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.

C1 of alpha - D- glucoseand C_(2) of beta - D-fructoseare linked togetherto form a molecule of sucrose .

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ANSWER :1
Discussion
2.

C — X bond is strongest in:

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`CH_3Cl`
`CH_3Br`
`CH_3F`
`CH_3I`

ANSWER :C
Discussion
3.

C - X bond is strongest in

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`CH_3Cl`
`CH_3Br`
`CH_3F`
`CH_3I`

ANSWER :C
Discussion
4.

Write the IUPAC name of

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SOLUTION : IUPAC NAME of the GIVEN COMPOUND is
Discussion
5.

(c) Why Zn+2 salts are white ?

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SOLUTION :(C) ABSENCE of UNPAIRED ELECTRON.
Discussion
6.

c) Write Nernst equation for the cell represented as: Mg(s)|Mg_((aq))^(+2)||Al_((aq))^(3+)|Al_((S)).

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Solution :The cell reaction according to the cell REPRESENTATION is `[Mg|Mg^(+2)||Al^(3+)|Al]`
`3Mg+2Al^(3+)rarr 3Mg^(2+)+2Al`
Nernst equation for the GIVEN cell is
`E_("cell")=E_("cell")^(Theta)-(0.059)/(6)log""([Mg^(2+)]^(3))/([Al^(3+)]^(2))`
Discussion
7.

(c) Why have the transition metal high enthalpy of hydration ?

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SOLUTION :(C) Due to their small size and LARGE NUCLEAR CHARGE.
Discussion
8.

Which is more basic among aqueous solutions of aniline and ammonia ? Give one reason.

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Solution :C) AMMONIA is more BASIC than aniline DUE to delocalisation of lone pair of electrons in the ring.
Discussion
9.

Which class of alcohols do not readily form turbidity with Lucas reagent?

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Solution :`1^(@)` ALCOHOLS donot readily form turbidity with LUCAS REAGENT.
Discussion
10.

(c) What is the effect of lanthoids contraction beyond lanthanoid ?

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Solution : (c) SIZE of RESPECTIVE 4d and 5d SERIES elements becomes comparable from fourth GROUP onwards (e.g., Zr and HF).
Discussion
11.

What major molecular shape does the tertiary structure of protein lead to?

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SOLUTION :Fibrous (or globular) SHAPES are FORMED by tertiary structure of protein.
Discussion
12.

What is Zeigler-Natta catalyst?

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SOLUTION :`TiCI_(4)` and triethyl aluminium is known as Zeigler-Natta CATALYST
Discussion
13.

What is the effect of a catalyst on DeltaG of a reaction?

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SOLUTION :CATALYST has no EFFECT on `DELTAG` of a REACTION.
Discussion
14.

(C) turns lime water milky which disappears on continuous passage of (C) forming a compound (F). Identify (A) to (F) and give chemical equations for the reactions at step (i) to (iv).

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ANSWER :`A = 2CuCO_(3).Cu(OH)_(2), B=CuO,C=CO_(2), D=Du, E=Cu_(2)[FE(CN)_(6)], F = Ca(HCO_(3))_(2)`. Azurite
Discussion
15.

Molar conductivities (Lambda_(m)^(@)) at infinite dilution of NaCl, HCl and CH_(3)COONa arc 126.4, 425.9 and 91.0 S cm^(2) mol^(-1) respectively. Lambda_(m)^(@) for CH_(3)COOH will be

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SOLUTION :`Lambda_(m)^(@)CH_(3)COOH=LAMBDA^(@)CH_(3)COONa+Lambda^(@)HCl-Lambda^(@)NaCl`
`=91.0+425.9-126.4`
`="390.5 cm"^(2)"MOL"^(-1)`
Discussion
16.

c) Resistance of a conductivity cell containing 0.1 M KCl solution is 100Omega. Cell constant of the cell is 1.29/cm. Calculate the conductivity of the solution at the same temperature.

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SOLUTION :Given
Resistance `=100Omega`
Cell constant `=1.29//cm`
To find conductivity of the cell we have the formula
`K=("Cell constant")/("RESISTANT")`
`K=(1.29)/(100)=1.29xx10^(-2)SCM^(-1)`
Discussion
17.

C+O_(2)rarrCO_(2), DeltaH=XCO+(1)/(2)O_(2)rarrCO_(2),DeltaH=YThen the heat of formation of CO is

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X-Y
Y-2X
X+Y
2X-Y

Solution :X-Y and FIND the REQUIRED RESULT.
Discussion
18.

C-O-C bond in ethers can be cleaved by

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`KMnO_(4)`
`LiAlH_(4)`
KOH
HI

Answer :D
Discussion
19.

C-O-C bond angle in ether is more than H-O-H bond angle in water although oxygen is sp^(3) hybridised in both the cases. Explain

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Solution :DUE to BIGGER size of ALKYL groups, there are repulsive interactions present.
Discussion
20.

C-O bond length in phenol is less than C-O bond length in methyl alcohol because

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partial double bond CHARACTER DUE to resonance
partial double bond character due to INDUCTIVE effect
more electronegativity of oxygen
oxygen contain two LONE pair of electrons

Answer :A
Discussion
21.

C-O bond in ether is not cleaved by

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HCI
Dil. `H_2SO_4`
HBr
HI

Answer :A
Discussion
22.

Name the final product of ammonolysis of an alkyl halide.

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Solution :The process of cleavage of the C-X bond of HALO alkane by AMMONIA is called ammonolysis. This reaction is carried out in a sealed vessel at 373 K.
`underset(1^(@)" amine")(R-NH_(2))OVERSET(R-X)rarr underset(2^(@)-" amine")(R_(2)-NH)overset(R-X)rarr underset(3^(@)" amine")(R_(3)-NH) overset(R-X)rarr underset("ammonium SALT")underset("Quaternary")(R_(4)-N^(+)H^(-))`
The final product of ammonolysis of an ALKYL halide is quaternary ammonium slat.
Discussion
23.

Name the enzyme that catalyses the reaction: H_(2)NCONH_(2)+H_(2)Orarr2NH_(3)+CO_(2).

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Solution :The enzyme that USED in this reaction is urease
`NH_(2)CONH_(2)+H_(2)O overset("urease")rarr 2NH_(3)+CO_(2)`
Discussion
24.

Name the base present only in DNA but not in RNA.

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SOLUTION :c) The BASE present only in DNA but not in RNA is Thymine.
Discussion
25.

c) Molecular mass of polymers are expressed as an average. Give reason

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Solution :MOLECULAR mass of POLYMERS are expressed as an average because polymers contains CHAINS of different lengths.
Discussion
26.

C is

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SOLUTION :
Discussion
27.

c) For the reaction, H_(2)+I_(2)rarr 2HI, the rate of disappearance of H2 is 1xx10^(-4)Ms^(-1). What is the rate of appearance of HI.

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Solution :Rate EXPRESSION for the reaction
`H_(2)+I_(2)rarr2HI` is
`-(Delta[H_(2)])/(Deltat)=-(Delta[I_(2)])/(Deltat)=(1)/(2)([HI])/(Deltat)"…(1)"`
Given that
`-(Delta[H_(2)])/(Deltat)="Rate "=1xx10^(-4)MS^(-1)`
SUBSTITUTE in the equation (1)
Rate of appearance will be equal to rate of disappearance
`-(Delta[H_(2)])/(Deltat)=(1)/(2)(Delta[HI])/(Deltat)`
`1xx10^(-4)=(1)/(2)(Delta[HI])/(Deltat)`
`2(1xx10^(-4))=(Delta[HI])/(Deltat)="Rate "=2xx10^(-4)ms^(-1)`
Rate of appearance of `HI` is `2xx10^(-4)ms^(-1)`
Discussion
28.

Fluorine does not exhibit positive oxidation state. Why?

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Solution :FLUORINE does not EXHIBIT positive oxidation STATE because it is the most electronegative element.
Discussion
29.

C electricity deposits

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10.8 G of AG
965000 G of Ag
electrochemical EQUIVALENT of Ag
half of electrochemical equivalent of Ag

Solution :Formfaraday's first LAW
mass W= Z.Q
whenQ = 1C
w=Z = electrochemical equivalent
Discussion
30.

C (diamond) +O_(2)(g)rarrCO_(2)(g), DeltaH-395 kJC (graphite) +O_(2)(g)rarrCO_(2)(g), DeltaH-393.5 kJ From the data, the DeltaH when diamond is formed from graphite, is

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`-1.5 kJ`
`+1.5 kJ`
`+3.0 kJ`
`-3.0 kJ`

SOLUTION :Aim: `C_(("GRAPHITE"))rarrC_(("DIAMOND")), eq. (ii)-(i)`GIVES the result.
Discussion
31.

C-COOH overset(NH_(3).Heat)toA overset(Br_(2)+KOH)toB overset(HONO)toC. The compound ‘C' is

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SOLUTION :
Discussion
32.

C Cl_(4) is used as a fire extinguisher, because

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its boiling point is low
its melting point is high
it gives incombustible vapour
it FORMS COVALENT bond.

Solution :`C Cl_(4)` is USED as fire extinguishers under the name pyrene. The dense vapours form protective layer on the burning objects and prevent the oxygen or AIR to come in contact with the burning objects and gives incombustible vapours.
Discussion
33.

C Cl_(4)+H_(2)overset(Fe)underset(H_(2)O)rarr (A)+HClThe product (A) is …….

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`CH_(2)Cl_(2)`
`CHCl_(3)`
`CHI_(3)`
`C Cl_(2)F_(2)`

ANSWER :B
Discussion
34.

C Cl_4 does not give a precipitate with AgNO_3 be because

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It FORMS COMPLEX with `AgNO_3`
`Cl_2` gas is evolved
Chloride IONS are not formed
`AgNO_3` does not give `Ag^+` ions

Answer :C
Discussion
35.

C Cl_3CH=CH_2overset(Cl_2H_2O)rarrA. A is :

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`HC Cl_3underset(OH)UNDERSET(|)C=CH_2Cl`
`HC Cl_3underset(OH)underset(|)C=CH_2OH`
`HC Cl_3underset(CL)underset(|)C-underset(Cl)underset(|)CH_2`
`HC Cl_3underset(OH)underset(|)C-underset(OH)underset(|)CH_2`

ANSWER :B
Discussion
36.

C Cl_3 is used as……………..

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SOIL STERILIZING agent
organic synthesis
good solvent
antioxidant

Answer :A
Discussion
37.

C CI_(4) and freons

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are geencompoundsbecausethey aregreencoloured
DEPLETE ozoneconcentration
causeincreasein ozoneconcentration
HAVENO EFFECTON ozoneconcentration

ANSWER :B
Discussion
38.

c) Calculate the pH of an aqueous solution obtained by mixing 50 ml of 0.2 M HCI with 50 ml 0.1 M" " NaOH.

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Solution :No. of moles of HCI `=0.2 xx50xx10^(-3)`
`=10xx10^(-3)`
No. of moles of NAOH `=0.1 xx50xx10^(-3)`
`=5xx10^(-3)`
No. of moles of HCI after MIXING
`=10xx10^(-3)-5xx10^(-3)`
`5xx10^(-3)`
after mixing total VOLUME `= 100` mL
`therefore` Concentration of HCI in moles per litre
`=(5xx10^(-3)"Mole")/(100xx10^(-3) L)`
`[H_3 O^+]=5xx10^(-2) M`
`pH=-log (5xx10^(-2))`
`=2-log5`
`=2-0.6990`
`=1.30`.
Discussion
39.

The C - C bond length of benzene is _____.

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`1.54overset@A`
`1.20overset@A`
`1.39overset@A`
`1.34overset@A`

ANSWER :C
Discussion
40.

c) Between which is an allylic chloride?

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SOLUTION :
Discussion
41.

c) Between t_(2g) and e_(g) sets of d-orbitals of a central metal in an octahedral complex, which set has higher energy?

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Solution :In octahedral COMPLEX the central metal ion, surrounded.by 6 ligands. The REPULSION is more when the metal d-orbital directed TOWARDS the ligand and less when it is away from the ligands. So `d_(x^(2)-y^(2)) and d_(z^(2))` orbitals are towards the ligand. Hence they EXPERIENCE more repulsion and will be raised in energy. So they forin eq set
Discussion
42.

C and Si belong to group 14. The maximum coordination numberof carbon in commonly occurring compounds in 4, whereas that of silicon is 6. This is due to :

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Large SIZE of silicon
Availability of vacant d-orbitals in silicon
More ELECTROPOSITIVE nature of silicon
Silicon being VULNERABLE to ATTACK by nucleophilic reagents

Answer :B
Discussion
43.

C-6 H_5 CH=CH-CHO is

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Benzaldehyde
Salicyldehyde
Cinnamaldehyde
None of the above

Answer :C
Discussion
44.

C - 14 is used in carbon dating of dead objects because

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Its half-life is `10^(3)` years
Its half-life is `10^(4)` years
It is FOUND in nature abundantly and in definite RATIO
It is found in dead ANIMALS abundantly

Solution :`C -14` is found in nature abundantly and in definite ratio
Discussion
45.

C-12 and C-14 isotopes are found as 98% and 2% respectively in any sample. Number of atoms of C-14 in 12 g of sample will be

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1.5 MOLE ATOMS
`1.03 xx 10^22` atoms
`3.06 xx 10^22` atoms
`3.08 xx 10^23 ` atoms

Solution :No. of C attoms of C -14 in 12g of sample
` = (12 xx 2)/(100 xx 14) xx 6.02 xx 10^23 =1.032 xx 10^22`
Discussion
46.

Byavais lattices are of

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10 TYPES
8 types
7 types
14 types

Solution : There are 14 Bravais LATTICES .
Discussion
47.

ByWurtz reaction, amixtureof methyl iodineand ethyliondinegives :

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Propane
ethane
butane
a mixtueof theabovethree

Answer :D
Discussion
48.

By which substance silver is leached ?

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`KCN` 
`NaCN`
`ZN(CN)_2`
`[Zn(CN)_4]^(2-)`

ANSWER :B
Discussion
49.

By which reagent the diethyl ether is dissociate at high temperature ?

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Water
HI
`KMnO_(4)`
HCl

Solution :`UNDERSET("Diethyl ether")(CH_(3)CH_(2)-O-CH_(2)CH_(3)+ 2HI ) OVERSET(Delta)to underset("Ethyl IODIDE")(2CH_(3)CH_(2)I+ H_(2)O)`
Discussion
50.

By which reagent n.propyl alcohol and iso propyl alcohol can be separated ?

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`PCI_(5)`
Reduction
Oxidation with POTASSIUM diformate
REACTION with `O_(3)`(ozonolysis)

SOLUTION :Oxidation with potassium diformate
Discussion