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.

In chemical behaviour, earlier members of lanthanoids are quite reactive similar to______________.

Answer»

SOLUTION :CALCIUM
Discussion
2.

In chemical adsorption how many layers are there on adsorbent?

Answer»

Infinite
TWO
Zero
One 

ANSWER :D
Discussion
3.

In chemical adsorption, gas molecules are held to the surface by formation of chemical bond and nearly ........... is given out as heat of adsorption

Answer»

SOLUTION :400 kJ/mole
Discussion
4.

In CHCl_(2)+OH^(-)toHCOO^(-), the intermediate (s) is/are

Answer»

:CC`l_(3)^(-)`
:CC`l_(2)`
`CH(OH)_(3)`
:CC`l_(3)`

ANSWER :A::B
Discussion
5.

In charging the lead accumulator battery

Answer»

`PbO_(2)` DISSOLVES
`H_(2)SO_(4)` is reproduced
`PbSO_(4)` deposits on lead electrode
Pb deposits on lead eletrode

Solution :Reactions taking place during charging i.e., when electrical energy is supplied to it from an external field.
`2PbSO_(4(s))+2H_(2)OtoPb_((s))+PbO_(2(s))`
`+4H_((aq))^(+)+2SO_(4(aq))^(2-)`.
Discussion
6.

In chain growth polymerization initiator is

Answer»

free RADICALS
CATIONS
anions
ONE of these

ANSWER :D
Discussion
7.

Between 1M CH_3COOH and0.1M CH_3COOH ,which has higher electrical conductivity?

Answer»

More ACIDIC
LESS acidic
EQUAL acidic
None

Answer :A
Discussion
8.

CH_3COOH is _______ than HCOOH.

Answer»

More ACIDIC
LESS acidic
EQUALLY acidic
None

Answer :A
Discussion
9.

In CH_3CH_2OH the bond which most readily undergoes heterolytic cleavage during its reaction with CH_3COOH//H_2SO_4 is :

Answer»

C-C
C-O
O-H
C-H

Answer :C
Discussion
10.

In CH_(3)CH_(2)OH, the bond that undergoes hyteroytic cleavage most readily id :

Answer»

C-C
C-O
C-H
O-H

Solution :N//A
Discussion
11.

In CH_(3)CH_(2)OH the bond that undergoes hetrolytic change most readily is

Answer»

C-C
C-O
C-H
O-H

Solution :O-H bond undergoes CLEAVAGE most readily because oxygen being more ELECTRONEGATIVE can ACCOMMODATE the negative charge more effectively after the cleavage.
Discussion
12.

In CH_(3)CH_(2)OH, the bond that undergoes heterolytic cleavage most readily is

Answer»

C-C
C-O
C-H
O-H

Solution :`CH_(3)CH_(2)OH OVERSET("Heterolytic cleavage")RARR CH_(3)CH_(2)^(+)+OH^(-)`
Discussion
13.

In CH_3CH_2OH the bond that undergoes heterolytic cleavage nost readily is:

Answer»

C - C
C - O
C - H
O - H

Answer :B
Discussion
14.

In CH_(3)CH_(2)OH, the bond that undergoes heterolytic cleavage most readily is:

Answer»

`C-C`
`C-O`
`C-H`
`O-H`

ANSWER :B
Discussion
15.

In CH_(3)-underset(..)(O)-CH_(3), oxygen atom has p^(3) hybridisation with two lone pair of electron. C-O-C bond angle is:

Answer»

`110^(@)`
`109^(@)28'`
`106^(@)51'`
`104^(@),31'`

Solution :The bonding ANGLE increases DUE to steric hinderance.
Discussion
16.

In CH_(3)-CH_(2)-CH_(2)-Br,C-Br bond is formed by the overlapping of

Answer»

`2sp^(3)-2p_(Z)`
`2sp^(3)-3p_(z)`
`2sp^(3)-3p_(z)`
`2sp^(3)-4p_(z)`

Answer :A
Discussion
17.

In CH_2CI_2, the oxidation number of C is :

Answer»

`-4`
`+2`
ZERO
`+4`

ANSWER :C
Discussion
18.

In certain matters lithium differs from other alkali metals,the main reason for this is

Answer»

Small SIZE of Li atom and `Li^(+)` ION
Extremely high electropositiviy of Li
Greater hardness of Li
Hydration of `Li^(+)` ion

Solution :Small ATOMIC and ionic size LEADS to high electronegativity and hydration energy.
Discussion
19.

In certain cases, achiral ligands can become chiral on coordination to a metal, leading to a complex that is chiral. Usually the non-chiral ligand contains a donor that rapidly inverts as a free ligand, but that becomes locked in one configuration on co-ordination. An example is MeNHCH_(2)CH_(2)NHMe, where two N atoms become chiral centres on co-ordination to a metal atom. Consider two bidentate ligands for square planar complexes only : A A to Me-NH-CH_(2)-CH_(2)-NHMe CD to Me-NH-CH_(2)-CH_(2)-NH_(2) In the complex solution [M(CD)_(2)], identify the complex isomer which does not have any mirror planes but is nor chiral.

Answer»




Solution :(b) has CENTRE of SYMMETRY.
(C ) has PLANE of symmetry, (a) and (d) are optically active.
Discussion
20.

In certain cases, achiral ligands can become chiral on coordination to a metal, leading to a complex that is chiral. Usually the non-chiral ligand contains a donor that rapidly inverts as a free ligand, but that becomes locked in one configuration on co-ordination. An example is MeNHCH_(2)CH_(2)NHMe, where two N atoms become chiral centres on co-ordination to a metal atom. Consider two bidentate ligands for square planar complexes only : A A to Me-NH-CH_(2)-CH_(2)-NHMe CD to Me-NH-CH_(2)-CH_(2)-NH_(2) Total number of chiral and achiral complexes obtained from M(A A)_(2) Will be respectively :

Answer»

2, 1
4, 2
4, 3
2, 2

Solution :
Discussion
21.

In certain cases, physisorption tansfers into chemisorption as temperature is increased. Explain with an example.

Answer»


Answer :`H_2` is adsorbed on nickel by vander WALL's forces (i.e., physisorption) at ORDINARY TERMPERATURE. But if temperature is increased, molecules of `H_2` DISSOCIATE to from H atoms which are HELD on the surface of nickel by chemisorption.
Discussion
22.

In certain areas where coal is cheap, artificial gas is produced for household use by the 'water gas' reaction C(s) + H_2O(g) overset(600^@C)(to) H_2(g) + CO(g) Assuming that coke is 100% carbon, calculate the maximum heat obtainable at 298 K from the combustion of 1.0 kg of coke and compare this value to the maximum heat obtainable at 298 K from burning the water gas produced from 1.0 kg of coke. Heat of combustion of C, H_2 and CO are -94.1, -68.4 and -68.0 kcal/mole respectively.

Answer»

SOLUTION :More ENERGY is obtainable from the water gas, 11366 kcal
Discussion
23.

In C Cl_4, the four valencies of carbon are directed towards the corners of a:

Answer»

Square
Rectangle
Rhombus
Tetrahedron

Answer :D
Discussion
24.

In Castner - Kellner cell for the manufacture of NaOH( caustic soda ), Hg acts as cathode as well as anode.

Answer»


Solution :This cell is also CALLED mercurry cathode cell. In this cell, `Hg` flows along the bottom of the cell and acts as cathode in the left HAND compartment fitted with graphite anodes in brine `(NACL)` solution, whereas the right hand compartment is fitted with `Fe` CATHODES dippped in very dilute `NAOH` solution and `Hg` acts as anode in this compartment.Thus `Hg` acts as an intermediate electrode by induction, cathode on `LHS` and anode in `RHS.`
Discussion
25.

In Castner-Kellner cell for production of sodium hydroxide

Answer»

BRINE is electrolyzed using GRAPHITE electrods
Molten sodium chloride is electrolysed
Sodium amalgam is FORMED at mercury cathode
Brine is electrolyzed with PT electrodes

Answer :C
Discussion
26.

In Castner-Kellner cell for production of sodium hydroxide :

Answer»

Brine is electrolyzed with Pt electrodes
Brine is electrolyzed using graphite electrodes
Molten magnesiumchloride is electrolysed
SODIUM AMALGAM is FORMED at MERCURY cathode

Answer :D
Discussion
27.

In case of zero order reactions

Answer»

`t_(1//2)=2t_(1//4)`
`t_(3//4)=3t_(1//2)`
`t_(OO)=1t_(1//2)`
All

Solution :`(C_(0)-C_(t))=x=Kt,(C_(0)-(C_(0))/2)=K.t_(1/2)impliest_(1/2)=(C_(0))/(2K),(C_(0)-(C_(0))/4)=K.t_(3/4)impliest_(3/4)=(3C_(0))/(4K)=3/2xx(C_(0))/(2K)`
`(C_(0)-(3C_(0))/4)=K.t_(1/4)impliest_(1/4)=(C_(0))/(4K)=1/2xx(C_(0))/(2K),C_(0)-0=Kt_(alpha)impliest_(alpha)=(C_(0))/K`
So `t_(3/4)=3/2xxt_(1/2),t_(1/4)=1/2xxt_(1/2),t_(1/2)=((t_(alpha))/2)`
Discussion
28.

In case of trisubstituted benzene , if the substituents are different, then the number of isomers will be

Answer»

5
8
6
10

Answer :D
Discussion
29.

In case of R,S configuration the group having highest priority is

Answer»

`-NO_(2)`
`-NH_(3)`
`-CN`
`-OH`

SOLUTION :Atomic mass of OXYGEN is more than that of C and N
Discussion
30.

In case of radioactive radiations :

Answer»

some are not deviated by electric and MAGNETIC field
some CARRY negative charge
all are electromagnetic WAVES
all produce X- rays with when suddenly stopped .

SOLUTION :`GAMMA`- ray = No deviation , `alpha` ray = (+) ve charge & deviated , `beta` - ray = (-) ve charge & deviated
Discussion
31.

In case of PCl_(5) lone pair electrons are present

Answer»

s-ORBITAL
p-orbital
`SP^(3)`-o-orbital
`sp^(2)`-orbital

Solution :
`sp^(3)` HYBRID orbital
Discussion
32.

In case of octahedral compex, if the e_(g) orbitals (d_(x^(2) - y^(2)) " and " d_(z^(2))) are asymmetricaaly filled, their degeneracy is destroyed and the ligands approaching along +Z and -Z directions experiences different amount of repulsions than the ligands approaching along the +X, -X, +Y " and " -Y directions. As a result, the symmetrical nature of such complexes is lost and either elongation or compression along Z-axis taken place. Answer the following three questions based on the above situation. In which of the following case, no such elongation or compressions are expected ?

Answer»

`d^(4)` - weak LIGANDS
`d^(5)`- weak ligands
`d^(7)`- STRONG ligands
`d^(9)`- strong or weak ligands

Answer :B
Discussion
33.

In case of octahedral compex, if the e_(g) orbitals (d_(x^(2) - y^(2)) " and " d_(z^(2))) are asymmetricaaly filled, their degeneracy is destroyed and the ligands approaching along +Z and -Z directions experiences different amount of repulsions than the ligands approaching along the +X, -X, +Y " and " -Y directions. As a result, the symmetrical nature of such complexes is lost and either elongation or compression along Z-axis taken place. Answer the following three questions based on the above situation. Select the coorect statement.

Answer»

`[NI (CN)_(4)]^(2-)` is tetrahedral
`[Ni (CN)_(4)]^(2-)` is square PLANAR
`[Cu (CN)_(4)]^(3-)` is square planar
`[CO (CN)_(4)]^(2-)` is tetrahedral

Answer :B
Discussion
34.

In case of octahedral compex, if the e_(g) orbitals (d_(x^(2) - y^(2)) " and " d_(z^(2))) are asymmetricaaly filled, their degeneracy is destroyed and the ligands approaching along +Z and -Z directions experiences different amount of repulsions than the ligands approaching along the +X, -X, +Y " and " -Y directions. As a result, the symmetrical nature of such complexes is lost and either elongation or compression along Z-axis taken place. Answer the following three questions based on the above situation. Which of the following is incorrect regarding K_(4) [Cr F_(6)] ?

Answer»

It has two long and FOUR short `CR-F` bonds
It has four long and two short `Cr-F` bonds
SPIN only magnetic moment of the COMPLEX is approximately `4.9 Bm`
If `Cr` is replaced by `Cu`, similar types of deformation in the regular octahedral geometry are observed

Answer :B
Discussion
35.

In case of nitrogen, NCl_(3) is possible but not NCl_(5) while in case of phosphorous, PCl_(5) as well as PCl_(5) are possible. It is due to

Answer»

Availability of vacant d-orbital in P but not in N
Lower electronegativity of P than N
Lower tendency of H bond formation in P than N
Occurrence of P in SOLID while N in GASEOUS STATE at room temperature

Answer :A
Discussion
36.

In case of nitrogen, NCl_(3) is possible but not NCl_(5)while in case phosphrous, PCl_(3) as well as PCl_(5) are possible. It is due to

Answer»

Availability of vacant d orbitals in P but not in N
Lower electronegativity of P than N
Lower tendency of H-bond formation in P than N
Occurrence of P is SOLID while N in gaseous STATE at room temperature

Answer :1
Discussion
37.

In case of nitrogen, NCl_(3) is possible but not NCI_5 gwhile in case of phosphorous, PCl_3 as well as PCI_5 IM are possible. It is due to

Answer»

Availability of VACANT d-orbital in P but not in N
Lower electronegativity of P than N
Occurrence of P in SOLID while N in gaseous state at ROOM temperature
Occurrence of P in solid while N in gaseous PREL T state at room temperature

Solution :Nitrogen does not have d-orbitals.
Discussion
38.

In case of [Mn(NH_(3))_(6)]^(2+):

Answer»

Relation in between `Delta_(0)" and "P " is " Delta_(0) lt P`
Two UNPAIRED electrons are present in axial d-orbital of `Mn^(2+)`
Hybridisation of central metal atom of complex is `d^(2)sp^(3)`
d-orbitals involved in hybridisation of central atom are `d_(x^(2))" and "d_(xy)`

Solution :`[Mn(NH_(3))_(6)]^(2+)` has `d^(5)` config. (High spin)
`d_(xy)^(1)d_(yz)^(1)d_(ZX)^(1)d_(x^(2)-y^(2))d_(Z^(2))^(1)` pairing does not occur.
`"Hyb"impliessp^(3)d^(2)`.
Discussion
39.

In case of meta-attack on aniline (where -NH_(2) group is o-, p- directing) and in case of ortho-attack on nitrobenzene, resonating structures are possible. Which of them are more stable ?

Answer»


SOLUTION :N//A
Discussion
40.

In case of hydrogen and helium the van der Waals' forces are :

Answer»

Strong
Very strong
Weak
None

Answer :C
Discussion
41.

In case of heteronuclear diatomics of the type AB, where A is mor electronegative than B, bonding molecular orbital resembles the character of A more that of B. The statement

Answer»

is false
is true
cannot be evaluated SINCE DATA is not sufficient
is true only for certain systems.

Solution :BMOs resemble with the more electronegative atom. More electromegative ATOMS are lower in ENERGY and this is due to the hold on the electrons by the nucleus.
Discussion
42.

In case of halogen family which trend occurs as the atomic number increases

Answer»

Ionization radius DECREASE
Ionization POTENTIAL decrease
Covalent CHARACTER in `MX_2` decrease ( where M=metal and X = hologen)
None

Answer :B
Discussion
43.

In case of group III elements, the inert pair effect is predominant in

Answer»

Boron
Aluminium
Indium
Thallium.

Solution :The HEAVIEST element SHOWS MAXIMUM TENDENCY to show INERT pair effect.
Discussion
44.

In case of first order reactions

Answer»

`t_(3//4)=2t_(1//2)`
`t_(7//8)=3t_(1//2)`
`t_(0.99)=10t_(1//2)`
All

Solution :`1to1/2to1/4to1/u,((2.303xxlog2)/(t_(1/2)))xxt_(0.99)=2.303log(100/0.01)`
`log2xx(t_(0.99))/(t_(2)^(1))=LOG10^(4),t_(0.99)=(4xxt_(1/2))/0.3010=13.289xxt_(1/2)=13.3xxt_(1/2)`
Discussion
45.

In case of Daniel cell Zn//Zn^(2+)////Cu^(2+)//Cu which of the followings is NOT true ?

Answer»

ZINC ACTS as an ANODE
Electrons MOVE from ZN to Cu
Copper acts as cathode
Zinc acts as cathode

Solution :Zinc is anode.
Discussion
46.

In case of d-block elements

Answer»

OUTERMOST and penultimate SHELLS are INCOMPLETE
 both penultimate and pre-penultimate shells are incomplete
outermost SHELL is incomplete
INNERMOST shell is incomplete

Answer :A
Discussion
47.

In case of chemisorption, why adsorption first increases and then decreases with temperature?

Answer»

Solution :(i) Chemisorption involves a high activation ENERGY so it is also referred to as activated adsorption.
(ii) It is found in chemisorption that it first INCREASES and then decreases with INCREASE in temperature. When adsorption is plotted, the graph first increaes and then decreases with temperature.
(III) The initial increase illustrates the requirement of activation of the surface for adsorption is DUE to fact that formation of activated complex requires certain energy. But later it decreaes at high temperature is due to desorption as the kinetic energy of the adsorbate increases (exothermic nature)
Discussion
48.

In case of condensation of polymers ?

Answer»

High MOLECULAR weight POLYMERS are formed all at once.
Lower molecular weight polymers are formed all at once.
Molecular weight of polymers rises throughout the reaction.
Have nospecific RELATION to their molecular weight .

ANSWER :C
Discussion
49.

In case of chemisorption, why adsorptionfirst increases and then decreases with temperature?

Answer»

Solution :Adsorption isotherms represents the variation of adsorption at constant temperature. When amount of adsorption is PLOTTED versus temperature at constant pressure, it is called adsorption isobar.

X is the amount of adsorbate adsorbed an "m" g of adsorbent.
(i) In physical adsorption, `x/m`decreases with increases in T, But in CHEMICAL adsorption, `x/m`increases with rise in temperature and then decreases.
(ii) The increase illustrates the requirement of ACTIVATION of the surface for adsorption is due to fact that FORMATION of ACTIVATED complex requires certain energy.
(iii) The decrease at high temperature is due to desorption, as the kinetic energy of the adsorbate increases
Discussion
50.

In case of chemisorption, why adsorption first increases and then decreases ?

Answer»

SOLUTION :The initial increase is DUE to the FACT that heat supplied acts as a ACTIVATION energy. The decrease afterwards is due to the exothermic NATURE of adsorption equilibrium.
Discussion