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.

The character of Fe-C bond in Fe(CO)_5 is _________.

Answer»

`PI` only
`SIGMA` only
ionic only
both `pi` and `sigma`

ANSWER :D
Discussion
2.

The changed name of Khurchatovium (Z=104) is

Answer»

Joliotium
Dubnum
RUTHERFORDIUM
Hatrium 

ANSWER :C
Discussion
3.

The change of Fe in (Fe(CN)_(6)]^(3-) is …………….

Answer»

`-6`
`+3`
`-3`
`+6`

ANSWER :B::C
Discussion
4.

The change of energy on vaporizing 1.00 kg of liquid water at 0^(@)C and 1 atm is

Answer»

2367 kJ `KG^(-1)`
`-2367 kJ kg^(-1)`
`-2367 kJ mol^(-1)`
`-2367 kJ g^(-1)`

ANSWER :A
Discussion
5.

The change of entropy (dS) is defined as

Answer»

`DS=deltaq//T`
`dS=dH//T`
`dS=deltaq_("eqv")//T`
`dS=(dH-dG)//T`

ANSWER :C
Discussion
6.

The change of energy on freezing 1.00kg of liquid water at 0^(@)C A0.001 molal solution of [Pt(NH_(3))_4 CI_4] in water had a freezing point depression of 0.005 4^(@)C.If K_(f) for water is 1.80, the correct formulation for the above molecule is

Answer»

`[Pt(NH_(3))_(4)CI_(3)]CI`
`[Pt(NH_(3))_(4)CI_(2)]CI_(2)`
`[Pt(NH_(3))_(4)CI]CI_(3)`
`[Pt(NH_(3))_(4)CI_(4)]`

Solution :`DELTA T_(F)= imk_(f), 0.0054=ixx1.8xx0.001`
i=3 so it is `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`.
Discussion
7.

The change is entropy for the fusion of 1 mole of ice is [M.P of ice = 273 K molar, enthalpy of fusion for ice = 60 kJ mol^(-1)]

Answer»

`11.73 JK^(-1) MOL^(-1)`
`18.84 JK^(-1) mol^(-1)`
`219.7 JK^(-1) mol^(-1)`
`24.47 JK^(-1) mol^(-1)`

Solution :Entropy change of fusion
`DeltaS_(f)=(DeltaH_(f))/(T)`
`DeltaH_(f)=60XX10^(3)J mol^(-1), T=273 K`
`DeltaS_(f)=(60xx10^(3))/(273)=219.7 J K^(-1)mol^(-1)`.
Discussion
8.

The change in the oxidation state of iodine when axcess chlorine water is added to an iodide salt is

Answer»


ANSWER :6
Discussion
9.

The change in the optical rotation (with time) of freshly prepared solution of sugar is known as

Answer»

SPECIFIC rotation
Inversion
Rotatory motion
Mutarotation

Answer :D
Discussion
10.

The change in the internal energy of a substance

Answer»

Increase with increase in temperature
Decrease with increase in temperature
Can be calculated the relation `E=mc^2`
REMAINS UNAFFECTED with CHANGE in temperature

Answer :A
Discussion
11.

The change in optical rotation with time of freshly prepared solutions of sugar is known as :

Answer»


ANSWER :MUTAROTATION
Discussion
12.

The change in optical rotation of a freshly prepared solution of glucose with time is called :

Answer»

SPECIFIC ROTATION
OPTICAL inversion
Muta rotation
RACEMISATION.

Answer :C
Discussion
13.

The change in internal energy for an isolated system at constannt volume is

Answer»

ZERO
`-2.18` erg
`+4.27` erg
`+2.78` erg

Solution :For an isolated system there is no transfer of energy as heat or work. So, ACCORDING to the first LAW of thermodynamics.
`DeltaU=q+W`
`implies DeltaU=0+0=0`
Discussion
14.

The change in Gibbs free energy for areaction is expanded by...........

Answer»

`DeltaG=DeltaH+TDeltaS`
`DeltaG=DeltaH-TS`
G = H - TS
`DeltaG=DeltaH-TDeltaS`

ANSWER :D
Discussion
15.

The change in Gibbs free energy (DeltaG) of the system along provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of asystem at constant temperature and pressure will be: DeltaG_("system")=DeltaH_("system")-TDeltaS_("system") At constant temperature and pressure: DeltaG_("system")lt0 (Spontaneous) DeltaG_("system")=0 (equilibrium) DeltaG_("system")gt0 (non-spontaneous) Free energy is related to the equilibrium constant as: DeltaG^(@)=2.303RTlog_(10)K_(e). Q. A reaction has positive values of DeltaH and DeltaS. From this you can deduce that the reaction:

Answer»

MUST be SPONTANEOUS at any TEMPERATURE.
cannot be spontaneous at any temperature.
Will be spontaneous only at low temperature
will be spontaneous only at high temperature.

Answer :D
Discussion
16.

The change in Gibbs free energy (DeltaG) of the system along provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of asystem at constant temperature and pressure will be: DeltaG_("system")=DeltaH_("system")-TDeltaS_("system") At constant temperature and pressure: DeltaG_("system")lt0 (Spontaneous) DeltaG_("system")=0 (equilibrium) DeltaG_("system")gt0 (non-spontaneous) Free energy is related to the equilibrium constant as: DeltaG^(@)=2.303RTlog_(10)K_(e). Q. For a reaction to be spontaneous at all temperature.

Answer»

`DeltaG-ve,DeltaH+ve and DeltaS+ve`
`DeltaG+ve,DeltaH-ve and DeltaS+ve`
`DeltaG-ve,DeltaH-ve and DeltaS-ve`
`DeltaG-ve,DeltaH-ve and DeltaS+ve`.

ANSWER :D
Discussion
17.

The change in Gibbs free energy (DeltaG) of the system along provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of asystem at constant temperature and pressure will be: DeltaG_("system")=DeltaH_("system")-TDeltaS_("system") At constant temperature and pressure: DeltaG_("system")lt0 (Spontaneous) DeltaG_("system")=0 (equilibrium) DeltaG_("system")gt0 (non-spontaneous) Free energy is related to the equilibrium constant as: DeltaG^(@)=2.303RTlog_(10)K_(e). Q. If b oth DeltaH and DeltaS are negative, the reaction will be spontaneous:

Answer»

at HIGH temperature
at LOW temperature
at all temperature
at ABSOLUTE zero

Answer :C
Discussion
18.

The change in entropy of an ideal gas during a reversible isothermal expansion is

Answer»

negative
positive
zero
infinite

Answer :B
Discussion
19.

The change in entropy of 2 moles of an ideal gas upon iosthermal expansion at 243.6 K from 20 litre until the pressure becomes 1 atm, is :

Answer»

1.385 cal/K
`-1.2` cal/K
1.2 cal/K
2.77 cal/K

Answer :D
Discussion
20.

The change in entropy, Delta Sis positive for an endothermic reaction. If enthalpy change Delta Hoccurs at the same temperature T, then the reaction is feasible

Answer»

at all temperatures
when `Delta H gt T Delta S`
when `Delta H LT T Delta S `
when `Delta H = T Delta S`

Solution :We know `DeltaG = Delta H - T Delta S`(GIBBS Helmholtz EQUATION)
For endothermic REACTION `Delta H` = +ve and also given `Delta S` = +ve. The reaction is feasible when `Delta G` = -ve. The `Delta G`value will be -ve only when `Delta H lt T Delta S`
Discussion
21.

The change in entropy (dS) is defined as

Answer»

`DS = deltaq//T`
`dS = dH//T`
`dS = delta q_(REV)//T`
`dS = (dH - dG) // T`

Solution :`dS = delta q_(rev)//T`
Discussion
22.

The change in entrophyof 2moles of ideal gas uponisothermalexpansion at243.6 Kfrom 20 liter untilthepressurebecomes1 atm, is :

Answer»

1.385 cal/K
`-1.2 cal//K`
`-30 J//mol.K`
None of these

Solution :`DeltaS= nRT In (V_(2))/(V_(1))`
`V_(2) = 40`
`V_(1) = 20`
`DeltaS = 2 XX 2 In 2`
` = 4 In 2 =2.77 cal`
Discussion
23.

The change : ""_(15)^(30)Pto""_(14)^(30)Si requires the emission of

Answer»

`ALPHA`-PARTICLE
`BETA`- particle
neutron
positron.

ANSWER :D
Discussion
24.

The change ""_(15)^(30)Pto""_(14)^(30)Si requires the emission of :

Answer»

`ALPHA-` PARTICLE
`BETA`- particle
Neutron
positron.

ANSWER :D
Discussion
25.

The chalogen containing equal number of 's' and 'p' electrons is

Answer»

O
S
MG
TE

ANSWER :A
Discussion
26.

The chamber acid contains____% H_2SO_4 :

Answer»

`10.20%`
`35.45%`
`67.80%`
`82.90%`

ANSWER :C
Discussion
27.

The chalcogen having same number of electrons both in penultimate and antipenultimate shells is

Answer»

O
S
Se
Te

Answer :D
Discussion
28.

The chair form is ______stable than boat form by potential energy _______ kJ/mol

Answer»

More, 44 KJ/mol
Less, 44 KJ/mol
More, 12.55 KJ/mol
Less, 4 KJ/mol

Answer :1
Discussion
29.

The CFSE is the highest for

Answer»

`[CoF_(4)]^(2-)`
`[Co(NCS)_(4)]^(2-)`
`[Co(NH_(3))]^(3+)`
`[CoCl_(4)]^(2-)`

ANSWER :B::C::D
Discussion
30.

The chain isomer of 2-methyl propanal is

Answer»

2-butanone
butanal
2-methyl propanol
but - 3-ene-2-ol

Answer :B
Discussion
31.

The chain length of silicone polymer can be controlled by adding-

Answer»

`MeSiCl_3`
`Me_2SiCl_2`
`Me_3SiCl`
`Me_4Si`

ANSWER :C
Discussion
32.

The CFSE for octahedral [CoCl_(6)]^(4-) is 18,000 cm^(-1). The CFSE for tetrahedral [CoCl_(4)]^(2-) will be

Answer»

18000`CM^(-1)`
16,000`cm^(-1)`
8000`cm^(-1)`
20000`cm^(-1)`

ANSWER :C
Discussion
33.

The CFSE for octahedral [CoCl_(6)]^(4-) complex is 18000 cm^(-1). Then, Delta for tetrahedral [CoCl_(4)]^(2-) will be :

Answer»

`18000 cm^(-1)`
`16000 cm^(-1)`
`8000 cm^(-1)`
`2000 cm^(-1)`

SOLUTION :`Delta_(t)=(4)/(9)Delta_(0)=8000" "cm^(-1)`
Discussion
34.

The CFSE for [CoCl_(6)]^(4-) is 18000cm^(-1). The Delta for [CoCl_(4)]^(2-) will be Xx10^(3)cm^(-1). What is the value of X

Answer»


Solution :`Delta_(t)=Delta_(0)xx(4)/(9)=18xx10^(3)xx(4)/(9)=8xx10^(3)CM^(-1)`
Discussion
35.

The CFSE for [CoCl_(4)]^(4-) is 18000cm^(-1) the value of y ios

Answer»


SOLUTION :`Delta_(t) = (4)/(9) Delta_(0)`
Discussion
36.

The centric structure of benzene was proposed by

Answer»

Dewar
Kekule
Landenberg
Armstrong and Baeyer

Answer :D
Discussion
37.

The central oxygen atom in ether is

Answer»

`sp^2`-hybridised
`sp^3` -hybridised
sp-hybridised
d`sp^2` -hybridised

Answer :B
Discussion
38.

The central oxygen atom in ether is :

Answer»

sp-hybridised
`sp^2`-hybridised
`sp^3`-hybridised
`sp^3d^2`-hybridised

Answer :C
Discussion
39.

The central nervous system depressants that reduce the nervous tension and promote relaxation without producing sleep are called :

Answer»

hypnotics
sedatives
anaesthetics
antibacterials.

Answer :B
Discussion
40.

The central metal ion in the complex makes available a number of empty orbitals for the formation of co-ordinate bonds with suitable ligands. The inner d-orbitals or outer d-orbitals are involved. The empty hybrid orbitals of metal ion overlap with the filled orbitals of ligand to from metal ligand co-ordinate bonds. For co-ordination number 6 octahedral geometry and sp^(3),d^(2)ord^(2)sp^(2) hybridisation involved and for cordiantion number 4 tetrahedral geometry, square planar geometry and sp^(3).dsp^(2) hybridisation involved. Geometry of [Ni(CN)_(4)]^(-2) ion is

Answer»

tetrahedral
Octahedral
SQUARE planar
none.

Solution :`CN^(-)` is a strong field ligand so rearrangement of ELECTRONS occurs.

After rearrangement :

`:.` Square planar geometry.
Discussion
41.

The central metal ion in the complex makes available a number of empty orbitals for the formation of co-ordinate bonds with suitable ligands. The inner d-orbitals or outer d-orbitals are involved. The empty hybrid orbitals of metal ion overlap with the filled orbitals of ligand to from metal ligand co-ordinate bonds. For co-ordination number 6 octahedral geometry and sp^(3),d^(2)ord^(2)sp^(2) hybridisation involved and for cordiantion number 4 tetrahedral geometry, square planar geometry and sp^(3).dsp^(2) hybridisation involved. Hybridisation present in [Cr(NH_(3))_(6)]^(+3)

Answer»

`SP^(3)d^(2)`
`sp^(3)`
`d^(2)sp^(3)`
`DSP^(2)`

SOLUTION :
Discussion
42.

The central metal ion in the complex makes available a number of empty orbitals for the formation of co-ordinate bonds with suitable ligands. The inner d-orbitals or outer d-orbitals are involved. The empty hybrid orbitals of metal ion overlap with the filled orbitals of ligand to from metal ligand co-ordinate bonds. For co-ordination number 6 octahedral geometry and sp^(3),d^(2)ord^(2)sp^(2) hybridisation involved and for cordiantion number 4 tetrahedral geometry, square planar geometry and sp^(3).dsp^(2) hybridisation involved. Type of complex formed in [Fe(H_(2)O)_(6)]^(+3) ion

Answer»

inner orbital complex
outer orbital complex
both a and b
none

Solution :
`H_(2)O` s a WEAK field lignad, So no regrouping of electrons. Outer orbitals participate in `SP^(3)d^(2)` hybridization.
Discussion
43.

The coordination number of central metalion in [Cu(H_2O)_4]^(2+)is:

Answer»

4
2
6
5

Answer :A
Discussion
44.

The central ion Co^+3 with coordination number 6 is bonded to the ligands NH_3 andBr to form a dipositive complex ion.Write the formula or IUPAC name of the complex.

Answer»

SOLUTION :`[CO(NH_3)_5Br]^2+`:Pentaamminebromidocobalt(III)ION
Discussion
45.

The central ion Ag^+ with coordination number 2 form a positive complex ion with NH_3 ligand.Also Ag^+ forms a negative complex ion with CN^- ligand. Give the denticity of NH_3 and CN^- ligands.

Answer»

SOLUTION :DENTICITY of both `NH_3` and `CN^-` is ONE(I)
Discussion
46.

Write the formula and name of a hexa-dentate ligand.

Answer»

SOLUTION :Ethylenediammine TETRA ACETATE ION
Discussion
47.

The central ion in [Cu(H_2O)_4]^(2+) ion is :

Answer»

`CU^(2+)`
`O^+`
`H^+`
NONE

ANSWER :A
Discussion
48.

The central ion Ag^+ with coordination number 2 form a positive complex ion with NH_3 ligand.Also Ag^+ forms a negative complex ion with CN^- ligand.Write the formula of above positive and negative complex ions Give the IUPAC name of each

Answer»

SOLUTION :`[AG(NH_3)_2]^+`:Diamminesiver(I)ION ,`[Ag(CN)_2]^-:Dicyanoargentate(I)ION
Discussion
49.

The central dogma of molecular genetics states that the genetic information flows from :

Answer»

`"DNA"rarr"RNA"rarr"Proteins"`
`"DNA"rarr"RNA"rarr"Carbohydrates"`
`"Amino ACIDS"rarr"Proteins"rarr"DNA"`
`"DNA"rarr"Carbohydrates"rarr"Proteins"`

Solution :`DNA"rarr"RNA"rarr"Proteins"`
Discussion
50.

The central dogma of molecule genetics states that the genetic information flows

Answer»

Amino acids `rarr` Proteins `rarr DNA`
`DNA rarr` Carbohydrates `rarr` Proteins
`DNA rarr RNA rarr `Proteins
`DNA rarr RNA rarr` Carbohydrates

Solution :The FOLLOWING concepts constitute the central dogma of molecular genetics:
(a) `DNA` molecules can synthesize other `DNA` molecular idential with the original (Replication).
(b) A single strand of `DNA` can act as a TEMPLATE on which a molecule of `RNA` is synthesized in a specific manner (Transcription)
(C ) `RNA` molecules in turn directs the synthesis of specific proteins which are charactersitic of each kind of organism (Translation).
They were summarized by Franicis CRICK as follows:
`DNA overset("Replication")rarr DNA overset(overset("Transcription")rarr)underset("REVERSE transcription")larr RNA overset("Translation")rarr "Protein"`
Discussion