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.

What is Boyle temperature ?

Answer»

Solution :The TEMPERATURE at which a REAL GAS behaves like an ideal gas over an appreciable pressure range is CALLED Boyle temperture or Boyle POINT.
Discussion
2.

Which pH value is more in the following ? (a)0.1 M HCl and 0.1 M NaOH (b) 0.1 M HCl and 0.01 M HCl (c) 0.1 M NaOH and 0.01 M NaOH

Answer»

SOLUTION :(a)0.1 M NAOH ,(B)0.01 M HCL , (b)0.1 M NaOH
Discussion
3.

Which of the following cannot act as Lewis acid ?

Answer»

`C Cl_(4)`
`SiCl_(4)`
`GeF_(4)`
NONE 

ANSWER :A
Discussion
4.

What is the oxidation state of CO in [CO(H_(2)O)_(5)CI]^(2+) ?

Answer»

`+2`
`+3`
`+1`
`+4`

Solution :Since `H_(2)O` is a NEUTRAL molecule sum of its O.N is zero WHIL O.N of CI is -1 now x is the O.N of Co then
`x+5xx0+(-1)=+2`
`THEREFORE x=+2+1=+3`
Discussion
5.

When copper is heated with conc. HNO_(3) it produces

Answer»

`Cu(NO_(3))_(2),NOandNO_(2)`
`Cu(NO_(3))_(2)andN_(2)O`
`Cu(NO_(3))_(2)andNO_(2)`
`Cu(NO_(3))_(2)andNO`

Solution :`Cu+4HNO_(3)(conc.)TOCU(NO_(3))_(2)+2NO_(2)+2H_(2)O`
Discussion
6.

Which of the following compunds exhibit gemetrical isomerism? Draw their cis-trans isomers (i) CH_3-CH=C Cl_2(ii) CH_3-CH=CHCl(iii)ClCH=CHCl(IV)CH_3-CH_2-CH=CH_2

Answer»

Solution :Compound (II) and (III) show GEOMETRICAL ISOMERISM.
Discussion
7.

The solubilityof PbSO_4 "in"0.01 M Na_2SO_4 solution is (K_(sp)of PbSO_4 is 1.25 xx 10^(-9) )

Answer»

` 1.25 XX 10 ^(-7) M`
` 1.25 xx 10^(-9) M`
` 1.25 xx 10^(-10) M`
` 0.1 M`

Solution :` PbSO_4 HARR Punderset(s) B^(+2)+ Sunderset( 0.01)Ooverset( 2-)`
` S( 0.01 )= 1.25 xx 10 ^(-9), S =1 .25 xx 10 ^(-7) `
Discussion
8.

Which of the following compounds will give a secondary alcohol on reaction with Grignard reagent followed by acid hydrolysis ? (I) HCHO (II) C_(2)H_(5)CHO (III) CH_(3)COCH_(3) (IV) HCOOC_(2)H_(5) Select the correct answer using the codes given below.

Answer»

II only
III only
I and IV
II and IV

Answer :D
Discussion
9.

Which of the following arrangements represent increasingg oxidation number of the central atom?

Answer»

`CrO_(2)^(-),ClO_(3)^(-),CrO_(4)^(2-),MnO_(4)^(-)`
`ClO_(3)^(-),CrO_(4)^(2-),MnO_(4)^(-),CrO_(2)^(-)`
`CrO_(2)^(-),ClO_(3)^(-),MnO_(4)^(-),CrO_(4)^(2-)`
`CrO_(4)^(2-),MnO_(4)^(-),CrO_(2)^(-),ClO_(3)^(-)`

ANSWER :A
Discussion
10.

What's the main difference between lone pair and bonded pair of electrons.

Answer»

Solution :a valence set of two electrons that EXISTS WITHOUT bonding or SHARING with other atoms is CALLED lone pair. Pair of electrons involved in a covalent bond is called bonding pair.
Discussion
11.

Which of the following is secondary alkyl halide?

Answer»

ISO BUTYL CHLORIDE
Isopentyl chloride
Neopentyl chloride
ISOPROPYL chloride

Solution :isopropyl chloride
Discussion
12.

When phenol is treated with CHCl_(3) and NaOH followed by acidification, salicyladehyde is obtained. Which of the following species are involved in the above mentioned reaction as intermediates?

Answer»




SOLUTION :SEE MACHANISM of REIMER Tiemann REACTION.
Discussion
13.

What iselectronconfigurationof atom?Explainitsmethodby example.

Answer»

Solution :Thedistributionof electronintoorbitalsof an atmin calleditselectronicconfiguration
method-1 :Thesubshellisrepresentedby therespectiveylettersymboland thenumber ofelectronspresentin thesubshellisdepicted assubshellrepresentedfor differentshellsisnumberbeforethe respectivesubshell.
Example : Electronconfigurationexpressbyorbitalnotationof B (z=5)(z= 6)are asunder

Method -2:(Electron configurationby orbitaleachorbitalof thesubshellisrepresentedbyboxis as under

the electrons isrepresentedby NARROW (i) a POSITIVE`(+(1)/(2))`SPINOR anarrow`(to)`1A negativespin `(-(1)/(2))`
Example: theorbitalpictureof theseelementscan berepresentedas follows :
Discussion
14.

Thesubstance usedpurefatsandoilsis _______.

Answer»

SODIUMCARBONATE
SODIUM chloride
sodiumhydroxide
SODIUMSULPHATE

ANSWER :C
Discussion
15.

Which one of the following reactions is not associated with the Solvay process of manufacture of sodium carbonate ?

Answer»

`NH_(3)+H_(2)O+CO_(2)toNH_(4)HCO_(3) `
`NaCl+NH _(4) HCO_(3)toNaHCO_(3)`
`2NaHCO_(3) toNa_(2)CO_(3)+H_(2)O+ CO_(2)`
`2NaOH+CO_(2)to Na_(2) CO_(2)+ H_(2) O `

ANSWER :D
Discussion
16.

What is the product of each reaction

Answer»


ANSWER :`(##RES_CHM_ROHR_E03_103_A01##)`
Discussion
17.

What is the main difference between ionic bond and covalent bond ?

Answer»

Solution :In Ionic bond POSSESSING molecule ions have - opposite change and because of opposite charge they are stabilize by electrostatic force of attraction and POSSESS low energy.
While the molecules having covalent bond the atoms combine by joint sharing of their valence ELECTRON and attain stable electron configuration like inert gas so energy decreases and molecule is stable.
The atoms having more difference in electronegativity form ionic bond. The atom having LESS difference in electronegativity form covalent bond.
Discussion
18.

What is one amu or u ?

Answer»



SOLUTION :`(1)/(12)` part of `""^(12)C` is CALLED 1 AMU or u.
1 amu `=1.66056xx10^(-24) GM`
Discussion
19.

Which colour is formed in the Lassaigne's test for nitrogen?

Answer»

PRUSSIAN BLUE colour
Blood RED colour
Purple colour
Canary YELLOW colour

Answer :B
Discussion
20.

The values of Kp_(1) and Kp_(2) for the reaction given below are in the ratio 2: 1. If degree of dissociation of X and A are equal, then total pressure of equilibrium (1) and cquilibrium (2) are in the ratio X hArr Y + Z , Kp_(1) : A hArr 2B : Kp_(2)

Answer»


Solution :`UNDERSET(1-alpha_(1))(X) HARR underset(alpha_(1))(Y)+underset(alpha_(1))(Z)`
`Kp_(1)=(((alpha_(1))/(1+alpha_(1)).p_(1)) xx ((alpha_(1))/(1+alpha_(1)).p_(1)))/(((1-alpha_(1))/(1+alpha_(1)) xx p_(1)))`
`KP=(alpha_(1)^(2) xx p_(1))/(1-alpha_(1)^(2))` ...(1)
`underset(1-alpha_(1))(A) harr underset(2alpha_(1))(2B), Kp_(2)`
`Kp_(2)=(((2alpha_(1))/(1+alpha_(1)) xx p_(2))^(2))/((1-alpha_(1))/(1+alpha_(1)) xx p_(2)), Kp_(2)=(4alpha_(1)^(2) xx p_(2))/(1-alpha_(1)^(2))` ...(2)
`((1))/((2))(Kp_(1))/(Kp_(2))=(1)/(4) xx (p_(1))/(p_(2)) IMPLIES (p_(1))/(p_(2))=4 xx (Kp_(1))/(Kp_(2))=4 xx 2=8`
Discussion
21.

Which of the following equation depicts reducing nature of H_2O_2 ?

Answer»

`2[Fe(CN)_6]^(4-) + 2H^(+) + H_2O_2 to 2[Fe(CN)_6]^(3-) + 2H_2O`
`I_2 + H_2O_2 + 2OH^(-) to 2I^(-) + 2H_2O + O_2`
`Mn^(2+) + H_2O_2 to Mn^(4+) + 2OH^(-)`
`Pbs+4H_2O_2 to PbSO_4 + 4H_2O`

Solution :`H_2O_2` ACTS as a oxidising agent and reducing agent in alkaline following REACTION indicate its reducing nature in basic medium.
`I_2 + H_2O_2+ 2OH^(-) to 2I^(-) + 2H_2O + O_2`
`2MnO_4^(-) + 3H_2O_2 to 2MnO_2 + 3O_2 + 2H_2O + 2OH^(-)`
Discussion
22.

What happens when slaked lime is treated with dry chlorine?

Answer»

SOLUTION :`2CA(OH)_2Cl_2 RARR CaCl_2+CaSiO_3`
Discussion
23.

What are the elements present in group 2? Give their general electronic configuration.

Answer»

SOLUTION :Group 2 contains beryllium, magnesium, calcium, STRONTIUM, BARIUM and radium. Their general ELECTRONIC configuration is (noble gas] `ns^(2)`
Discussion
24.

Which of the following statements is notcorrect ?

Answer»

The fraction of the total VOLUME OCCUPIED by the atoms in a PRIMITIVE cell is 0.48.
Molecular solids are generally volatile.
The number of carbon atoms in a unit cell of DIAMOND is 4.
The numberof Bravais LATTICES in which a crystal cell of diamond is 14.

Solution :The fraction of the total volume occupied by the atoms in a primitive cell is 0.524
Discussion
25.

The values of x and y in the following redox reaction xCI_(2)+6 OH^(-) rarr CIO_(3)^(3) rarr CIO_(3)^(-)+y CI^(-)+3 H_(2)O are

Answer»

x=2, y=4
x=5,y=3
x=3,y=5
x=4,y =2

Answer :B
Discussion
26.

The study of diple moment of a molecule is useful to explain the shape of a molecule and also to predict a The net dipole moment of a polyatomic molecule is the resltant of the different bond moments present in that molecule . The values are generalluy expressed in Debye or in the S.I. units in terms of Coulomb meter (C m) A covalent molecule , X-Y is found to have a dipole moment of1.5 xx10^(-29) C m and a bond length of 150 pm . The per cent ionic character of the bond will be

Answer»

`50%`
`62.5%`
`75%`
`90%`

Solution :`mu_("calculated")=(1.602 xx10^(-19) C) xx (150 xx 10^(-12)m)`
` = 2.4 xx 10^(-29) `C m
`mu_("calculated")=1.5 xx10^(-29) C m`
`therefore % `ionic CHARACTER`=(1.5 xx10^(-29) ) / (2.4xx 10^(-29)) XX100 = 62.5%`
Discussion
27.

What is borax glass?

Answer»

SOLUTION :`Na_2 B_4 O_7`
Discussion
28.

The study of diple moment of a molecule is useful to explain the shape of a molecule and also to predict a The net dipole moment of a polyatomic molecule is the resltant of the different bond moments present in that molecule . The values are generalluy expressed in Debye or in the S.I. units in terms of Coulomb meter (C m) Which our of the following will have maximum dipolemoment ?

Answer»

`NF_(3)`
`NCl_(3)`
`NBr_(3)`
`NH_(3)`

Solution :
F is more electronegative than N ,N-F bond
OPPOSES the bond moment of N with the LONE pair .
In all order cases , it favours but electronegativity
differenceis maximum between N and H . Hence,
`NH_(3)` has maximum DIPOLE moment . (Electronegativity of
Cl = N = 3.0)
Discussion
29.

Which of the following statement(s) is (are) correct?

Answer»

The `p^(H)` of `10^(-8)` M SOLUTION of `HCl` is `8`
The conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`
Auto protolysis constant of water increases with temperature
When a solution of weak monoprotic acid is titrated against a STRONG base, at half of neutraliastion POINT `p^(H) = 1//2p^(Ka)`

Solution :N//A
Discussion
30.

The study of diple moment of a molecule is useful to explain the shape of a molecule and also to predict a The net dipole moment of a polyatomic molecule is the resltant of the different bond moments present in that molecule . The values are generalluy expressed in Debye or in the S.I. units in terms of Coulomb meter (C m) 1 Debye is equivalent to

Answer»

`3.33 xx10^(-30) ` C m
`1.602xx10^(-27)` C m
`10^(-20)` C m
`3.33 xx10^(-12)` C m

Solution :1 Debye =` 3.335 XX 10^(30)` C m
Discussion
31.

When 50 gm of a sample of sulphur was burnt in air 4% of the sample was left over. Calculate the volume of air required at STP containing 21% oxygen by volume.

Answer»

Solution :Weight of sample of sulphur TAKEN =50g
PERCENTAGE of impurity =4%
Weight of impurity in sample `=(4 xx 50)/(100)=2gr`
Weight of sulphur in sample =50-2=49grams
Combustion of sulphur GIVES sulphurdioxide
`S+O_(2) to SO_(2)`
1 mole of S=1 mole of `O_(2)`
32grams of S=22.4 L of `O_(2)` at STP
48 GRAMS of S=?
Volume of oxygen required at STP `=(48)/(32) xx 22.4=33.6 L`
21 L of `O_(2)` is present in 100L of air
33.6 L of `O_(2)` present in ?
Volume of air required at STP `=(33.6)/(21) xx 100=160L`
Discussion
32.

The value of Delta_(f) H^( Theta ) for NH_(3) is -91.8 "kJ mol"^(-1). Calculate enthalpy change for the following reaction: 2NH_(3(g)) to N_(2(g)) + 3H_(2(g))

Answer»

SOLUTION :`(1)/(2) N_(2(g)) + (3)/(2) H_(2(g)) to NH_(3(g)) , Delta_(f) H^( Theta ) = -91.8 "kJ mol"^(-1)`
(`Delta_(f) H^( Theta )` means ENTHALPY of formation of 1 mole of `NH_(3)` )
`therefore` Enthalpy change for the formation of 2 moles of `NH_(3)`
`N_(2(g)) + 3H_(2(g)) to 2NH_(3(g)) , Delta_(f) H^( Theta )= 2 xx -91.8 =-183.6 "kJ mol"^(-1)`
for the reverse REACTION,
`2NH_(3(g)) to N_(2(g)) + 3H_(2(g)) , Delta_(f) H^( Theta ) = + 183.6 "kJ mol"^(-1)`
Hence, the value of `Delta_(f) H^( Theta ) ` for `NH_(3)` is `+183.6 "kJ mol"^(-1)`.
Discussion
33.

Which is the principle of paper chromatography?

Answer»

Adsorption
Absorption
Solubility
Partition

Answer :D
Discussion
34.

Which of the following compound does not undergo addition reaction?

Answer»

Ethyne
Ethane
Ethanal
Ethene

Solution :Ethene , `C_(2)H_(6)` is a saturated hydrocarbon. It has no `PI`-bond. It does not undergo ADDITION REACTION
Discussion
35.

What happens to the probability of finding the electron in the m.o.'s after the combination of two atomic orbitals ?

Answer»

SOLUTION :Probability (electron DENSITY) increases in the INTERNUCLEAR REGION of bonding m.o., WHEREAS it decreases in
the antibonding m.o.
Discussion
36.

Which of the following disproportionation reaction

Answer»

`2Cu^(+) rarrCu^(2+) + Cu`
`3I_(2) rarr5I^(-) +I^(5+)`
` H_(2)O + Cl_2 rarrCl^(-)+ CIO^(-) + 2H^(+)`
All of these

Solution :All are disproportionation REACTIONS.
Discussion
37.

When 30 ml of 5.93 millimolar solution of AgNO_(3) was added to 2.0 ml of 8.89 millimolar solution of KCl, the mixture turns turbid. The solubility product of AgCl is

Answer»

`1.96xx10^(-6) "mol"^(2)L^(-2)`
`3.92xx10^(-6) "mol"^(2)L^(-2)`
`1.96xx10^(-6) "mol"^(-1)L^(-1)`
`3.92xx10^(-6) "mol"^(-1)L^(-1)`

Solution :After mixing `[Ag^(+)]`:
`M_(1)V_(1)=M_(2)V_(2), i.e., 30xx5.93xx10^(-13)=32xxM_(2) "or" M_(2)=5.56xx10^(-3)M`
`[Cl^(-)] : 2.0xx8.89xx10^(-3)=32xxM_(2) " or" M_(2)=5.56xx10^(-4)M`
`:.` Ionic product of AGCL `=[Ag^(+)][Cl^(-)]`
`=(5.56xx10^(-3) "mol" L^(-1))(5.56xx10^(-4) "mol" L^(-1))`
`=3.09xx10^(-6) "mol"^(2) L^(-2)`
which exceeds (a) but not(b).
UNITS in (C) and (d) are not correct.
Discussion
38.

Two oxides of copper contain respectively 88.8% and 79.91 % of copper. Calculate equivalent masses of copper in each of the oxide.

Answer»


ANSWER :63.4, 31.8
Discussion
39.

Which of the following is incorrect statement ?

Answer»

`SnF_4` is ionic in NATURE
`PbF_4` is COVALENT in nature
`SiCl_4` is EASILY hydrolysed
`GeX_4`(X=F, Cl,Br,I) is more stable than `GeX_2` .

Solution :`SnF_4` and `PbF_4` are ionic in nature .
Discussion
40.

What is the molarity of H_2SO_4 solution which contains 98% H_2SO_4 by weight and whose specific gravity is 1.84 ?

Answer»

1.84
18.4
9.2
9.6

Answer :B
Discussion
41.

Which of the following is the main product in the ozonolysis of 1,4-dimethylcyclohexene followedby a reductive workup with Zn and ethanoic acid?

Answer»




Solution :The products of ozonolysis followed by a REDUCTIVE workup are aldehyde(s) and/or KETONE(s). Note the POSITION of the METHYL substituent in the ring-opened product
Discussion
42.

When 1 mole of N_2 and 1 mole of H_2 is enclosed in a 5 L vessel and reaction is allowed to attain equilibrium . It is found that at equilibrium there is x mole of H_2 . The number of moles of NH_(3) formed would be

Answer»

`(2X)/(3)`
`(2 (1 + x))/(3)`
` (2 (1 - x))/(3)`
`((1- x))/(2)`

ANSWER :C
Discussion
43.

What mass of NHO_3 is needed to convert 5g of iodine into iodic acid according to the reaction ?

Answer»

12.4 g
24.8 g
0.248 g
49.6 g

Solution :`I_2+4HNO_(3) rarr2HFO_(3)+4NO_2+H_2O`
eq. `HNO_3=I_2,OVERSET(+5)(NO_3^(-))overset((1))RARROVERSET(+4)(NO_2)`
`w_1/E_1"" =w_2/E_2,overset(0)I_2rarroverset(5)(2IO_3^(-))+5`
`w_1 = (5xx10xx63)/(254) = 12.4 ` GM
`1 + 14 + 48 = 63`
Discussion
44.

Which of the following reactions will yield 2,2-dibromopropane ?

Answer»

`CH_3-CH=CH_2+HBr to `
`CH_3 -C -=CH+ 2HBR to `
`CH_3 CH=CHBr + HBr to `
`CH-=CH+ 2HBr to `

Solution :`CH_3C-=CH+2HBr UNDERSET"addition"overset"Mark"to CH_3C(Br)_2-CH_3`
Discussion
45.

Which of the following compound is known as oil of winter gree ?

Answer»

PHENYL benzoate
Phenyl salicylate
Phenyl acetate
Methyl salicylate

Answer :D
Discussion
46.

Which is not a s- block element ?

Answer»

`[Ar]4s^(2)3d^(10)4p^(6)5s^(1)`
`1s^(2)2S^(2)2p^(1)`
`[He]2s^(2)2p^(6)3s^(1)`
None of these

Solution :The block of an element depends on the type of sub- shell which receives the last ELECTRON. As last electron enters in p- subshell of outmost shell according to AUFBAU RULE, the element of optin `(b)` belongs to p-block.
Discussion
47.

Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides ?

Answer»

`MgO LT K_(2)O lt Al_(2)O_(3) lt Na_(2)O`
`Na_(2)O lt K_(2)O lt MgO lt Al_(2)O_(3)`
`K_(2)O lt Na_(2)O lt Al_(2)O_(3) lt MgO`
`Al_(2)O_(3) lt MgO lt Na_(2)O lt K_(2)O`

Solution :CORRECT order of basic strength
`Al_(2)O_(3) lt MgO lt Na_(2)O lt K_(2)O`
Group 1 oxides are more basic than group 2 oxides. Basic character of oxides increases down the group.
Discussion
48.

What is theheat of formation of C_(6)H_(6), given that the heats of combustion of benzene, carbon and hydrogen are 782, 94 and 68K. Cal respectively

Answer»

`+14` KCAL
`-14` kCal
`+28` kCal
`-28` kCal

Solution :`C_(6)H_(6) + (15)/(2)O_(2) rarr 6CO_(2) + 3H_(2)O`
`DELTA H_("comb") = 6 Delta H_(f_(CO_(2))) + 3 Delta H_(f_(H_(2)O)) - Delta H_(f_(C_(6)H_(6)))`
`-782 = 6 XX (-94) + 3 xx (-68) - Delta H_(f_(C_(6)H_(6)))`
`rArr Delta H_(f_(C_(6)H_(6))) = + 14` Kcal
Discussion
49.

What are hybridisation states of each carbon atom in the following compounds? (i) CH_(2) = C = O (ii) CH_(3)CH = CH_(2) (iii) (CH_(3))_(2)CO (iv) CH_(2) = CHCN (v) C_(6)H_(6)

Answer»

SOLUTION :
Discussion
50.

When alpha-D glucose is dissolved in water , it undergoes mutarotation to form an equilibrium mixture of alpha - D glucose and beta - D glucose containing 63*6 % of the latter. Calculate K_(c) for the mutarotation.

Answer»

Solution :`{:(,alpha -" D GLUCOSE ",hArr,beta - " D glucose "),(" At EQUILIBRIUM " ,36*4%,,63*6 %):}`
` K_(c)(63*6)/(36*4) = 1* 747`
Discussion