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.

Based VB theory explain why[Cr(NH_3)_6]^(3+) is paramagnetic , while [Ni(CN_4) ^(2-)is diamagnetic.

Answer»

Solution :(a) `[Cr(NH_(3))_(6)]^(3+)`
In this complex Cr is in the +3 oxidation STATE.
Electronic configuration of Cr atom
Electronic configuration of `Cr^(3+)` ion
Hybridisation and formation of `[Cr(NH_(3))_(6)]^(3+)` complex
* Due to the presence of three unpaired electrons in `[Cr(NH_(3))_(6)]^(3+)` it behaves as a PARAMAGNETIC substance
* The spin magnetic moment, `mu s= sqrt(3(3+2)) = sqrt15 = 3.87BM`
* `[Cr(NH_(3))_(6)]^(3+)` is an inner orbital octahedral complex.

(b) `[Ni(CN)_(4)]^(2-)`
In this complex Ni is the +2 oxidation state.
Electronic configuration of Ni atom
Electronic configuration of `Ni^(2+)` ion

Hybridisation and formation of `[Ni(CN)_(4)]^(2-)` Complex

Since `CN^(-)` is strong field LIGAND, hence the electrons in 3d orbitals for forced to PAIR up and there is no unpaired electron in `[Ni(CN)_(4)]^(2-)`, hence it should be DIAMAGNETIC substance.
Discussion
2.

Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid.

Answer»

Solution :Associated colloid : SOAP solution. It may contain as many as 100 molecules or more.
Multimolecular colloid : Sulphur sol. It CONSISTS of PARTICLES containing a THOUSAND or more `S_8` molecules.
Discussion
3.

Based on the type of particles of dispersed phase and dispersion medium give an example each of associated colloid and mutimolecular colloid.

Answer»
Discussion
4.

Based on the structures how is the reduction ability of H_(3)PO_(2) or H_(3)PO_(3) accounted for?

Answer»

Solution :`H_(3)PO_(2)` has TWO P-H bonds. The two H atoms linked directly to central P atom IMPARTS reducingcharacter to the acid.
SIMILARLY `H_(3)PO_(3)` has one P-H bond, imparting reduction asbility.
Discussion
5.

Based on the nature of order present in the arrangement of the constituent particles, solid are classified into two crystalline and amorphous. List out any four points of difference between crystalline and amorphous solids.

Answer»

SOLUTION :
Discussion
6.

How will you prepare Nylon 6,6 ?

Answer»

SOLUTION :REFER to THEORY SECTION
Discussion
7.

Classify the following as addition and condensation polymers: Terylene, Polyvinyl chloride, polythene

Answer»

Solution :ADDITION polymers-Polyvinyl chloride, polythene
CONDENSATION polymers-Bakelite, TERYLENE
Discussion
8.

Based on the given diagram, which of the following statements regarding the solutions of two miscible volatile liquids are correct? 1. Plots AD and BC show that Raoult's law is obeyed for the solution in which B is a solvent and A is the solute and as well as for that in which A is solvent and B is solute. 2. Plots CD shows that Dalton's law of partial pressures is observed by the binary solutions of components A and B. EF + EG=EH,and AC and BD corresponds to the vapour pressures of the pure solvents A and B, respectively. Select the correct answer using the codes given below:

Answer»

1 and 2
2 and 3
1 and 3
1,2, and 3

Solution :N//A
Discussion
9.

Based on the given diagram, which of the following statements regarding the solutions of two miscible volatile liquids are correct ?

Answer»

Plots AD and BC show that Raoult's LAW is obeyed for the solution in which B is a solvent and A is the solute and as well as as for that in which A is solvent and B is solute.
Plots CD SHOWS that Dalton's law of partial pressure is obeyed by the binary SOLUTIONS of components A and B.
`EF + EG = EH` and AC and BD correspond to the VAPOUR pressure of the pure solvent A and B respectively.
As B is added to the component A, vapour pressure of A as well as of B changes.

Answer :A::B::C::D
Discussion
10.

Based onthe following thermochemical equations H_(2)O(g)+C(s)rarrCO(g)+H_(2)(g), DeltaH=131 kJCO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),DeltaH=-282 kJC(s) +O_(2)(g)rarrCO_(2)(g),DeltaH=X kJThe value of X is

Answer»

`-393 KJ`
`-655 kJ`
`+393 kJ`
`+655 kJ`

Solution :eq. (i) + eq. (II) + eq. (III) gives
X=131-282-242=-393 kJ.
Discussion
11.

Based on the following graph, mark the correct option for the reaction: A rarr P

Answer»

II order reaction
Rate CONSTANT is `0.5 (mol//L)^(1-n) sec^(-1)`
`[A]_(0) = 0.5M`
Reaction COMPLETE in finite time

Answer :C
Discussion
12.

Based on the following data for a reaction what is its order (Ato products) "Conc.A" ""2M""0.2M""0.02M""0.00 "Time in min"""0""10""20""prop

Answer»

`1ST`
`2ND`
`3RD`
zero

Answer :A
Discussion
13.

Based on the first law of thermodynamics, which one of the following is correct ?

Answer»

For an ISOTHERMAL process,`q=+omega`
For an isothermal process ,`DeltaU`=-q
For an adiabatic process,`DeltaU`=-w
For an adiabatic process,q =-w

Solution :BASED on the first law of a cyclic process is zero,i.e., `DeltaU`=0.
`therefore` q=-w
Discussion
14.

Based on the first law of thermodynamics, which one of the following statement is correct?

Answer»

For an ISOTHERMAL PROCESS, q=+W
for an ISOCHORIC process, `DeltaU=-q`
For an ADIABATIC process, `DeltaU=-W`
For a CYCLIC process, q=-W

Solution :`DeltaU=q+w`
For an isothermal process, `DeltaU=0,""q=-W`
For an isochoric process, `W=0,""DeltaU=q`
For an adiabatic process, `q=0,""DeltaU=W`
For a cyclic process, `DeltaU=0,""thereforeq=-W`
Discussion
15.

Based on the data given below, the correcy order of reducing power is: Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V

Answer»

`BR^(-) lt Fe^(2+) lt Al`
`Fe^(2+) lt Al lt Br^-`
`Al lt Br^(-) lt Fe^(2+)`
`Al lt Fe^(2+) lt Br^(-)`

Solution :Lower the REDUCTION potential, more is the REDUCING power .
`Br^-` lt `Fe^(2+)` lt Al
Discussion
16.

Based on the data given below , the correctorder of reducing power is : Fe_((aq))^(3+) + e to Fe_((aq))^(2+) , E^(@)= 077V Al_((aq))^(3+) + 3e to Al_((s)) , E^(@) = -1.66V Br_(2(aq)) + 2e to 2Br_((aq))^(-) : E^(@) = + 1.08 V

Answer»

`Br^(-) lt Fe^(2+) lt AL `
`Fe^(2+) lt Al lt Br^(-) `
`Al lt Br^(-) lt Fe^(2+) `
`Al lt Fe^(2+) lt Br^(-)`

ANSWER :A
Discussion
17.

Based on the data given below , the correct order of reducing power is : Fe^(3+)(aq) +erarrFe^(2+)(aq),E^@=+0.77V AI^(3+)(aq)+3erarrAI(s),E^@=-1.66V Br_2(aq) +2e rarr2Br^-(aq),E^@ =+1.08V

Answer»

`Br^- LT Fe^(2+) lt AI`
`Fe^(2+) lt AI lt Br^-`
`AL lt Br^- lt Fe^(2+)`
`AI lt Fe^(2+) lt Br^-`

Answer :A
Discussion
18.

Based on solute - solvent interactions, arrange the following in order of increasing solubility in n - octane and explain.Cyclohexane, KCl, CH_(3)OH, CH_(3)CN.

Answer»

Solution :n - octane is a non - polar solvent. Therefore, the solubility of a non - polar SOLUTE is more than that of a polar solute in the n - octane as like dissolves in like.
The order of increasing POLARITY is :
Cyclohexane `lt CH_(3)CN lt CH_(3)OH lt KCl`
Therefore, the order of increasing solubility is :
`KCl lt CH_(3)OH lt CH_(3)CN lt` Cyclohexane.
Discussion
19.

Based on solute-solvent interactions, arrange the following in order of increasing sobulity in n-otane and explain. Cyclohexane, KCl, CH_(3)OH, CH_(3)CN

Answer»

Solution :(i) Cyclohexane and n-octane both are non-polar. Hence, they mix completely in all proportions.
(II) KCL is an ionic compound while n-octane is non-polar. Hence, KCl will no dissolve at all in n-octane.
(iii) `CH_(3)OH and CH_(3)CN`both are polar but `CH_(3)CN` is less polar than `CH_(3)OH`. As the SOLVENT is non-polar, `CH_(3)CN` will dissolve more than `CH_(3)OH` in n-octane.
THUS, the order of solubility will be `KCl lt CH_(3)OH lt CH_(3)CN lt "Cyclohexane."`
Discussion
20.

Based on solute-solvent interactions, arrange the following in order of increasing solubilityin n-octane and explain : Cyclohexane, KCI, CH_3OH, CH_2CN .

Answer»

Solution :(i) Cyclohexane and n-octane both are non-polar. Hence, they MIX completely in all PROPORTIONS. LIKE dissolves like. (II) KCL is an ionic compound while n-octane is non-polar. Hence, KCl will not dissolve at all inn-octane. (iii) `CH_3OH` and `CH_3CN `both are polar but `CH_3CN`is less polar than `CH_3OH. CH_3CN`willdissolve more than `CH_3OH ` in n-octane.
Thus, the order of solubility will be `KCl lt CH_3OH lt CH_3CN lt ` Cyclohexane.
Discussion
21.

Based on particles of dispersed phase, collodidal systems are classified into multimolecular, macro molecular and associated colloids. Which of the following colloidal system is an example for multimolecular colloidal system?

Answer»

STARCH in water
Soap solution
Ferric HYDROXIDES in water
Polyvinyl ALCOHOL in water

Answer :C
Discussion
22.

The bond order depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statement is/are correct about bond order ?

Answer»

Can have a NEGATIVE quantity
Has always an integral value
Is a NON ZERO quantity
Can assume any POSITIVE or integral or fractionalvalue including zero

Answer :D
Discussion
23.

Based on molecular forces what type of polymer is neoprene?

Answer»

SOLUTION :ELASTOMER.
Discussion
24.

Based on molecular forces what type of polymer is neoprene ?

Answer»

SOLUTION : ELASTOMER.
Discussion
25.

Based on Langmuir adsorption isotherm, the intercept in the graph (m/x "versus" 1/p) is equal to

Answer»

`1/a`
`B/a`
`a/b`
`a/("SLOPE")` 

ANSWER :B
Discussion
26.

Based on electron gas theory explain why metals are malleable.

Answer»

Solution :When force (stress) is applied to a metal Kernel (positive ions or lattice points LAYERS slide over one another. The shape of the metal gets deformed.
Freely MOVING electrons also move around the deformed layers and the environment around the metal ions remains unchanged.
OR
According to electron gas THEORY a metal CONTAINS an ARRAY of positively charged metal ions in a Sea or mobile valence electrons and the bonding is non directional.
When force (stress) is applied to a metal the Kernel layers slide one over ther other. The shape of the metal gets deformed.
Discussion
27.

Based on data given below, match column-I and column II 3X(g)overset("250K,1atm")toX_3(g), DeltaH=-200 KJ//mol , DeltaS=-300 J//K//mol X(l)overset("1atm,250K")toX(g), DeltaH_(vap)=10 KJ//mol X_3(l)overset("500K,1atm")toX_3(g), DeltaH_(vap)=80 KJ//mol X_3(g,250 K , 1 atm )toX_3(g, 500 K, 1 atm) DeltaH=100KJ/mol,DeltaS=100 J/K/mol , DeltaG=60 KJ/mol Normal boiling point of X=250 K normal boiling point of X_3=500 K. Enthalpy change and entropy change of cooling X_3(l) to lower temperature is neglible.

Answer»


SOLUTION :
OVERALL
`DeltaH=3xx10-200+100-80+0=300-200+100-80=-150` kJ/mol
`DeltaS=20-300+100-160+0=-240` J/mol/K
`DeltaG=0+(-125)+60+0+0=-65` kJ/mol
Discussion
28.

Based on Arrhenius concept, define acid and bases and give an example for each.

Answer»

Solution :According to Arrhenius, an ACID is a substance that dissociates to give hydrogen ions in water For example, `HCI, H_2SO_4` etc. are acids. Their DISSOCIATION in aqueous solution is expressed as
`HCI_((g))overset(H_2O)(hArr) Ca_((aq))^(2+)+2OH_((aq))^(-)`
Similarity a base is a substance that dissociates to give hydroxyl ions in water. For example, substance LIKE `NAOH,Ca(OH)_2` etc. are bases.
`Ca(OH)_2 overset(H_2O)(hArr) Ca_((aq))^(2+)+2OH_((aq))^(-)`
Discussion
29.

Base uracil is not present in DNA.

Answer»


ANSWER :1
Discussion
30.

Base sugar phosphate unit of nuclic acid is known as …………….....

Answer»

NUCLEOTIDE
NUCLEOSIDE
phosphotide
NONE of these

SOLUTION :nucleotide
Discussion
31.

Base strength of

Answer»

`i gt iii gt ii`
`i gt ii gt iii`
`ii gt i gt iii`
`iii gt ii gt iii`

Solution :BASE strength decreases as the electronegativity of the CARBON carrying the negative charge INCREASES from `sp^(3)-sp^(2)-sp`. Thus(B) is CORRECT
Discussion
32.

Base present in RNA and not in DNA is :

Answer»

Uracil
Cytosine
Guanine
Thymine.

Answer :A
Discussion
33.

Base-catalyzed aldol condensation occurs with

Answer»

PROPIONALDEHYDE
benzaldehyde
2-methylpropanal
2,2-dimethylpropanal

Answer :A::C
Discussion
34.

Base-catalysed aldol condensation occurs with

Answer»

Propionaldehyde
Benzaldehyde
2-methylpropionaldehyde
2,2-dimethylpropionaldehyde.

Solution :CARBONYL COMPOUND having a-hydrogen ATOM SHOWS this REACTION.
Discussion
35.

Base catalysed aldol condensation occurs with:

Answer»

Propionaldehyde
Benzaldehyde
2, 2-dimethylpropionaldehyde
None

Answer :A
Discussion
36.

Base catalysed aldol condensation occurs with

Answer»

propanal
butanal
ETHANAL
all of these

Solution :Aldolcondensation reactionshown by those aldehydes and ketones containing `alpha` hydrogen atoms or aldehydes and ketones having methyl or methylene GROUP in `alpha` POSITION.
Compounds a,B, c have `alpha` - hydrogen atoms.
Discussion
37.

Baryta water is:

Answer»

BAO
`CA(OH)_2`
`Ba(OH)_2`
`BaSO_4`

ANSWER :C
Discussion
38.

Barytes ore is :

Answer»

`BeSO_4`
`BeCl_2`
`BaSO_4`
`BaCl_2`

ANSWER :C
Discussion
39.

Bartlett equation.

Answer»

Solution :The first orange COLOURED COMPOUND of noble GSES was prepared by Bartlett.
`X_(e)+PtF_(6)RARR X_(e)^(+)PtF_(6)^(-)`
This is KNOWN as Bartlett equation.
Discussion
40.

Barium is extracted with its ore:

Answer»

Dolomite
Witherite
Carnallite
Gypsum

Answer :B
Discussion
41.

Barium ions, CN^(-) and Co^(2+) form an ionic complex.If this complex is 75% ionised in aqueous solution with Vant Hoff factor (i) equal to four and paramagnetic moment is found to be 1.73 BM (due to spin only) then the hybridisation state of Co (II) in the complex will be :

Answer»

`sp^3d`
`d^2sp^3`
`sp^3d^2`
`dsp^3`

Solution :`{:(Ba_(alpha-2)[Co(CN)_x]_2 to,(X-2) Ba^(2+),+2[Co(CN)_x]^(-(x-2))),(1-alpha,(x-2)alpha, ""2alpha):}`
`i=1-alpha+xalpha-2alpha+2alpha`
`(x-1)alpha=3 " " x-1=3/alpha`
`alpha=0.75 , " " x=5`
`:.` FORMULA of COMPLEX is `Ba_3[Co(CN)_5]_2` and HYBRIDISATION is `dsp^3` as it is a inner orbital complex with strong field ligand
Discussion
42.

bar(C)H_(2) - underset(underset(..)O":")underset(||)(C) - CH_(3) harr CH_(2) = CH_(2) underset(underset(..)":O:"^(-))underset(|)(C) - CH_(3)

Answer»

RESONATING structures
tautomers
geometrical isomers
optical isomers.

Solution :`bar(C)H_(2) - UNDERSET(underset(..)O":")underset(||)(C) - CH_(3) harr CH_(2) = CH_(2) underset(underset(..)":O:"^(-))underset(|)(C) - CH_(3)`
These structures INVOLVE only movement of electrons but not of atoms and hence are resonating structures .
Discussion
43.

Barbituric acid is used as a …………..

Answer»

SOLUTION :TRANQUILIZER
Discussion
44.

Barbituric acid is used as

Answer»

An antipyretic
An antiseptic
An antibiotic
A tranquillizer

Answer :D
Discussion
45.

Barbituric acid is a condensation product of A and B. A, B are :

Answer»

`1`. `H_2NCONH_2`, `HO_2C-CO_2H`
`2`. `H_2NCONH_2`, `ClCOCH_3`
`3`. `H_2N-NH_2`, `ClCOCH_2COCl`
`4`. `H_2NCONH_2`, `CLOC CH_2COCl`

SOLUTION :malonyl chloride + UREA ` to` malonyl urea ( Barbituric acid )
Discussion
46.

Barbituric acid and its derivativesare well known as

Answer»

tranquilizers
antiseptics
analgesics
antipyretics.

Solution :The drug which acts on the central nervous system (CNS) and help in reducing stress and fatigue by INDUCING a sense of well being are called tranquilizers. They are PSYCHOTHERAPEUTIC drugs. The most commonly used tranquilizersare BARBITURIC acid and its 5, 5-disubstituted derivatives such as veronal LUMINAL and SECONAL.
Discussion
47.

Barbituric acid and its derivatives are well known as

Answer»

TRANQUILLIZERS
Antiseptics
Antibiotics
ANALGESICS

ANSWER :A
Discussion
48.

Barbituric acid and its derivatives are well known

Answer»

ANALGESICS
antipyretics
tranquillizers
antiseptics

Solution :Barbituric ACID and its DERIVATIVES are tranquilizers and precisely to SAY hypnotic tranquilizers.
Discussion
49.

Barbiturates exert deppressant action on …...

Answer»

GUT
HEART
CEREBROSPINAL axis
All the above

Answer :C
Discussion
50.

Barbiturates are used as :

Answer»

ANTIBIOTIC
Analgesic
Tranquilizer
Antiseptic

Answer :C
Discussion