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.

Geometrical shape of XeF_6 is

Answer»

HEXAGONAL
DISTORTED octahedral
Octahedral
Square pyramidal

Solution :
GEOMETRICAL SHAPE of `XeF_6` is as
Discussion
2.

Froth floatation process is based on :

Answer»

WETTING PROPERTIES of ORE particles
Specific GRAVITY of ore particles
Magnetic properties of ore particles
Electical properties of ore particles

Answer :A
Discussion
3.

Geometrical isomersim is possible in:

Answer»

isobutene
acetone-OXIME
benzophenone-oxime
acetophenone -oxime

ANSWER :D
Discussion
4.

Geometricalisomersm isshownby :

Answer»

2-methyl-1-pentene
3-Hexene
2-Pentyne
2,3- Dimethyl-2- BUTENE

ANSWER :B
Discussion
5.

Froth flotation process is suitable for concentrating …………… ores.

Answer»

OXIDE ORES
Carbonate ores
Chloride ores
SULPHIDE ores

Answer :D
Discussion
6.

Froth floatation process involves the……

Answer»

TREATMENT of the ore with WATER and pine oil
Washing of the ore with a steam of water
Owing off the ore over a conveyor BELT rolling over magnetic roller
melting of ore

Answer :A
Discussion
7.

Geometrical isomers differ in:

Answer»

POSITION of FUNCTIONAL group.
position of atoms.
spatial ARRANGEMENT of atoms.
length of CARBON chain.

Answer :C
Discussion
8.

Geometrical isomerism may be possible with

Answer»

TETRAHEDRAL complex
square PLANAR complex
pentagonal PYRAMIDAL complex
square pyramidal complex

Answer :B
Discussion
9.

Froth floatation process for the concentration of ores is carried out in the case of the following ore

Answer»

oxides
sulphides
sulphates
nitrate

Answer :B
Discussion
10.

Geometrical isomerism is shown by ?

Answer»




`Ph-CH=CH-CH_3`

SOLUTION :Conditions for GEOMETRICAL ISOMERISM is not FILLED in C.
Discussion
11.

Froth floatation process for the concentration of ores is an illustration of the practical application of

Answer»

Adsorption
Abdorption
Coagulation
Sedimentation

Answer :A
Discussion
12.

Geometrical isomerism is show by

Answer»




ANSWER :B
Discussion
13.

Froth floatation processfor the concentration of ores is a practical application of :

Answer»

Adsorption
Absorption
Coagulation
Sedimentation

Answer :A
Discussion
14.

From williamsons synthesis, which one of the following is most desirable to prepare ether?

Answer»

`3^@` R-X and ALKOXIDE of `1^@` ALCOHOL
`3^@` R-X and alkoxide of `2^@` alcohol
`2^@` R-X and alkoxide of `1^@` alcohol
`1^@` R-X and alkoxide of `3^@` alcohol

Answer :D
Discussion
15.

Geometrical isomerism is not shown by:

Answer»

`CH_3CH_2C(CH_3)=C(CH_3)CH_2CH_3`
`C_2H_5CH=CHCH_2I`
`CH_2=C(CL)CH_3`
`CH_2=CH_2`

ANSWER :C
Discussion
16.

Froth floatation method is successful in seperating impurities from ores because _________.

Answer»

the pure ore is LIGHTER than water CONTAINING additives LIKE pine oil , fatty pineoil , fatty acid ETC
the pure ore is soluble in water containing additives like pine oil,fatty acid, etc.
the impurities are soluble in water containing additives like pine oil , fatty acid, etc.
the pure ore is not as easily WETTED by water as by pine oil , fatty acid, etc.

Answer :D
Discussion
17.

Geometrical isomerism is found in coordination compounds having coordination number :

Answer»

2
3
4 (TETRAHEDRAL)
6

Answer :D
Discussion
18.

Geometrical isomerism is exhibited by:

Answer»

2-chlorobut-2-ene
but-2-ene
3-methylpent-2-ene
2-methyl but-2-ene

ANSWER :A::C::D
Discussion
19.

Froth floatation method may be used to increase the concentration of mineral in

Answer»

Calamine
Bauxite
Haematite
Chalcopyrite

ANSWER :D
Discussion
20.

Froth floatation:

Answer»

is a physical METHOD of separation MINERAL from the gangue
is a method of concentration of ore depending on the difference in wetability of gangue and the ore particles
is USED for the concentration of sulphide ores
is a method in which IMPURITIES sink to be bottom and ore particles pass on the surface with froth

Answer :A,B,C,D
Discussion
21.

Geometrical isomerism is found in co-ordination compounds having co-ordination number:

Answer»

2
3
4 (TETRAHEDRAL)
6

Answer :D
Discussion
22.

what happens when Formaldehyde reacts with ammonia

Answer»

Solution :When formaldehyde is treated with AMMONIA it GIVES Hexamethylene tetramine (HMT) which is also KNOWN as urotropine .
Discussion
23.

Geometrical isomerism is exhibited by ………….

Answer»

TETRAHEDRAL complex
Linear complex
Square PLANAR complex
All of the above

Answer :A::C
Discussion
24.

From which of the following, tertiary butyl alcohol is obtained by the action of methyl magnesium iodide?

Answer»

HCHO
`CH_3COCH_3`
`CH_3CHO`
`CO_2`

SOLUTION :`CH_3COCH_3`
Discussion
25.

From which of the following the hydration energy of Mg^(2+) is larger

Answer»

`Na^(+)`
`AL^(3+)`
`Be^(2+)`
`Cr^(3+)`

Solution :HYDRATION energy INCREASES along the period.
Discussion
26.

Geometrical isomerism in square planar complexes is given by

Answer»

`Ma_4` TYPE COMPLEX
`Ma_3b` type complex
`Ma_2b_2` type complex
`Mb_4` type complex

ANSWER :C
Discussion
27.

From which of the following Nylon 6 is prepared ?

Answer»

1,3-butadiene
CHLOROPRENE
ADIPIC acid
Caprolectum

SOLUTION :Caprolectum
Discussion
28.

From which enzymes are made ?

Answer»

Vitamin
Lipid
Carbohydrates
Protein

Answer :D
Discussion
29.

Geometrical isomerism in coordination compounds is exhibited by :

Answer»

SQUARE PLANAR and TETRAHEDRAL complexes
Square planar and OCTAHEDRAL complexes
Tetrahedral and octahedral complexes
Square planar, tetrahedral and octahedral complexes

Answer :B
Discussion
30.

From which from of iron, other forms of iron can be produced ?

Answer»

CAST iron
Wrought iron
Pig iron
Steel

Answer :C
Discussion
31.

Geometrical isomerism arises in heteroleptic complexes due to different possible geometrical arrangement of the ligands. Important examples of this behaviour are found with coordination number 4 and 6. Such isomerism is not possible for a tetrahedral geometry but it is possible for square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical isomers are mirror images that cannot be superimposed on one another. These are called enantiomers. Optical isomerism is common in octahedral complexes involving bedentate ligands. Select the correct statement.

Answer»

Both `[NiCl_(2)(PPh_(3))_(2)]` and its analogous Pd(II) show geometrical isomerism.
`[CoBrCl(en)_(2)]` will show geometrical isomerism but is chiral compound.
`cis-[CO(NH_(3))_(4)Br_(2)]^(+)` can exist as enantiomer
A complex of palladium (II) with TWO CHLORIDE ions and two thiocyanate ions willshow linkage as well as geometrical isomerism.

Answer :D
Discussion
32.

From the thermochemical reactions, C_ graphite+1//2O_2=CO, triangleH=-110.5 kJ CO+1//2O_2=CO_2, triangleH=-283.2 kJtriangleH for the reaction, C_graphite+O_2=CO_2 is:

Answer»

`-393.7 KJ`
`+393.7 kJ`
`-172.7 kJ`
`+172 .7 kJ`

ANSWER :A
Discussion
33.

Geometrical isomerism can exist only when the molecule:

Answer»

Has a centre of symmetry
Has plane of symmetry
Has TWO different GROUPS attached to both the CARBON atoms of the double bond
Rotates the plane polarised light to the particular direction

Answer :C
Discussion
34.

From the values of 1 faraday anf Avogadro constant, show that 1 faraday may be called 1 mole of electricity.

Answer»

Solution :Since 1 faraday = 96500 COULOMBS
`therefore (1F)/("AV, no.") = (96500)/(6.022 xx 10^(23))`
`= 1.6 xx 10^(-19)` coulomb
= charge of on electron.
`therefore` 1 F = charge of an electron `xx` Av. no.
Since 1 F of electricity is the charge of Av.no. of electrons, 1 faraday may be CALLED 1 mole of electricity.
Discussion
35.

Geometrical isomerism arises in heteroleptic complexes due to different possible geometrical arrangement of the ligands. Important examples of this behaviour are found with coordination number 4 and 6. Such isomerism is not possible for a tetrahedral geometry but it is possible for square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical isomers are mirror images that cannot be superimposed on one another. These are called enantiomers. Optical isomerism is common in octahedral complexes involving bedentate ligands. Which of the following complexes will show geometrical isomerism ?

Answer»

`CS[FeCl_(4)]`
`CrCl_(3)(PY)_(3)`
`[Co(en)_(2)]^(2+)`
`[NI(CO)_(4)]`

ANSWER :B
Discussion
36.

From the unit cell dimension, we can accurately calculate

Answer»

GAS constant.
Henry's LAW constant.
Avogadro constant.
None of these.

Answer :C
Discussion
37.

Geometrical isomerism can be shown by :

Answer»

`[AG(NH_(3)(CN)]`
`Na_(2)[Cd(N0_(2))_(4)]`
`[PtCl_(4)I_(2)]`
`[Pt(NH_(3))_(3)Cl][Au(CN)_(4)]`

Solution :(C ) `Ma_(4)b_(2) to [M (a a)(b b)(a a)]" "[M(AB)(ab)(a a)]`
Discussion
38.

From the standard potentials shown in the fillowing figure, calculate the potentials E_(1)^(@) and E_(2)^(@).

Answer»


SOLUTION :`E_(1)^(@)=(-[4xx1.47+1xx1.6])/(-5)=1.496V`
`E_(1)^(@)=(-[5xx1.496+1.07])/(-6)=1.425V`
Discussion
39.

Geometrical isomerism arises in heteroleptic complexes due to different possible geometrical arrangement of the ligands. Important examples of this behaviour are found with coordination number 4 and 6. Such isomerism is not possible for a tetrahedral geometry but it is possible for square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical isomers are mirror images that cannot be superimposed on one another. These are called enantiomers. Optical isomerism is common in octahedral complexes involving bedentate ligands. Which one of the following statements is false ?

Answer»

`[CIS-Pt(NH_(3))_(2)Cl_(2)]` will have some dipole moment
`[cis-Pt(NH_(3))_(2)Cl_(2)]` will show geometrical as well as optical isomerism.
`cis-[CrCl_(2)("ox")_(2)]` is a chiral molecules.
(a) and (B) both

Answer :B
Discussion
40.

Geometrical isomerism arises in heteroleptic complexes due to different possible geometrical arrangement of the ligands. Important examples of this behaviour are found with coordination number 4 and 6. Such isomerism is not possible for a tetrahedral geometry but it is possible for square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical isomers are mirror images that cannot be superimposed on one another. These are called enantiomers. Optical isomerism is common in octahedral complexes involving bedentate ligands. For which of the following complexes, are optical isomers possible ? (P) [Cr("ox")_(3)]^(3-)""(Q) [Cr(NH_(3))_(4)(ox)^(+) (R ) [Co("ox")_(2)(NH_(3))_(2)]^(-) Select the correct answer using the codes given below :

Answer»

<P>P only
P and Q only
P and R only
P, Q and R

ANSWER :C
Discussion
41.

Geometrical isomerism arises in heteroleptic complexes due to different possible geometrical arrangement of the ligands. Important examples of this behaviour are found with coordination number 4 and 6. Such isomerism is not possible for a tetrahedral geometry but it is possible for square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical square planar as well as octahedral complexes. Optical isomers are mirror images that cannot be superimposed on one another. These are called enantiomers. Optical isomerism is common in octahedral complexes involving bedentate ligands. The total number of isomers possible for the complex [Co(en)_(2)Cl_(2)]^(+) is :

Answer»

3
4
5
2

Answer :A
Discussion
42.

Geometric isomerism exists in ……….. Complexes due to different possible three dimensional spatial arrangement of ligands around the central metal atom.

Answer»


ANSWER :HETEROLEPTIC
Discussion
43.

From the stability constant (hypothetical values), given below, predict which is the most stable complex ?

Answer»

`Cu^(2)+4NH_(3) to [Cu(NH_(3))_(4)]^(2+), K=4.5xx10^(11)`
`Cu^(2)+4CN^(-) to [Cu(CN)_(4)]^(2-), K=2.0xx10^(27)`
`Cu^(2)+2en to [Cu(en)_(2)]^(2+), K=3.0xx10^(15)`
`Cu^(2)+4H_(2)O to [Cu(H_(2)O)_(4)]^(2+), K=9.5xx10^(8)`

Answer :B
Discussion
44.

Geomerical isomerism is found in co-ordination compounds having minimum co-ordination number.

Answer»


Solution :SQUARE planar COMPLEXES with `[Ma_(2)b_(2)]` formulas only will EXHIBIT GL. But tetrahedral complexes do not exhibit GI.
Discussion
45.

From the stability constant (hypothetical values) given below, predict which is the strongest ligand:

Answer»

`CU^(2+)+4NH_3iff[Cu(NH_3)_4]^(2+),(K=4.5xx10^11)`
`Cu^(2+)+4CNiff[Cu(CN)_4]^(2-),(K=2.0xx10^27)`
`Cu^(2+)+4CNiff[Cu(EN)_2]^(2+),(K=3.0xx10^15)`
`Cu^(2+)+4H_2Oiff[Cu(H_2O)_4]^(2+),(K=9.5xx10^8)`

Answer :B
Discussion
46.

From the stability constant (hypothetical value given below predict which is the strongest ligand

Answer»

`Cu^(2+)+4NH_(3)HARR[Cu(NH_(3))_(4)]^(2+),K=4.5xx10^(11)`
`Cu^(2+)+4CN^(-)hArr[Cu(CN)4)]^(2-),K=2.0xx10^(27)`
`Cu^(2+)+2enhArr[Cu(en)_(4)]^(2-),K=3xx10^(15)`
`Cu^(2+)+4H_(2)OhArr[Cu(H_(2)O)_(4)]^(2+),K=9.5xx10^(8)`

Solution :Strength of ligand `prop` K
Discussion
47.

From the solution of an electrolyte , one mole of electrons will deposit at cathode ,

Answer»

63.5 gm of Cu
24 gm of Mg
11.5 gm of NA
9.0 gm of Al

SOLUTION :1 g eq. of Al = 9 g = 1 mole of electrons
Discussion
48.

Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in the solid state ?

Answer»

`Ag_(2)SO_(4)`
`CuF_(2)`
`ZnF_(2)`
`Cu_(2)Cl_(2)`

SOLUTION :`CuF_(2)` is COLOURED due to the presence of UNPAIRED ELECTRONS.
Discussion
49.

From the rate expressions for the following reactions, determine their order of reaction and the dimensions of the rate constants : (i) 3" NO "(g)toN_(2)O(g)+NO_(2)(g)," Rate "=k[NO]^(2) (ii) H_(2)O_(2)(aq)+3" I"^(-)(aq)+2" H"^(+)to2H_(2)O(l)+I_(3)^(-)," Rate "=k[H_(2)O_(2)][I^(-)] (iii) CH_(3)CHO(g)toCH_(4)(g)+" CO "(g)," Rate "=k[CH_(3)CHO]^(3//2) (iv) C_(2)H_(5)Cl(g)toC_(2)H_(4)(g)+HCl(g) ," Rate "=k[C_(2)H_(5)Cl].

Answer»

Solution :(i) ORDER = 2, Dimensions of `k=("Rate")/([NO]^(2))=(molL^(-1)s^(-1))/((molL^(-1))^(2))=Lmol^(-1)s^(-1)`
(ii) Order = 2, Dimensions of k = Same as in (i)
(III) Order `=(3)/(2)`, Dimensions of `k=("Rate")/([CH_(3)CHO]^(3//2))=(molL^(-1)s^(-1))/((molL^(-1))^(3//2))=L^(1//2)mol^(-1//2)s^(-1)`
(iv) Order = 1, Dimensions of `k=("Rate")/([C_(2)H_(5)Cl])=(molL^(-1)s^(-1))/(molL^(-1))=s^(-1).`
Discussion
50.

Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

Answer»

`Ag_(2)SO_(4)`
`CuF_(2)`
`ZnF_(2)`
`Cu_(2)Cl_(2)`

Solution :`CU^(2+) "has " d^(9)` configuration i.e., it has one UNPAIRED ELECTRON.
Discussion