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.

Metals like Pt and Pd can adsorb large volume of hydrogen under specific conditions. Such adsorbed hydrogen by the metal is known as

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OCCLUDED hydrogen
Absorbed hydrogen
Reactive hydrogen
Atomic hydrogen

Answer :A
2.

Metals having ns' as the valence electronic configuration

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act as STRONG oxidising agents
are HIGHLY electronegative
are highly electropositive
have a FIRST IONIZATION potential of more than

Answer :C
3.

Metals have high thermal conductivity. Give reason.

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Solution :High themal CONDUCTIVITY of METALS is due to thermal excitation of many ELECTRONS from the valence band to the CONDUCTION band.
4.

Metals have _________ density .

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SOLUTION :The ELECTROSTATIC attraction between the metal ions and the free electrons yields a three DIMENSIONAL close packed crystal with a large number of NEAREST metal ions. So metals have HIGH density.
5.

Metal(s) generally obtained by the reduction of their oxides with H_(2) are

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Al, Zn
MO, W
Na, K
Ca, Mg.

Solution :Molybdenum (Mo) and trugesten (W) are OBTAINED by the reduction of their OXIDES with hydrogen.
6.

Metals form basic hydroxides. Which of the following metal hydroxides is the least basic?

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`MG(OH)_(2)`
`CA(OH)_(2)`
`Sr(OH)_(2)`
`Ba(OH)_(2)`

SOLUTION :BASIC character of the hydroxide increases down the GROUP as the size of metal increases. Solubility of hydroxide increases and hydroxide of Be and Mg are almost insoluble.
7.

Metals form basic hydroxides . Which of the following metal hydroxide is the least basic ?

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`Mg(OH)_(2)`
`CA(OH)_(2)`
`SR(OH)_(2)`
`BA(OH)_(2)`

Solution :As the ionization ENTHALPY increases from `Mg to Ba`, the M-O bond becomes weaker and weaker down the group and hence basicity increases down the group . Thus , `Mg(OH)_(2)` is least basic .
8.

Metals commonly extracted by Gold Schmidt's alumino-thermic process is/are

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GOLD
Chromium
Iron
Maganese

Solution :Gold and iron are not extracted by Goldschmidt's aluminothermic PROCESS. Gold never exists as oxide, iron if extracted by this METHOD will most many TIMES more than its cost in the market.
9.

Thecrystals which are good conductors of electricity and heat are

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Solution :(i) The bonding in METAL is explained by molecular ORBITAL theory. As per this theory, the atomic ORBITALS of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band GAP.
(ii) The bonding molecular orbitals are completely filled with an ELECTRON pair in each, and the anti-bonding molecular orbitals are empty.
(iii) Absence of band gap accounts for high clectrical conductivity of metals.
10.

Metals are ductile in nature. Why?

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SOLUTION :In the close. packed structure of metallie erystal, it contains many slip planes along which MOVEMENT can occur during MECHANICAL loading, so the metal ACQUIRES ductility.
11.

Metalloid elements are placed in

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<P>S - BLOCK 
p - Block 
d - Block 
F - Block 

ANSWER :B
12.

Metallic tin in the presence of HCl is oxidised by K_(2)Cr_(2)O_(7) solution to stannic chloride. What volume of decinormal dichromate solution would be reduced by 1 g of Sn ?

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Solution :`underset(3xx118.7 g)(3SN)+underset(2xx294 g)(2K_(2)Cr_(2)O_(7))+28 HCl to 3SnCl_(4)+4KCl+4CrCl_(3)+14H_(2)O`
`K_(2)Cr_(2)O_(7)` required for 1 g of Sn`=(2xx294)/(3xx118.7)=1.65 g`
13.

Metallic magnesium has a hexagonal close packed structure and a density of 1.74 g//cm^3. Assuming magnesium atoms to be spherical, calculate the volume of eachatom and atomic radius of Mg atom (Atomic mass of Mg =24)

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Solution :As MG has HEXAGONAL close packed structure , no. of atoms per unit cell = 6 , i.e., Z=6
`rho=(ZxxM)/(a^3xxN_0)=(ZxxM)/(VxxN_0)` (V=`a^3`=volume of the unit cell)
`therefore 1.74 =(6xx24)/(Vxx(6.023xx10^23))` or `V=1.374xx10^(-22) cm^3`
Now , as in hexagonal close packing , 74% of the space is occupied by atoms, therefore, volume occupied by atoms =`74/100xxV`
`=74/100xx1.374xx10^(-22) =1.018xx10^(-22) cm^3`
As a unit cell contains 6 Mg atoms, therefore , volume of each Mg atom
`=(1.018xx10^(-22))/6=1.697xx10^(-23) cm^3`
As Mg atom is ASSUMED to be spherical,
`therefore 4/3pir^3 =1.697xx10^(-23)`
On SOLVING, we GET `r=1.594xx10^(-8)` cm =1.594 Å
14.

Metallic magnesium has a hexagonal close packed structure and a density of 1.74 g//cm^(3) . Assuming magnesium atoms to be spherical, calculatethe volume of each atom and atomic radium of Mg atoms (Atomic mass ofMg= 24)

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<P>

Solution :As MG has hexagonal close packed structure , no. of atoms per unit cell = 6 ,i.e,Z = 6
`p= (Z XX M)/( a^(3)xxN_(0)) = (Z xxM)/(V xx N_(0)) "" (V =a^(3)`= VOLUME of the unit cell)
` 1.74 = ( 6xx24)/( V xx (6.023 xx10^(23))) orV = 1.374 xx 10^(-22)cm^(3)`
Now ,as in hexagonal close packing, 74%of the space is occupiedby atoms, therefore, volume occiped by atoms ` = 74/100 xx V`
` = 74/100 xx 1.734 xx 10^(-22) = 1.018 xx 10^(-22)cm^(3)`
As a unit cell contains 6 Mg atom, therefore, volume of each Mg atom.
`( 1.018 xx 10^(-22))/6= 1.697 xx 10^(-23) cm^(3)`
As Mg is assumed to be spherical.
` 4/3 pir^(3) = 1.697 xx 10^(-23)`
on solving we get ` r =1.594 xx 10^(-8)cm= 1.594Å`
15.

Metallic magensium has a hexagoanl close-packed structure and a density of 1.74 g //cm^(3). Assume magnesium atoms to be sphere of radius r. 74.1 % of the space is occupied by atoms. Calculate the volume of each atom and the atomic radius r.(Mg = 24.31)

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Answer :`1.72 XX 10^(-23) CM^(3), 1.60 Å`
16.

Metallic lustre is explained by

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Diffusion of metal ions
Oscillation of looseelectrons
Excitation of FREE protons
Existence of BCC lattice

Solution :Metallic lustre is explained by oscillation of LOOSE ELECTRONS.
17.

Metallic Iusture is explained by

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Diffusion of metal IONS
Oscillations of loose electrons
Excitation of FREE protons
EXISTENCE of BCC LATTICE

Solution :Oscillating free electrons explain the metallic lusture.
18.

Metallic________ is usedfor makingwindowsofXraytubes .

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MAGNESIUM
beryllium
sulphur
Magnalium

ANSWER :B
19.

Metallic hydrides are otherwise called ................

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SALT HYDRIDES
Saline hydrides
Covalent hydrides
INTERSTITIAL hydrides

Solution :Interstitial hydrides
20.

Metallic hydride is formed by reaction of hydrogen with

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Iron
MANGANSE
PALLADIUM
Molybdenum

Solution :Metallic HYDRIDE is formed by palladium.
21.

Metallic elements are described by their standard electrode potential, fusion enthalpy, atomic size, etc. The alkali metals are characterised by which of the following properties?

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High boiling point
High NEGATIVE standard electrode potential
High density
Large atomic SIZE

Solution :Alkali metals have LARGEST size and low density and have low boiling point. These metals lose ELECTRONS easily DUE to less effective nuclear charge and have high negative standard electrode potential.
22.

Metallic elements are described by their standard electrodepotential , fusion enthalpy , atomic size , etc . The alkali metals are characterised by which of the following properties ?

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High boiling point
High negative STANDARD ELECTRODE POTENTIAL
High density
LARGE atomic size

SOLUTION :Alkali metals have high negative standard electrode potential and large atomic size .
23.

Metallic elements are described by their standard electrode potential, fusion enthalpy, atomic size, etc. The alkali metals are characterised by which of the following properties? (i) High boiling point (ii) High density (iii) Large atomic size

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I & ii
I & III
only ii
only iii

Answer :D
24.

Metallic calcium is prepared by.....

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DISPLACEMENT of calcium by IRON from calcium SULPHATE SOLUTION.
electrolysis of MOLTEN calcium chloride.
reduction of lime by coke.
electrolysis of aqueous solution of calcium nitrate.

Answer :C
25.

Metalic ti in resence of HCI is oxidised by K_(2)Cr_(2)O_(7) to stannic chloide whast volume of decinormal in aicdic medium?

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Solution :BALANCED chemical equation is
`2K_(2)Cr_(2)O_(7)+3Sn+28HCIrarr4KcI+4 CrCI_(3)+3 SnCI_(4)+14 H+_(2)O`
`2(2xx39+2xx52+7xx16)=356.1 g`
Now 356.1 Sn react with `K_(2)Cr_(2)O_(7)=(588)/(356.1)g=1.651 g`
Eq .wt of `K_(2)CrO_(7)=("mol.wt")/(6)=(294)/(6)=49`
Decinormak `K_(2)Cr_(2)O_(7)` slution means 1000 mL solution CONTAINS 4.9 g`K_(2)Cr_(2)O_(7)i.e 4.9 g K_(2)Cr_(2)O_(7)` are present in 1000 mL solution
`therefore` 1.651 g willbe present in =`(1000)/(4.9)xx1.651=336.9 mL`
26.

Metaldehyde is the product of the following

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`4CH_(3)-CHOoverset(Conc.H_(2)SO_(4),0^(@)C)to`
`CH_(3)-overset(O)overset(||)C-CH_(3)overset(Conc.H_(2)SO_(4), ro omtemp.)to`
`3CH_(3)CHOoverset(Conc.H_(2)SO_(4),ro omtemp.)to`
`C_(6)H_(5)-CHO overset("conc.H_(2)SO_(4))to`

ANSWER :3
27.

Metallic radius of Ca is 200pm. Covalent radius of Ca is

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200 PM
230 pm
280 pm
174 pm

Answer :D
28.

Metal which obtained by reduction with carbon is

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Fe
Zn
Pb
Cu

Answer :D
29.

Metal which cannot be refined by electrolytic refining process is

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Cu
Cr
Ni
Al

Answer :B
30.

Metal which cannot be extracted by smelting process is

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Pb
Fe
Zn
Al.

Answer :D
31.

Metal used for dying organic solvents is

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Fe
Pt
Mg
Na.

Solution :Because it REACTS with WATER and REMOVES it.
32.

Metal that escapes into atmosphere by the fumes of automobile vehicles

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Hg
Pb
Na
Cu

ANSWER :B
33.

Metal used as catalyst in the hydrogenation of vegetable oils

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Iron
Molybdenum 
Nickel
Sodium 

ANSWER :C
34.

Metal protected by a layer of its own oxide is

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AL
Ag
Au
B

Solution :Al is very REACTIVE towards oxygen GIVES an OXIDE LAYER
`4 Al + 3O_(2)rarr2Al_(2)O_3`
35.

Metal ions are activatorsand increase catlyticactivity of enzymemolecular. If water vapours is assumed to be a perfect gas, molar enthalpychangefor vaporisationof 1 mole of water1 bar 100^(@)C is 41 kJ//mol. Choose the correct statement(s).(TakenR= 8.3 J//"mole"//K)

Answer»

`DeltaU_("VAPORISATION")`of1 MOLEOF WATER at 1 barand `100^(@)C= 37.904 kJ//"mol"`
`(DeltaU=DeltaH)` forconversion of thewaterinto ice at `0^(@)C`
In theisothermalprocessof (b), `DeltaH=0`
`(DeltaH=DeltaU)` for conversion of 1 mole of water into steam`100^(@)`C

Answer :a,b
36.

Metal ion M^(+3) loses three electrons than what will be its oxidation number ?

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SOLUTION :`M^(+3)TOM^(+6)+3E^(-)`
37.

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH , CsH, the correct order of increasing ionic character is

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`LiH GT NaH gt CSH gt KH gt RbH`
`LiH LT NaH lt KH lt RbH lt CsH`
`RbH gt CsH gt NaH gt KH gt LiH`
`NaH gt CsH gt RbH gt LiH gt KH`

Solution :IONIC characterincreases as the size of the ATOM increases or the electronegativity of the atom decreases, i.e., `LiH lt NaOH lt KH lt RbH lt CsH`.
38.

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RBH, CSH, the correct order of increasing ionic character is ......

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LiH > NaH > KH > RbH > CsH
LiH < NaH < KH < RbH < CsH
RbH < CsH < LiH < KH < NaH
NaH > CsH > LiH > RbH > KH

Solution :LiH < NAL
39.

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH and CsH, the correct order of increasing ionic character is

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`LiH gt NaH gt CsH gt KH gt RBH`
`LiH LT NaH lt KH lt RbH lt CsH`
`RbH gt CsH gt NaH gt KH gt LiH`
`NaH gt CsH gt RbH gt LiH gt KH`

Solution :Ionic character INCREASE as the size of the atom increases or the ELECTRONEGATIVITY of the atom decreases : `LiH lt NaH lt KH lt RbHlt CsH`
40.

Metal carbonates decompose on heating to give metal oxide and carbon dioxide . Which of the metal carbonates is most stable thermally ?

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`MgCO_(3)`
`CaCO_(3)`
`SrCO_(3)`
`BaCO_(3)`

SOLUTION :Thermal stability of metal carbonates increases as the ELECTROPOSITIVE character of the metal or the BASICITY of the metal HYDROXIDE increases from `Be(OH)_(2)` to `Ba(OH)_(2)` . Thus `BaCO_(3)` is most stable .
41.

Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates is most stable thermally?

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`MgCO_(3)`
`CaCO_(3)`
`SiCO_(3)`
`BaCO_(3)`

Solution :Thermal stability of `MCO_(3)` Group-2 depends on the MO. If MO is stable `MCO_(3)`is thermally unstable and vice versa.
`MCO_(3) to MO+CO_(2)`. Thermal stability of MO decreases as the size of M increases. Therefore, BaO is least stable and `BaCO_(3)`is most stable.
42.

Metal carbonate on reaction with ........ acid gives nitrate.

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NITROUS acid
Nitric acid
Picric acid
Sulphuric acid

Answer :B
43.

Metal atom with atomic number 21 has the differentiating electron enetering in 3d but not in 4s. However, in the formation of divalent cation 4s electrons are removed. Why?

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Solution :While filling ELECTRONS 4s is filled FIRST and then 3D, based on Auf-bau PRINCIPLE.
During the FORMATION of divalent cation, it is easy to remove lectrosn from 4s compared to that of 3d.
44.

Metal A when burnt in air forms only an oxide while metal B When brunt in air forms a mixture of oxide and nitride. A and B are

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MG and Ca
K and Mg
Be and Ca
Na and Ca

Solution :K can FORM only OXIDES when reacts with air and Mg can form oxide and nitride.
45.

Metal A reduces silica converting itself into B. B absorbs moisture an converts into C. When C is heated with the reduction product of silica liberates a gas. Then A,B,C and gas are

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`Na, Na_(2)O, NAOH, H_(2)`
`Na, Na_(2)O_(2), Na_(2)CO_(3), O`
`Na, Na_(2)O, Na_(2)CO_(3), CO_(2)`
`Na, Na_(2)O_(2), NaOH, H_(2)`

SOLUTION :`SiO_(2)+NararrSi+Na_(2)Ooverset(H_(2)O)rarrNaOH`
46.

Meta-directing and deactivating group in aromatic electrophilic substitution is

Answer»

`-CH_(3)`
`-OH`
`-NO_(2)`
`-CL`

SOLUTION :`-CH_(3) ""` Activaing and o, p-directing
`-OH ""` ACTIVATING and o, p-directing
`-NO_(2) ""` DEACTIVATING and m-directing
`-Cl ""` Deactivating and o, p-directing
`-OCH_(3) ""` Activating and o, p-directing
47.

Meta directing group in electrophilic substitution in benzene nucleus is

Answer»

`-O-UNDERSET(O)underset(||)C-CH_3`
`-OVERSET(O)overset(||)C-O-CH_3`
`-NH-overset(O)overset(||)C-CH_3`
`-NH_2`

SOLUTION :`-overset(O)overset(|)C - O-CH_3`
48.

Met-enkephalin, an endorphin, serves as natural pain reliver that changes or removes the perception of nerve signals, How many types of functional groups are present in Met-enkephalin.

Answer»


ANSWER :6
49.

Mesoxalic acid is obtained when glycerol reacts with

Answer»

Dil. `HNO_(3)`
`BI(NO_(3))_(2)`
`H_(2)O_(2)+FeSO_(4)`
`KMnO_(4)`

SOLUTION :`{:(COOH),("|"),(C=O),("|"),(COOH):}`
Mesoxalic acid
50.

Mesomeric effect is a permanent effect in which pi electrons are transferred from a

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Multiple bond to an atom
Multiple bond to a single COVALENT bond
Both 1 and 2
None

Answer :C