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 among the following, does not conduct electric current and why? Molten NaCl, Solid Pb,AgNO_3 solution and methanol.

Answer»

SOLUTION :METHANOL will not CONDUCT ELECTRIC CURRENT because it does not ionise.
Discussion
2.

Which of the following is a redox decomposition reaction ?

Answer»

`CaCO_(3)(s)overset(Delta)toCaO(s)+CO_(2)(G)`
`2KClO_(3)(s)overset(Delta)to2KCl(s)+3O_(2)(g)`
`2H_(2)O(l)+2F_(2)(g)to4HF(aq)+O_(2)(g)`
`2Na(s)+Cl_(2)(g)to2NaCl(s)`

Answer :B
Discussion
3.

What happens to an wquilibrium in a reversible reaction if a catalyst is added to it ?

Answer»

Solution :When CATALYST is added, the state of EQUILIBRIUM is not disturbed but equilibrium is attained quickly. This is because the backward the satalyst increases the RATE of FORWARD and backward reaction to the same extent.
Discussion
4.

Tlaloalkanes undergo nucleophilic substitution reaction whereas haloarenes undergo clectrophilic substitution reaction. Comment.

Answer»

Solution :Haloalkanes undergo nueleophilic substitution reactions due to high clectronegativity of the halogen atom. The C-X bond in haloalkanes is SLIGHTLY polar, thereby the C atom acquires a slight positive charge in C - X bond. Hence C atom is a good targetfor attack for nucleophiles. Therefore- X atom of haloalkanes is replaced by a nucleophile easily.
`NU: + overset(DELTA+)R - overset(delta-)XTO R - Nu + X`
On the other hand in HALOARENES, the halogen atom releases electron to the benzene nucleus relatively electron rich with respect to the halogen atom. As a result the electrophilic attack occursat ortho and para positions. Hence haloarenes undergo electrophilic substitutionreactions.
Discussion
5.

Write the structural formulae of the following compounds : (i) Picric acid (ii) Cumene (iii) Styrene (iv) Phthalimide (v) Hydroquinone (vi) Pyrogallol (vii) 1-Phenylpropanone (viii) 3-Bromo-1-Phenylpropene (ix) Anisole (x) Benzophenone (xi) Cinnamic acid (x) Benzophenone (xi) Cinnamic acid (xii) p-Toluidene (xiii) 2,6-Dimethylphenol (xiv) p-Methoxybenzaldehyde.

Answer»

SOLUTION :
Discussion
6.

The standard enthalpies of formationof A(NH_(3)), B(CO_(2)), C(HI)and D(SO_(2))are respecitvely -46.19,-393.4, + 24.94 and -296.9 kJ mol^(-1) respectively. The increasing order of their stability is

Answer»

`Blt D lt A lt C`
`C lt A lt D lt B`
`D lt B lt C LTA`
` A lt C lt D lt B`

Solution :More`-ve` is`Delta_(f)H^(@)` ,greateris the stability.
Discussion
7.

Which of the followingreaction does not stick to the cleassical idea of redox reactions ?

Answer»

` 3Fe_(3)O_4(s) +8Al(s) rarr 9 Fe(s) +4Al_2O_2(s)`
`2Na(s) +H_2(g) rarr 2NaH(s)`
` H_2S(g) +Cl_2(g) rarr 2HCL(g) +S(s)`
None of these

Solution :In reaction (1), aluminim is oxidizes because oxygen is added to it while ferrous ferric oxide`(Fe_3O_4)` is reduced because oxygen has been removed from it. In reaction (3), chlorine isn reuced due to the addition of hydrogen toit whilen ` H_2S` is oxidized due to the REMOVAL of hydrogen from S or we can say due to the addition of chlorine (an electronegative element ) to hydrogen. In reaction (2) , the classical concept of redox reactions fails necause both Na and `H_2` are of similar nature, i.e., both are electropositive elements (possessing the tendency to form positive ions ). Accoring to the classical concept , Na is reduced to `NaH` due to the addition fo hydency to it and `H_2` is also reduced due to the addition fo electropositive element to it, However, with the caraful APPLICATION of the concept of electronegativity, we conclude that Na is oxidized while hydrogen is reduced. Thus, reaction (3) PROMPTS us to think in terms of another way to define redox reactions. We face similar problem in the FOLLOWING reaction:
`Cl_(2)+3F_(2) rarr 2ClF_(3)`
as both `Cl_(2)` and `F_(2)` are electronegative elements.
Discussion
8.

Which of the following carbocations would have the greatest stability ?

Answer»

`CH_(3)-S-CH_(2)^(+)`
`(CH_(3))_(2)NCH_(2)^(+)`
`CH_(3)-O-CH_(2)^(+)`
`F-CH_(2)^(+)`

Solution :The LONE pair of electrons on S is PRESENT in a 3p-orbital while those an N, O and F are present in a 2-p-orbital. Further, the carbon atom bearing the +ve charge has an empty 2p-orbital. Since EFFECTIVE RESONANCE stabilization occurs between orbitals of similar sizes, therefore, least effective resonance will be between S and C atoms. Among N, O and F, since is least electronegative, therefore N, O and F, since N is least electronegative, therefore, most effective resonance will be between N and C and hence `(CH_(3))_(2)NCH_(2)^(+)` carbocations would be the most stable.
Discussion
9.

What is octet rule?

Answer»

SOLUTION :The tendency of atoms of electrons to ATTAIN octet STRUCTURE of the nearset noble gas during FORMATION of a chemical bond is KNOWN as octet rule.
Discussion
10.

Write two uses of interstitial hydrides .

Answer»

Solution :(i) Storing `H_(2)` and (II) CATALYSTS for HYDROGENATION REACTIONS.
Discussion
11.

What effect does a catalyst have on the equilibrium position of a reaction ?

Answer»

a catalyst favours the FORMATION of products
a catalyst favours the formation of reactants
a catalyst does not change the equilibrium POSITION of a REACTION
a catalyst may favour reactants or PRODUCT formation , depending upon the direction in which the reaction is written.

Solution :a catalyst does not change the equilibrium position of a reaction
Discussion
12.

The total number of orbitals associated with the principal quantum number n=3 is……………..

Answer»

9
8
5
7

Solution :`n=3,1=0,m_1=0-`one s orbiital
n`=3,1=1,m_1=-1,0,1-`three p orbitals
`n=3,1=2,m_1=-2,-1,0,1,2-`FIVE d oritals, overrall NINE orbitals are POSSIBLE.
Discussion
13.

The stability of exactly half-filled and completely filled configuration is due to.........and .........

Answer»

SOLUTION :symmetrical DISTRIBUTION and LARGE EXCHANGE energy
Discussion
14.

Toluene undergoes nitration easilypolymerisation process

Answer»

Solution :(i)toluenehas amnethylgroupon thebenzeneringwhichis electronreleasinggroupandhenceactivatethe benzene ringby pushingthe electronon the benzenering.
(ii) `CH_(3)` GROUPIS ORTHO -paradirectorand ringactivator . thereforein tolueneorthoandsubstitutionreaction
(ii) The methylgrouphencemakesit around25 timesmorereactivethanbenzene . Thereforeit undergoesnitrationeasilythanbenzene.
Discussion
15.

What is the equilibrium expression for the reaction, P_(4(s)) + 5O_(2(g)) hArr P_(4)O_(10(s)) ? .

Answer»

`K_c = ([P_4O_(10)])/([P_4][O_2]^5)`
`K_c = 1/([O_2]^5)`
`K_c = [O_2]^5`
`K_c = ([P_4O_(10)])/(5[P_4][O_2])`

SOLUTION :`K=(1)/([O_(2)]^(5) xx 1)`
Discussion
16.

Which statement(s) about oxidation number is(are) correct?

Answer»

The oxidation number is the number of electrons lost (+ve) or GAINED (-ve) by an atom during the FORMATION of ionic compounds
For COVALENT compounds, the oxidation number is indicated by the charge that an atom of element would have acquired if the substance would have been ionic
Oxidation number may have fractional VALUES
Oxidation number is ALWAYS negative.

Solution :Oxidation number is charge of an atom in a molecule which can be any value.
Discussion
17.

V and P can be

Answer»

`CAO,CaCO_(3)`
`Na_(2)O,Na_(2)CO_(3)`
`NACL,Na_(2)CO_(3)`
`NaOH,Na_(2)CO_(3)`

Answer :A
Discussion
18.

What type of hybridisations are possible in the following geometeries? (a) octahedral (b) tetrahedral © square planar.

Answer»

Solution :(a) OCTAHEDRAL geometry is POSSIBLE by `SP^(3)d^(2)` (or) `d^(2)sp^(3)` hybridisation.
(b) Tetrahedral geometry is possible by `sp^(3)` hybridisation.
( C) Square planar geometry is possible by `dsp^(2)` hybridisation.
Discussion
19.

What is bond order ? Give example.

Answer»

Solution :It is the NUMBER of covalent BONDS holding the atoms in the MOLECULE. Example: If the BOND is formed by the sharing of two electron pairs, then the bond oder is 2. O=O or C=C bond in alkenes.
Discussion
20.

What is effective nuclear charge ?

Answer»

Solution :The net nuclear charge experienced by valence electrons in the outermost SHELL is CALLED the effective nuclear charge.
`Z_(eff)=Z-S`
Where,
Z - Atomic NUMBER S = SCREENING constant calculated by using Slater.s RULES.
Discussion
21.

The weight ratio of elements carbon, nitrogen and hydrogen in a compound of molecular mass 108 is 18:2:7. What is its molecular formula?

Answer»

Solution :`C:H:N:=(18)/(12):2/1:7/(14)=3:4:1`
Empirical formula is `C_(3)H_(4)N`
Empirical formula weight =54
`n=(108)/(54)=2`
Molecular formula `=2 XX (C_(3)H_(4)N)=C_(6)H_(8)N_(2)`
Discussion
22.

Which colorless gas turns brown in air?

Answer»

NO
`NO_2`
`N_2O_4`
`N_2O_5`

ANSWER :A
Discussion
23.

To neutralize completely 20 ml of 0.1 M phosphorus acid, 40 mol of KOH was required. What volume of this KOH solution will be required to neutralize 0.66 g of H_(3)PO_(2)?

Answer»

100 ml
200 ml
300 ml
66.7 ml

SOLUTION :Eqts of `H_(3)PO_(3)` = Eqts of KOH
`20xx0.1xx2=1xx40xxMimpliesM=0.1`
Now Eqts of KOH = Eqts of `H_(3)PO_(2)`
`0.1xx1xxV=(0.66)/(66)impliesV=0.1" lit"`
`V=100ml`
Discussion
24.

The structure of Vinyl iodide is ………………. .

Answer»

SOLUTION :`CH_2=CHI`
Discussion
25.

Two elements P and Q form the compounds P_(2),Q_(3) , and PQ_(2). If 0.15 mole of P_(2),Q_(3), weighs 15.9 g and 0.15 mole of PQ_(2) weighs 9.3 g, find the atomic weights of P and Q.

Answer»

Solution :Molecular weight of `P_(2),Q_(3)` = wt. of 1 mole `(15.9xx1)/(0.15) = 106.`
Molecular weight of `PQ_(2)` = wt. of 1 mole =`(9.3xx1)/(0.15) = 62.`
SUPPOSE that the atomic WEIGHTS of P and Q are RESPECTIVELY x and y. Thus,
2x + 3y = 106 `(P_(2)Q_(3) =106)`
x + 2y = 62. `(PQ_(2) = 62)`
On solving: x = 26
y = 18.
Discussion
26.

What is properties related to single particle of matter ?

Answer»

Solution :Properties related to SINGLE particle of MATTER : (i) Atomic VOLUME (ii) Ionisation enthalpy (III) Electronic charge density (iv) MOLECULAR shape (v) Molecular polarity etc.
Discussion
27.

When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. On the basis of this information mark the correct answer. {:([Co(H_(2)O)_(6)]^(3+)(aq)+4Cl^(-)(aq)hArr[CoCl_(4)]^(2-)(aq)+6H_(2)O(l)),("(pink)" "" "(blue)"):}

Answer»

`Delta H gt 0` for the REACTION
`Delta lt 0` for the reaction
`Delta H = 0` for the reaction
The sign of `DELTAH` cannot bepredicted on the basis of this information.

Solution :On cooling , the equilibrium SHIFTS backward or on HEATING , the equilibrium shifts FORWARD. Hence, reaction is endothermic, i.e., `Delta H gt 0`.
Discussion
28.

Which of the following is not an aromatic compound:

Answer»




SOLUTION :is not an AROMATIC COMPOUND DUE to Hukele.s RULE
Discussion
29.

Which of the following is used for cutting and welding?

Answer»

 ATOMIC HYDROGEN and OXY hydrogen torches
Liquid hydrogen
 CALCIUM hydride
Sodium BORO hydride

Answer :a
Discussion
30.

Thespliting of spectral linein magnetic andelectricalfieldis known as .

Answer»

Photoelectronandstarkeffect
Zemanand starkeffect
STARK and zemaneffect
PHOTOELECTRON and zemaneffect

ANSWER :B
Discussion
31.

Which system possess eqilibrium from the following ?(i) Saturated solution of sugar in open cup.(ii) At constant temperature mercury in thermometer and its vapour(iii) boiling water in open vessel(iv) floating ice in water at 0^@C(v) ice in water at 15^@C.

Answer»

Solution :(iii) and (v) is not EQUILIBRIUM.
(i) Sugar solid `hArr` sugar `sol^n` equilibrium
(ii)`Hg_((s)) hArr Hg_((g))`(con.temp) equilibrium
(iv)`H_2O_((s)) hArr H_2O_((L))` (`0^@` C, 1 atm) equilibrium
(v) It is not equilibrium as in short time ice will DISSOLVE in WATER.
Discussion
32.

Which of the following will oxidise Hydrogen peroxide?

Answer»

ACIDIFIED KMnO4
Cu
`dil.HNO_(3)`
`CrO_(2)CI_(2)`

ANSWER :a
Discussion
33.

Whybenzene undergoes electrophilic substitution reaction whereas alkenes undergoes additonreaction

Answer»

Solution :(i) Benzenepossesan unlybridisedp-orbitalcontainingone ELECTRON. Thelateraloverlapof theirp-orbitalsproduces 3`pi`- bond.
(II) Due todelocalisation strongs `pi`- BONDIS formedwhichmakesthe moleculestable
thereforebenzeneundergoes electrophilicsubstitutionreactionwheresalkenesundergoesadditionreaction .
Discussion
34.

What is Bronsted Lowry base ?

Answer»

SOLUTION :Accordic to Bronsted-Lowry concept the ACID and base is defined of following:
Acid : An acid is defined as a substance which has the tendency to give a PROTEN and acid is a proten donor.
Base : A base is defined as a substance which has a tendency to accept a proten and base is a proten ACCEPTOR.
Discussion
35.

What are meta directorsexplain with suitable example

Answer»

Solution :The group which increasethe electron density at meta position are CALLED as meta directorlet us consider the directive influcence of aldehydic (CHO) group BENZALDEHYDE is resonence hybrid of the following structures

In these resonance structures the postivie charge residue is presenton the ring structure it is quite evident that resonance delocalisles the positive charge on the atoms of the ringattack however resonating structures show that electron densityh is more at meta positionas compard to ORTHO and PARA positions hence CHO group is a meta director and a deactivator
Discussion
36.

What is the name of the reaction in which bromoethane is converted to iodoethane by reacting with Nal in acetone?

Answer»

Hunsdicker REACTION
DOW's PROCESS
Finkelstein reaction
SWARTS reaction

Solution :Finkelstein reaction
Discussion
37.

Why carbon forms covalent compounds whereas lead forms ionic compounds ?

Answer»

Solution :Carbon cannot lose all the four valence electrons to form `C^(4+)` ions because the sum of the first four ioniatization enthalpies of carbon is very large. It alsocannot GAIN four electronto form `C^(4-)`ion as their anionwould be highlyunstablebecause it would be very difficultfor a nucleuswith only 6 protons to hold on family 10 electrons. Therefore,carbon forms only covalentcompounds by SHARING of electrons:
In contrast due to inert pair effect, +2 oxidationstate of PB is more stablethanits +4 oxidationstate.
Further on movingdown the group, the ionization enthalpy decreases. As a result , the sum of the first two IONIZATIONENTHALPIES of Pb is quitelow and hence Pb caneasily lose two electronsto form `Pb^(2+)` ions.
Consequently, Pb form ionic compounds.
Discussion
38.

What is oxidation number of Cr and N in (underline(N)H_(4))_(2)underline(Cr)O_(4)?

Answer»

`+6,+5`
`-6,-3`
`+6,-3`
`-3,6`

ANSWER :C
Discussion
39.

What is the mole fraction of the solute in a 1.00m aqueous solution ?

Answer»

0.0354
0.0177
0.177
1.77

Solution :1.00 m solution MEANS 1 mole solute is present in 1000 G WATER.
`n_(H_2O)=1000/18` = 55.5 mol `H_2O`
`X_"solute"=n_"solute"/(n_"solute" + n_(H_2O))`
`=1/(1+55.5)`=0.0177
Discussion
40.

Which of the following is not a greenhouse gas ?

Answer»

Methane
Ozone
Carbondioxide
Nitrogen

Answer :D
Discussion
41.

Use the molecular orbital energy level diagram to show that N_(2) would be expected to have atriple bond, F_(2) a single bond and Ne_(2), no bond.

Answer»

Solution :Formation of `N_(2)` molecule :
Electronic configuration of N-atom `""_(7)N =1s^(2), 2s^(2) , 2p_(x)^(1) 2p_(y)^(1) , 2p_(z)^(1)`
`N_(2)` molecule = `sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2) , sigma^(**) 2s^(2), PI 2sp_(x)^(2) = pi 2p_(y)^(2), sigma 2p_(z)^(2)`

Bond ORDER value of 3 means that `N_(2)` CONTAINS a triple bond
Formation of `F_(2)` molecule :
`""_(9)F = 1s^(2), 2s^(2, 2p_(x)^(2), 2p_(y)^(2), 2p_(z)^(1)`
`F_(2) " molecule " = sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2) , sigma^(**) 2s^(2), sigma 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2)`

Bond order = `(1)/(2) [N_(b) - N_(a) ]= (1)/(2) (10 - 8) =1 `
Bond order value 1 means that `F_(2)` contains SINGLE bond.
Formation of `Ne_(2)` molecule :
`""_(10)Ne = 1s^(2), 2s^(2) , 2p_(x)^(2) , 2p_(y)^(2) , 2p_(z)^(2)`
`Ne_(2) "molecule" = sigma 1s^(2), sigma^(**) 2s^(2), 2 sigma 2s^(2) , sigma^(**) 2s^(2) ,sigma 2p_(z)^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) = sigma 2p_(z)^(2)`
Bond order = `(1)/(2) [ N_(b) - N_(a) ]= (1)/(2) (10 - 10) =0 `

Bond order value zero means that there is no formation of bond between two Ne-atoms. Hence, `Ne_(2)` molecuule does not exist,
Discussion
42.

The unequal sharing of the bond pair of electrons between two atoms in a molecule causes

Answer»

NON POLAR covalen bond
Radical formation
Polar COVALENT bond
Decomposition of molecule

Answer :C
Discussion
43.

Which of the following alcohol is most acidic ?

Answer»

`CH_(3)OH`
`C_(2)H_(5)OH`

`(CH_(3))_(3)C-OH`

SOLUTION :Order of acidic strength in ALCOHOLS is `CH_(3)OH gt 1^(@) gt 2^(@) gt 3^(@)`.
Discussion
44.

The tendency of BF_3, BCl_3 and BBr_3 to behave as Lewis acid decreases in the sequence:

Answer»

`BCl_3 GT BF_3 gt BBr_3`
`BBr_3 gt BCl_3 gt BF_3`
`BBr_3 gt BF_3 gt BCl_3`
`BF_3 gt BCl_3 gt BBr_3`

Solution :p-p overlap between B and F is maximum due to identical SIZE and energy of p-orbitals, so electron deficiency in BORON of `BF_3` is neutralized partially to the maximum extent by BACK DONATION. Hence `BF_3` is least acidic.
Discussion
45.

Which one of the following statement is correct ?

Answer»

Melting point of and BOILING point of HI are
GREATER thanthose of HF
Boiling point of HI is less than that of HF but
melting point of HI is greater than that of HF
Boiling point of HI is greater than that of HF
but melting point of HI is less than that of HF
Melting point and boiling point of HI are less
than that of HF

ANSWER :B
Discussion
46.

Which of the following is an amphoteric ?

Answer»

`SnO_2`
`CO_2`
`P_2O_5`
MgO

Answer :A
Discussion
47.

Which of the following is not a monomer for ahigh molecularmass silicone polymer ?

Answer»

`Me_(3)SiCl`
`PhSiCl_(3)`
`MeSiCl_(3)`
`Me_(2)SiCl_(2)`

Solution :`Me_(3)SiCl` on hydrolysisgives `Me_(3)SiOH`whichhasonly oneOH group and hencecannot ACT as a MONOMERIN condensationpolymerization to GIVE high molecularmass silicone polymer.
Discussion
48.

Which of the fallowing alkyl halides is not suitable for Corey-house synthesis of alkanes?

Answer»

`CH_3I`


Solution :TERTIARY alkyl HALIDE cannot SUITABLE for Corey HOUSE synthesis.
Discussion
49.

Thermonuclear reaction is related to

Answer»

FUSION of HYDROGEN
Source of ENERGY from the SUN
Hydrogen bomb
All the above

ANSWER :D
Discussion
50.

The solubility of AgCl is 1.435xx10^(-5) gL^(-1) at 30^@C temp. then calculate its solubility product .

Answer»

SOLUTION :`1.0xx10^(-14)`
Discussion