidised by `O_3` to give sticking nature on GLASS
`2Hg+O_3rarrHg_(2)O+O_2`
`Hg_(2)O+H_2O_2rarrHg_(2)+H_(2)O+O_2`
15.

Which of the following options is correct w.r.t. ideal gas

Answer»

<P>`((delH)/(delT))_(P)-((delH)/(delT))_(V)=R`
`((delH)/(delT))_(P)LT((delH)/(delT))_(V)`
`((delH)/(delT))_(P)/((delU)/(delT))_(V)=gamma("Poisson's RATIO")`
`((delU)/(delT))_(P)-((delH)/(delT))_(p)=R`

Answer :C
16.

Which of the following solutions have the same concentration ?

Answer»

20 g of NAOH in 200 mL of solution
0.5 mol of KCl in 200 mL of solution
40 g of NaOH in 100 mL of solution
20 g of KOH in 200 mL of solution

Solution :(A) MOLARITY (M) `=("Weight of NaOH"xx1000)/("MOLECULAR weight of NaOH"xx V (mL))`
`= (20xx1000)/(40xx200)=2.5M`
(B) `M=(0.5xx1000)/(200)=2.5M`
(C) `M=(40xx1000)/(10xx100) = 10M`
(D) `M = (20xx1000)/(56xx200) = 1.785 M`
HENCE, 20 g NaOH in 200 mL of solution and 0.5 mol of KCl in 200 mL have the same concentration.
17.

Which of the following is the best reducing agent and why? Li,Cu, Br_2, F_2, H_2, K.

Answer»

SOLUTION :Lithium (Li) is best REDUCING AGENT because it has the lowest STANDARD reduction potential, i.e., `Li^(+)` is the most stable AMONGST these.
18.

What are enantiomers?

Answer»

Solution :(i) An optically active substance may exist in two or more ISOMERIC forms which have same physical and chemical properties but DIFFER in terms of DIRECTION of rotation of plane POLARISED light, such OPTICAL isomers which rotate the plane polarised light with equal angle but in opposite directions are known as enantiomers and the phenomenon is known as enantiomerism.
(ii) Isomers which are non-super imposable mirror images of each other are called enantiomers.
19.

Which of the following oxides can act as amphoteric ?

Answer»

`SiO_2`
`GeO_2`
`SnO_2`
`PbO_2`

ANSWER :B::C
20.

Which of the following pairs are positive isomers ?

Answer»

I and II
II and III
II and IV
III and IV

Solution :II and III REPRESENT position isomers. They differ in the position of carbonyl groups.
21.

Which of the following aromatic compound will undergo nitration most easily?

Answer»




SOLUTION : will UNDERGO NITRATION most EASILY
22.

Which oxidizing agent required to convert sulphur dioxide into sulphur trioxide ?

Answer»

DUST
`H_2 O_2`
`NO_2`
All of these

Solution :All of these
23.

Which of the following oxide is basic ?

Answer»

`Na_(2)O`
`P_(4)O_(10)`
`CO_(2)`
`Al_(2)O_(3)`

Solution :As we move from left to RIGHT in a period the basicity of oxide decreases.
`:. Na_(2)O gt Al_(2)O_(3) gt CO_(2) gt P_(4)O_(10)`
24.

Which of the folllowing attain the linear structrue :

Answer»

`Be Cl_(2)`
`NCO^(+)`
`NO_(2)`
`CS_(2)`

Solution :`BeCl_(2) (Cl-Be-Cl) and CS_(2) (S=C=S)` both are linear `NCO^(+)` is non-linear . HOWEVER, remember that
`""^(-)NCO(""^(-)N= C = O )`is linear because it is isoelectronic with `CO_(2)`.
`NO_(2)`is angular with angle `132^(@)`and each O-N BOND length of `1.20Å` (intermediate between single
and double bond) .
25.

Which gas is released on reaction of alkali metal hydride with water ?

Answer»

Oxygen
Hydrogen
Nitrogen
Sulphur dioxide

Answer :B
26.

When the screening effect increases, ionisation energy

Answer»

DECREASES
INCREASES
First increases and then decreases
Remains constant

Answer :A
27.

Which ordering of compounds is according to the decreasing order of the oxidationstate of nitrogen ?

Answer»

`HNO_3,NO,NH_4,Cl,N_2`
`HNO_3,NO,N_2,NH_4Cl`
`HNO_3,NH_4Cl,NO,N_2`
`NO,HNO_3,NH_4Cl,N_2`

Solution :Oxidation STATE of nitrogen in compounds is `HNO_3(+5),NO(+2), N_2(0), NH_4Cl(-3)`.
28.

What is orbital? Explainpointand boundarysurfacediagramof orbitals

Answer»

Solution :Orbital : The areaaround the nucleuswhereorbital . In eachorbitalsuch AREA is maximumis calledprobabilityfindelectronis maximum approx`90%`Forthis `w^(2)`is smallbut itgivedefintevaluefordefinitedistancefromnucleus. it isnotthat enclosea region of 100%probability.so wedrawthe figurefor 90%PROBABILITYOF findingelectron.
Pointand surfacediagramof orbital: in pointdiagrampointsshownforprobability .Bycombiningpointsshapeof orbitalis given. inboundarysurfaceor countersurfaceis drawn foranorbitalin WHICHTHE valueandprobabilitydensity`w^(2)`constant.
e.g. the boundarysurfacediagramof 1s,2s ,`2p_(x ) 2p_(y)2p_(Z)` orbitalare GIVE .
29.

Which of the following statements aboutCO_(3)^(2-) ion are correct ?

Answer»

The C-O bond order is 1.33
The formal charge on each oxygen atom is
0.67 units
It has two C-O single bonds and ONE C=O
double bond
The HYBRIDIZATION of central atom is `sp^(3)`

Solution :
Bond order `=(4)/(3) =1.33`
Formal charge `=(2)/(3) =0.67`
All bonds are equivalent due to resonance . Hence ,
(c) is wrong . Hybridization of central ATM is`sp^(2)` .
Hence (d) is wrong .
30.

Which of the following has a net dipole moment ?

Answer»

`NO_(3)^(-)`
`C Cl_4`
`BeF_2`
`SO_2`

ANSWER :D
31.

trans-Pent-2-ene is polar with trans-but-2-ene is non polar. Explain.

Answer»

Solution :In trans-but-2-ene, the diple momens of the C-`CH_3` bonds are equal and opposite and HENCE they EXACTLY cancel out each other. Thus, trans-but-2-ene is non-polar.

However, in trans-pent-2-ene, the +I-effect of `CH_3CH_2`-group is higher than of `CH_3` group, therefore, the DIPOLE moment of `C-CH_3` and `C-CH_2CH_3` bonds are unequal . Although these two DIPOLES oppose each other , yet they do not exactly cancel out each other and hence trans-pent-2-ene has a small but finite dipole moment and thus is polar.
32.

What happens when ferrimagnetic Fe_3O_4 is heated to 850 K and why ?

Answer»

SOLUTION :FERRIMAGNETIC `Fe_3O_4` on heating to 850 K BECOMES paramagnetic . This is due to greater alignment of domains (spins ) in ONE direction on heating
33.

What is the formula of a compound in which element P forms ccp lattice and atoms of Q occupy 2/3rd of tetrahedral voids ?

Answer»


Solution :Suppose oxide ions =n. Then octahedral VOIDS= n. HENCE, `Al^(3+)` ions `=2/3xxn="2N"/3`
`therefore` Ratio `Al^(3+) : O^(2-)="2n"/3:n =2:3` , i.e., formula is `Al_2O_3`
34.

What is value of compressibility factor (Z) ?

Answer»

`(PV)/(nRT)`
`(RT)/(pV)`
`(2)/(3)RT`
`(RT)/(NPV)`

ANSWER :A
35.

The solubility of BaSO_(4) in water is 2.42xx10^(-3)g L^(-1) at 298 K. The value of its solubility product (K_(sp)) will be (Given molar mass of BaSO_(4)=233 g L^(-1))

Answer»

`1.08xx10^(-10) "mol"^(2)L^(-2)`
`1.08xx10^(-12) "mol"^(2)L^(-2)`
`1.08xx10^(-14) "mol"^(2)L^(-2)`
`1.08xx10^(-8) "mol"^(2)L^(-2)`

Solution :Solubility (S) `=2.42xx10^(-3) g L^(-1)`
`=(2.42xx10^(-3))/(233) "mol" L^(-1)`
`=0.0104xx10^(-3) "mol" L^(-1)`
`=1.04xx10^(-5) "mol" L^(-1)`
`{:(BaSO_(4) ,hArr,Ba^(2+),+,SO_(4)^(2-)),(,,S,,S):}`
`K_(sp)=[Ba^(2+)][SO_(4)^(2-)]=SxxS=S^(2)`
`=(1.04xx10^(-5) "mol" L^(-1))^(2)`
`=1.08xx10^(-10) "mol"^(2)L^(-2)`
36.

Which one of the elements with the following outer orbital configurations exhibits the largest number of oxidation states?

Answer»

`3d^3, 4s^2`
`3d^5, 4s^1`
`3d^5, 4s^2`
`3d^2, 4s^2`

ANSWER :C
37.

What is thebasicdifference inapproach betweenMendeleev'sPeriodicLaw andthe Modern PeriodicLaw ?

Answer»

Solution :MendeleevPeriodic Lawstatesthat thephysicaland chemicalpropertiesof the ELEMENTS are aperiodicfunction of theiratomicweigths while MODERN PeriodicLawstatesthat physicaland chemicalPropertiesof ELEMENTSARE aperiodicfunctionof theiratomicnumbers . thusthe basicdiffereceinapproachbetweenMendeleev's PeriodicLaw andModern PeriodicLaw is thechangein BASIS ofclassificationof elementsfromatomicweight to atomicnumber.
38.

What are isoelectronic ions? Give an example.

Answer»

Solution :There are some ions of DIFFERENT ELEMENTS having the same number of ELECTRONS are called isoelectronic ions.
EXAMPLE : `Na^+ , Mg^(2+) , Al^(3+) , F^(-) , O^(2-) and N^(3-)`
39.

What are the ozonolysis products of ortho xylene?

Answer»

SOLUTION :
`{:(CH_3-C=O , , ,CHO ,CH_3-C=O),(""|,""+,""," "|," "+""|),(CH_3-C=O,,,""CHO,""CHO):}`
1,2-Dimethylbenzene (o-Xylene) on ozonolysis GIVES glyoxal, methyl glyoxal and dimethyl glyoxal in 3:2:1 molar RATIO. It proves the existance of TWO resonance structures for benzene and its derivatives
40.

Which of the following angle corresponds to sp^(2)hydridisation ?

Answer»

`90^(@)`
`120^(@)`
`180^(@)`
`109^(@)`

SOLUTION :The angle CORRESPONDS to `SP^(2)` hybridisation, triangular planar is `120^(@)`.
41.

What two changes on the equilibrium, N_(2) (g) + 3 H_(2) (g)hArr 2 NH_(3) (g) , Delta H= -92.4" kJ." can keepitsstate undisturbed ?

Answer»

SOLUTION :Increase of TEMPERATURE ALONG with suitable increase of PRESSURE or increase of pressure along with suitable increase of temperature.
42.

Two particles A and B are in motion. If the wavelength associated with particle A is 5 xx 10^(-8)m, calculate the wavelength associated with particle B if its monentum is half of A

Answer»

Solution :By de Broglie equation, `lamda_(A) = (H)/(p_(A)) and lamda = (h)/(p_(B))`
`:. (lamda_(A))/(lamda_(B)) = (p_(B))/(p_(A))`
But `p_(B) = (1)/(2) p_(A)` (Given)
`:. (lamda_(A))/(lamda_(B)) = (1//2p_(A))/(p_(A)) = (1)/(2) or lamda_(B) = 2xx lamda_(A) = 2 XX 5 xx 10^(-8) m= 10^(-7)m`
43.

Which of the following statement is correct about alkanes?

Answer»

Alkanes react with acids or BASE in normal conditions
Alkanes react with oxidising or REDUCING TEMPERATURE in normal conditons
Alkanes are inert compounds
Alkanes are unsasturated compunds

Answer :C
44.

Which of the following hydrocarbons is the most reactive towards addition of H_(2)SO_(4)?

Answer»

ETHENE
PROPYLENE
3-methyl but- 1 - ene
1 - butene

Answer :C
45.

What is the co-ordination number of an atom in its own layer in ABAB type arrangement

Answer»


ANSWER :6
46.

Which of the following statements is not correct for BF_3 ?

Answer»

It can form adduct
It ACTS as a LEWIS base
It FORMS an ionic bond
It also forms dative BONDS with compounds like `NH_3`, etc.

Solution :It acts as a Lewis base
47.

Which of the following compound has incorrect IUPAC nomenclature?

Answer»

`underset("Ethyl butanoate")(CH_(3)CH_(2)CH_(2) -overset(overset(O)(||))(C )-OC_(2)H_(5))`
`underset("3-methyl butanal")(CH_(3)underset(underset(CH_(3))(|))(C )HCH_(2)CHO)`
`underset("2-methyl-3-pentanone")(CH_(3) underset(underset(CH_(3))(|))(C )H- overset(overset(O)(||))(C )-CH_(2)CH_(3))`
`underset("2-methyl-3-butanol")(CH_(3)-underset(underset(CH_(3))(|))(C H)-underset(underset(OH)(|))(C )HCH_(3))`

Solution :`underset("3-methyl-2-butanol is correct")(overset(4)(C )H_(3) - underset(underset(CH_(3))(|))overset(3)(C H)-underset(underset(OH)(|))overset(2)(CH)-overset(1)(CH_(3)))`
48.

When solid melts there is an ____ in entropy

Answer»

SOLUTION :an INCREASE
49.

Which of the following properties of atom could be explained correctly by Thomson model of an atom?

Answer»

Overall NEUTRALITY of atom.
Spectra of HYDROGEN atom
Position of protons, electrons and NEUTRONS in atom
Stability of atom

ANSWER :A
50.

Which one is viable particulate pollutant ?

Answer»

ALGAE
SMOKE
DUST
MIST

SOLUTION :algae