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.

SiO_(2) reacts with _____ to form water glass

Answer»

`Na_(2) CO_(3)`
`Na_(2)O`
`NaOH`
All

Solution :`SiO_(2) + Na_(2) O rarr Na_(2) SiO_(3)` (water gas)
`SiO_(2) + 2NaOH rarr Na_(2) SiO_(3) + H_(2)O`
`SiO_(2) + Na_(2) CO_(3) rarr Na_(2) SiO_(3) + CO_(2) uarr`
Discussion
2.

SiO_2 is a solid while CO_2 is a gas - explain.

Answer»

`SiO_(2)` contains WEAK vander Waal attraction while `CO_(2)` contains strong covalent BONDS 
Solid `SiO_(2)` has a three dimensional net WORK STRUCTURE whereas `CO_(2)` contains discrete molecules. 
Both CONTAIN strong covalent bonds 
Both contain weak vander Waal attraction 

Answer :B
Discussion
3.

SiO_(2) is solid while CO_(2) is a gas at ordinary temperature. Explain.

Answer»

`SiO_(2)`contains weak vander Waal ATTRACTION while `CO_(2)`contains strong covalent bonds
SOLID `SiO_2` has a three dimensional net WORK STRUCTURE whereas `CO_(2)`contains discrete molecules.
Both contain strong covalent bonds
Both contain weak vander Waal attraction

Answer :B
Discussion
4.

SiO_(2) + A rarr x + y . In this reaction 'Y' is one of the global warming gases.'A' is the water soluble alkali metal carbonate . Whose molecular weight is 106.The common name of 'x' is

Answer»

FLINT glass
water glass
Baking soda
wasing soda

Solution :`SiO_(2) + Na_(2) CO_(3) RARR Na_(2) SiO_(3) + CO_(2) uarr`
Discussion
5.

SiO_(2) does not react with

Answer»

`HF`
`H_(2)SO_(4)`
`Na_(2)CO_(3)`(FUSED)
`NAOH`

ANSWER :B
Discussion
6.

Simplest alkene (A) reacts with HCl to form compound (B). Compound (B) reacts with ammonia to form compound (C) of molecular formula C_(2)H_(7)N. Compound (C) undergoes carbylamine test. Identify (A), (B), and (C).

Answer»

Solution :`UNDERSET("(A) Ethene")(CH_(2)=CH_(2))+HCl to underset("(B) Chloroethane")(CH_(3)-CH_(2)Cl)`
`underset("(B) Chloroethane")(CH_(3)CH_(2)Cl)+NH_(3) to underset("(C) amine")(CH_(3)CH_(2)NH_(2))`
A `CH_(2)=CH_(2)`Ethene
B `CH_(3)CH_(2)Cl`Chloroethane
C `CH_(3)CH_(2)NH_(2)`ETHYLAMINE.
Discussion
7.

Simplest alkene (A) reacts with HCl to form compound (B).Conmpound (B) reacts with ammonia to form compound (C) of molecular formula C_2H_7N. Conpound (C) undergoes carbylamine test. Identify (A). (B), and (C).

Answer»

Solution :
(i)The simplest alkene (A) is `CH_2 = CH_2`, ethene.
(II) Ethene reacts with HCl to GIVE Chloroethane `CH_3 - CH_2Cl` as (B) by addition reaction.
`CH_2 = CH_2 + HCl to underset("Chloroethane (B)")(CH_3 - CH_2Cl)`
(iii) Chloroethane reacts with ammonia to give Ethylamine `CH_3 - CH_2" " NH_2` as (C ). It is a primary amine and Carbylamine test is the characteristic test for `1^@` amine.
`CH_3-CH_2Cl + NH_3 to underset("ETHYL amine (or) Aminocthane")(CH_3 - CH_2NH_2) + HCl`
Discussion
8.

Simple distillation of liquids involves simultaneouly

Answer»

VAPORISATION and condensation
heating and sublimation
vaporisation and sublimation
boiling and filtration

Answer :A
Discussion
9.

similar to % labelling of oleum ,amixture of H_(3)PO_(3) and P_(4)O_(6)is labelled as (100 +x)% where x is maximum mass of water which can reacts with P_(4)O_(6) present in 100 gm mixture of H_(3)PO_(3) and P_(4)O_(6) P_(4)O_(6) + H_(3)

Answer»

1.25 atm
2.23 atm
0.25 atm
None of the above

Answer :A
Discussion
10.

Similarity in the radius of Zr and Hf is explained on the basis of

Answer»

Lanthanide contraction
INERT PAIR effect
Same outershell configuration
ANOMALOUS configuration

ANSWER :A
Discussion
11.

Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of the molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Answer»

Solution :MASS of neutron `= 1.675 xx 10^(-27) KG`
`lamda = (H)/(mv) or v = (h)/(m xx lamda) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(1.675 xx 10^(-27)kg xx (800 xx 10^(-12) m)) = 4.94 xx 10^(4) ms^(-1)`
Discussion
12.

Silverpropionate when refluxed with Bromine in carbotetrachloride gives

Answer»

propionic acid
chloro ETHANE
BROMO ethane
chloro propane

Solution :bromo ethane
Discussion
13.

Silver salt of which of the following acids on treatment with bromine will form 2-bromopropane ?

Answer»

Propanoic acid
Butyric acid
Isobutyric acid
Gluteric acid

Solution :`CH_(3) underset(CH_(3))underset(|)(CH) - COOAg overset(+Br_(2), C Cl_(4))rarr underset(2-"BROMOPROPANE")(CH_(3) underset(CH_(3))underset(|)(CH) - Br) + CO_(2) + AgBr`
Discussion
14.

Silver salt of fatty acid is converted to bromo alkane by …………….. .

Answer»

SOLUTION :Hunsdicker REACTION
Discussion
15.

Silver propionate when refluxed with Bromine in carbon tetrachloride give ………………. .

Answer»

propionic acid
chloroethane
BROMOETHANE
CHLOROPROPANE

SOLUTION :bromoethane
Discussion
16.

Silver nitrate test is used to detect the presence of ............... .

Answer»

SOLUTION :HALOGENS
Discussion
17.

Silver metal crystallises with a face-centred cubic lattice. The length of the unit cell is found to be 3.0 xx 10^(-8) cm. Calculate the atomic radius and density of silver. (Molar mass of Ag = 108 g mol^(-1), N_A =6.02xx 10^23mol^(-1))

Answer»


SOLUTION :`a=3.0xx10^(-8)` cm. For fcc, `r=a/(2sqrt2)=(3.0xx10^(-8))/(2xx1.414)=1.06xx10^(-8)` cm
`rho=(ZxxM)/(a^3xxN_A)=(4xx108)/((3.0xx10^(-8))^3xx(6.02xx10^23))=26.6 "G cm"^(-3)`
Discussion
18.

Silver, mercury (our) and lead are grouped toether in the same group of qualitative analysis because they form:

Answer»

nitrates
carbonates which dissolve in DILL. `HNO_(3)`
insoluble chlorides
colourless compounds

Solution :All belong to group I and FORM insoluble chlorides with DILUTE HCL
Discussion
19.

Silver is not produced when:

Answer»

`AgNO_(3)` solution is treated with CU rod
`AgNO_(3)` (ammoniacal solution) reacts with glucose
`AgNO_(3)` solution + excess KCN is reacted with Zn dust
`AgNO_(3)` solution is treated with `Na_(2)S_(2)O_(3)`, FOLLOWED by heating

Answer :D
Discussion
20.

Silver ions react with chloride ions Ag^(+)(aq)+Cl^(-)(aq)toAgCl(s) 5cm^(3) of a 0.1mol/cm^(3) solution of the chloride of metal X needs 10cm^(3) of 0.1 mol/cm^(3) silver nitrate for complete reaction. What is the formula of the chloride-

Answer»

`XCl_(4)`
`XCl_(2)`
`XCl`
`X_(2)CL`

ANSWER :B
Discussion
21.

Silver iodide is used for producing artificial rains because Agl:

Answer»

is easy to SPRAY at HIGH altitude
is insoluble in WATER
is EAST to synthesize
has crystals SIMILAR to ice

Answer :A
Discussion
22.

Silver iodide is used for producing artificial rain because Ag_(2)I

Answer»

is easy to spray at high altitudes
is easy to synthesize
has CRYSTAL STRUCTURE similar to ice
is INSOLUBLE in water.

Answer :C
Discussion
23.

Silver iodide is used for producing artifical rain because AgI

Answer»

is easy tospray at HIGH altitude
is easy to synthesize
has CRYSTAL structure SIMILAR to ice
is insoluble in water

Answer :C
Discussion
24.

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6pm. Calculate the density of silver (Atomic mass =107.9 u)

Answer»

Solution :`rho=(ZxxM)/(a^3xxN_0)`
For CCP lattice (which is EQUIVALENT to fcc lattice ), Z=4 ATOMS/unit cell
Also we are GIVEN that `M=107.9 g mol^(-1)`
a=408.6 pm=`408.6xx10^(-12)` cm `therefore rho=(4 "atoms"xx107.9g mol^(-1)) /((408.6xx10^(-12)cm)^3 (6.022xx10^23 "atoms mol"^(-1))=10.5 g cm^(-3)`
Discussion
25.

Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is 408.6 pm, Calculate the density of sliver ( Atomic mass = 107.9 u)

Answer»

Solution :`p= (Z xx M)/(a^(3) xxN_(0))`
For ccp lattic ( whichh is equivalent to fcc lattice) , Z= 4 atoms/ unit cell.
Also we are given that M= 107.9 g ` mol^(-1)`
a = 408. 6 pm = ` 408.6 xx 10^(-12)cmtherefore p = (4 "atoms" xx 107.9"g mol"^(-1))/((408.6 xx 10^(12)CM)^(3) (6.022 xx 10^(23) " atoms mol"^(-1))) = 10.5 " g cm" ^(-3)`
Discussion
26.

Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of sliver ? ( Assume that each face atom is touching the four corner atoms).

Answer»

SOLUTION :For face-centred cubic unit cell of an ELEMENT , ` r = a/(2SQRT2) = 0.3535a= 0.3535 xx 409 "PM" = 144.6 "pm" `
Discussion
27.

Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm . What is the radius of an atom of silver ? (Assume that each face atom is touching the four corner atoms )

Answer»

SOLUTION :For face-centred cubic UNIT of an ELEMENT , `r=a/(2sqrt2)` = 0.3535 a = 0.3535 x 409 PM =144.6 pm
Discussion
28.

Silver crystallises in fcc lattice. If the edge length of the cell is 4.077xx10^(-8) cmand density is 10.5 "g cm"^(-3), calculate the atomic mass of silver.

Answer»

Solution :`M=(rhoxxa^3xxN_0)/Z=(10.5 "G CM"^(-3) XX (4.077xx10^(-8) cm)^3 xx(6.02xx10^23 "mol"^(-1)))/4=107.09 "g mol"^(-1)`
Discussion
29.

Silver crystallises in fcc lattice. If the edge length of the cell in4.077 xx 10^(-8) cm and densityis 0.5 g cm^(-3) , calculate the atomic mass of silver.

Answer»

SOLUTION : ` M =( p xx a^(3) xx N_(0))/ Z = ( 10.5 " g cm"^(-3) xx ( 4.077 xx 10^(-8) cm)^(3) xx(6.02 xx 10^(23) "MOL"^(-1)))/4 = 107.09 " g mol"^(-1)`
Discussion
30.

Silver containing lead asan impurity is purified by

Answer»

Poling
CUPELLATION
Levigation
DISTILLATION

Solution :Silver containing lead as IMPURITY is REMOVED by cupellation .
Discussion
31.

Silver chloride was prepared in two ways : (i) 0.5 g of a silver wire was dissolved in concentrated nitric acid and the excess of hydrochloric acid was added to it. The precipitate of silver chloride was separated by filtration, washed, dried and then weighed. The weight of the residue was found to be 0.66 g (ii) In another experiement, 1g of silver metal was heated in a current of chlorine till it was completely converted into its chloride which was found to weight 1.32 g Show that the data is according to the Law of Constant composition.

Answer»

Solution :In the first experiment
Mass of SILVER = 0.5 G , Mass of silver chloride = 0.66 g
PERCENTAGE of silver `= ((0.50g))/((0.66g))xx100=75.76%`
Percentage of CHLORINE `= 100 -75.76 = 24.24 %`
In the second experiment
Mass of silver = 1.0 g , Mass of silver chloride = 1.32 g
Percentage of silver `= ((1.0g))/((1.32g))xx100=75.76%`
percentage of chlorine `= 100 - 75.76 = 24.24 %`
The results are in accordance with the LAW of Constant Composition.
Discussion
32.

Silver acetate +I_(2) overset(CS_(2))rarr the main prodcut of this reaction is

Answer»

`CH_(3)I`
`CHI_(3)`
`CH_(3)COI`
`CH_(3)COOCH_(3)`

SOLUTION :`CH_(3)COOAg + I_(2) overset(CS_(2))RARR CH_(3)I + AgI + CO_(2)`
`CH_(3)I + AgOCOH_(3) rarr UNDERSET("Methyl ACETATE")(CH_(3)COOCH_(3))`
Discussion
33.

Silicones contain silicon strongly bonded to _ and __atoms.

Answer»

C, O
C, H
H, O
H, Cl

Answer :A
Discussion
34.

Silicones are used

Answer»

as CONDUCTORS 
as insulators 
as semiconductors 
to PREPARE GRAPHITE 

Answer :B
Discussion
35.

Silicones are used as water proof materials because they have

Answer»

hydrophobic ALKYL groups
hydrophilic alkyl groups
strong Si-O bonds
Weak Si-O bonds

Solution :Silica have hydrophobic alkyl groups ( ORGANIC PART)
Discussion
36.

Silicones are the polymers formed by hydrolysis of

Answer»

Silicondioxide
Silanes
Silicates
Chlorosilanes

Solution :chlorosilanes `OVERSET("HYDROLYSIS")to` SILICONES
Discussion
37.

Silicones are also referred to as ............ polymers because they have very high ......... .

Answer»

SOLUTION :HIGH TEMPERATURE THERMAL STABILITY
Discussion
38.

Silicone oil is obtained from the hydrolysis and polymerization of

Answer»

trimethychlorosilane and dimethyldichloro-silane
trimethylchlorosilane and methyltrichloro-silane
methyltrichlorosilance and dimethyldichloro-silane
triethylchlorosilance and diethyldichlorosilance

Solution :HYDROLYSISOF `(CH_(3))_(2)SiCl_(2)` FOLLOWEDBY condensation gives straightchain POLYMER , WHOSE CHAIN length is controlledby `(CH_(3))_(3)SiCl` .
Discussion
39.

Silicon tetrachloride readily undergoes hydrolysis but carbon tetrachloride does not undergo hydrdolysis undergo normal conditions. Explain.

Answer»

Solution :Since carbon (of second PERIOD) has no vacant d-orbital, its maximum covalency is 4, on the other hand, silicon (of third period) has vacant d-orbitals and its maximum covalency is 6. as Si-atom can extend its covalency to 6, `SiCl_(6)` UNDERGOES ready hyrolysis to YIELD `SiO_(2)`. A lone pair fo electrons from the O- atom of `H_(2)overset(..)(O)` is donated to the empty di-orbitals of Si, forming a co-ordinate intermediate which has a trigonal bipyramidal structural. the intermediate `[SiCl_(4)(H_(2)O)]` loses a molecule of HCl to form `SiCl_(3)(OH)`. in the same way, the other 3Cl-atoms are replaced by 3-OH groups to form ortho-silicic acid `[Si(OH)_(4)]` which finally loses 2 molecules of water to give `SiO_(2)`.

C-atom having no d-orbitals in its valence in its valence shell cannot extend its covalency beyond 4 and so it does not undergo hydrolysis undre normal conditions.
Discussion
40.

Silicon is use as …….

Answer»

seal
electrical resistant
grease
all of above

Answer :D
Discussion
41.

Silicon has a strong tendency to form polymerslike silicones. The chainlength of silicanepolymer can becontrolled by adding

Answer»

`MeSiCl_(3)`
`Me_(2)SiCl_(2)`
`Me_(3)SICL`
`Me_(4)Si`

Solution :Chainof siliconpolyme can be controlled by adding `(CH_(3))_(3)SiCl`.
Discussion
42.

Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding ……

Answer»

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

Solution :The chain length of the polymer can be CONTROLLED by adding `(CH_3)_3SiCl` which BLOCKS the ENDS as shown below:
Discussion
43.

Silicon has 4 electrons in the outermost orbit. In forming the bonds.

Answer»

It GAINS electrons
It loses electrons
it SHARES electrons
None of the above

Answer :C
Discussion
44.

Silicon has 4 electrons in the outermost orbit. In forming the bonds,

Answer»

It gains electrons
It loses electrons
It shares ELECTRON
NONE of the above

ANSWER :C
Discussion
45.

Silicon dopend with electron-rich impurity forms…….

Answer»

p-type SEMICONDUCTOR
n-type semiconductor
INTRINSIC semiconductor
INSULATOR

ANSWER :B
Discussion
46.

Silicon doped with electron-rich impurity forms ______

Answer»

p-type SEMICONDUCTOR
n-type semiconductor
INTRINSIC semiconductor
insulator

ANSWER :B
Discussion
47.

Silicon carbide is used as

Answer»

dehydrating AGENT
abrasive
solvent
catalyst

Solution :SILICON CARBIDE is used as abrasive
Discussion
48.

Silicon can be considered as

Answer»

REDUCTANT 
ACID 
OXIDANT 
BASE 

ANSWER :A
Discussion
49.

Silicanes are few in numberwhereasalkanes are large in number. Explain

Answer»

Solution :Carbon has the MAXIMUM tendency for catenation due to stronger `C-C (348kJ "mol"^(-1))`bonds. Asa RESULT, it formsa largenumber of ALKANES.
Silicon, on the other hand, due to other hand, due to weaker `Si-Si(297 KJ mol^(-1))`bonds has muchlesser tendency for catenationand henceforms only a few silanes.
Discussion
50.

Silica is insoluble

Answer»

`HF`
`NaOH`
`KOH`
`HNO_(3)`

ANSWER :D
Discussion