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.

Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so, why?

Answer»

SOLUTION :Due to smaller size Be and MG have relatively HIGHER ionisation energies.
Discussion
2.

Be and Mg are inert because ......... layer is formed on their surface.

Answer»

Sulphide
Nitrite
Hydroxide
Oxide

Answer :D
Discussion
3.

Be and / have following resemblance due to diagonal relationship :

Answer»

habe nearly EQUAL electronegatively
form amphoteric OXIDES
have same charge / radius ratio
both form dimeric halides

Solution :Be and AI SHOW diagonal relationship due to
A) Equal electro negativity
B) Both can form amphoteric oxides
C) Both have same charge / radius ratio
D) Both form dimeric halides
Discussion
4.

Be and Al exhibit many properties which are similar, but the two elements differ in

Answer»

exhibiting amphoteric NATURE in their oxides
forming POLYMERIC hydrides
exhibiting maximum covalency in compounds
forming COVALENT halides

Solution :Polymeric hydrides FORMATION
Discussion
5.

Be and Al exhibit diagonal relationship. Which of the following statements about then is//are not true ? (i) Both react with HCl to liberate H_(2) (ii) They are made passive by HNO_(3) (iii) Their carbides gives acetylene on treatment with water (iv) Their oxides are amphoteric

Answer»

(iii) and (IV)
(i) and (iii)
(i) only
(iii) only

Solution :The carbides of .Be. and .AL. liberates `CH_(4)` on treatment with water
`Be_(2)C+4H_(2)O to 2Be (OH)_(2)+CH_(4)`,
`Al_(4)C_(3)+6H_(2)Oto 2Al_(2)O_(3) +3CH_(4)`
Discussion
6.

Be and Al exhibit diagonal relationship which of the following statements about them is /are not true ? (i) Both react with HCl to liberate H_(2). (ii) They are made passive by HNO_(3). (iii)Their carbides give acetylene on treatment with water . (iv) Their oxides are amphoteric .

Answer»

(III) and (iv)
(i) and (iii)
(i) only
(ii) and (iii)

SOLUTION :Their carbides on treatment with `H_(2)O` gives methane . Aluminium CARBIDE also reacts with `H_(2)` to give methane .
`Al_(4) C_(3) + 12 H_(2)O to 4 Al(OH)_(3) + 3 CH_(4)`
Discussion
7.

Be and Al exhibit diagonal relationship. Which of the following statements about them is/are not true? (i) both react with HCl to liberate H_(2). (ii) Theey are made passive by HNO_(3). (iii) Their carbides give acetylene on treatment with water. (iv) Their oxides are amphoteric.

Answer»

(iii) and (iv)
(i) and (iii)
(i) only
(iii) only

Solution :The CARBIDES of both Be and Al liberate methane with WATER.
`Be_(2)C+2H_(2)O to 2BeO+CH_(4)`
`Al_(4)C_(3)+6H_(2)O to 2Al_(2)O_(3)+3CH_(4)`
Discussion
8.

Be and Al are diagonally related pair of elements. However, they don't have similarity in the following property.

Answer»

METALLIC nature of ELEMENTS 
amphoteric nature of OXIDE 
hydrolysis PRODUCT of carbide
electrical CONDUCTIVITY 

Answer :D
Discussion
9.

Be^(+3) and a proton are accelerated by the same potenatial, their de-Broglie wavelengths have the ratio (assume mass of proton = mass of neutron)

Answer»

`1:2`
`1:4`
`1:1`
`2:1`

SOLUTION :`(lambda_1)/(lambda_2)= SQRT((m_2)/(m_1))`
Discussion
10.

BCl_(3)is planar but anhydrous AlCl_(3)is tetrahedral . Explain .

Answer»

Solution :In `BCl_(3)`, B atom is `sp^(2)` HYBRIDIZED . The three half-filled HYBRID orbitals overlap with the half-filled orbitals
of CL atoms giving a triangular planar structure .
Anhydrous `AlCl_(3)`exists as DIMER with the formula `Al_(2) Cl_(6)`whose structure is shown below :

Thus, each Al atom is surrounded by four BOND paris and hence it assumes tetrahedral structure .
Discussion
11.

BCl_(3)is trigonal planar while AlCl_(3) is tetrahedral in dimeric state. Explain .

Answer»

Solution :Both `BCl_(3)` and `AlCl_(3)` ar electrondeficientmolecules havingsix electronin the valenceshell of their respective central atoms.To completetheir octets the central atoms in each case can ACCEPTA pair of electronsfrom the chlorineatoms ofanothermoleculeformingdimeticstrcutures. However,because ofsmall size of B, itcannotaccomodatesfour big sized Cl atoms around it.THEREFORE,`BCl_(3)` refers to exact as a monomeric planarmoleculein which B atom is `sp^(2)` HYBRIDIZED.
Onthe other HAND, Al because of it BIGGER size can easilyacommdate fourCl atomos around it. As a result, `AlCl_(3)` existsas a dimer.In this dimer, sincethe covalencyof Al has increasedto 4. Therefore, Al is`sp^(3)`- hybridisedand the four Clatoms are held tetrahedrallyaround it. for strucutre.
Discussion
12.

What is Z in the following reactions? BCl_(3)+H_(2)underset(450^(@)C)overset(Cu-Al) to X+HCl""X overset("methylation")toZ

Answer»

`CH_(3)BH_(2)`
`(CH_(3))_(4)B_(2)H_(2)`
`(CH_(3))_(3)B_(2)H_(3)`
`(CH_(3))_(6)B_(2)`

Answer :B
Discussion
13.

BCl_3 exists as monomer whereas AlCl_3 is dimerized through halogen bridging. Give reason. Explain the structure of the dimer of AlCl_3 also.

Answer»

Solution :`BCl_3` does not dimerize because of its SMALL size. `BCl_3` do not exist as a DIMER. BORON cannot ACCOMMODATE FOUR large sized chloride ions. `AlCl_3` exist as a dimer in which Al uses its vacant 3p-orbital by coordinate with Cl to complete their octet by forming dimer. `AlCl_3` achieves stability by forming a dimer.
Discussion
14.

BCl_(3) exists as monomer whereas AlCl_(3) is dimerised through halogen bridging. Give reason, Explain the structre of the dimer of AlCl_(3) also.

Answer»

SOLUTION :\
Discussion
15.

BCl_3 does not exist as dimer but BH_3 exist as dimer because

Answer»

Cl is more electropossitive than H
There is `P pi- P pi` back BONDING in `BCI_3` but `BH_3` does not CONTAIN such multiple bonding
Large CHLORINE atoms do not FIT between small sized boron atoms whereas small hydrogen atoms get FITTED between boron atoms
None of these

Answer :B
Discussion
16.

BCl_(3) doesnot exist as dimer but BH_(3) exist as dimer (B_(2)H_(6)) because

Answer»

chlorine is more electronagative than hydrogen
there is `p pi- d pi` back BONDING in `BCl_(3)` but `BH_(3)` doesnot contain such multiple bonds
LARGE size chlorine atom DONOT fir in between SMALL boran ATOMS where as small sized hydrogen atom get fitted in between boran atoms
none of the above

Solution :Small sized boran atom cannot coordinate to four large size chlorine atims.
Discussion
17.

BCl_(3) does not exist as a dimer but BH_(3) exists as B_(2)H_(6) because:

Answer»

<P>`Cl_(2)` more electronegatve than hydrogen
LARGE size of chlorine atom does not fit between small sized boron atoms while small sized hydrogen atoms OCCUPY the space between boron atoms
there is `p pi-d pi` back BONDING in `BCl_(3)`
both (B) and (c).

Answer :B
Discussion
18.

Bayer's reagent is used for testing ………

Answer»


ANSWER :UNSATURATION
Discussion
19.

Bayer's reagent is :

Answer»

ALKALINE PERMANGANATE solution
acidified permanganate solution
neutral permanganate solution
aqueous BROMINE solution

Answer :A
Discussion
20.

Bauxite is made up of Al_(2)O_3 + SiO_2 + TiO_2 + Fe_(2)O_3 . This ore is treated with conc. NaOH solution at 500 K and 35 bar pressure for a few hours and filtered hot. In the filtrate the species present are

Answer»

`NAAL(OH)_4` only
`Na_(2)Ti(OH)_6` only
`NaAl(OH)_4 and Na_(2)SiO_3` both
`Na_(2)SiO_3` only .

SOLUTION :N//A
Discussion
21.

Bauxite is an ore of

Answer»

Zinc
Copper
ALUMINIUM
Iron

Solution :BAUXITE `(Al_(2)O_(3).2H_(2)O)` is an ORE of aluminium.
Discussion
22.

Bauxite has composition

Answer»

`Al_(2)O_3`
`Al_(2)O_(3)H_(2)O`
`Al_(2)O_(3)2H_(2)O`
NONE of these

Solution :N//A
Discussion
23.

Basicity of H_3 BO_3 is

Answer»

1
2
3
0

Answer :A
Discussion
24.

Basicity of boric acid is

Answer»


SOLUTION :BORIC ACID `(H_(3)BO_(3))` is a weak MONOBASIC Lewis acid.
Discussion
25.

Basic strength of H_2"CC"H_3 H_2C= CH_(2) H-C-= C-H

Answer»

`(i) gt (III) gt (II)`
`(i) gt (ii) gt (iii)`
`(ii) gt (i) gt (iii)`
`(iii) gt (iii) gt (i)`

SOLUTION :Basic character decrease as PERCENTAGE of s character in hybrid ORBITAL increases.
Discussion
26.

Basic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen. Which property of hydrogen may be useful for this purpose ? Support your answer with the chemical equation if required.

Answer»

Solution :Hydrogen is a gas at room temperature. Because of its bulk, it is difficult to transport it either by rail or by road. So by cooling and APPLYING high pressure, GASEOUS `H_2` can be converted into liquid `H_2` which has much smaller volume and hence can be transported EASILY. THUS, the BASIC property of hydrogen which is useful for hydrogen economy is that it can be converted into a liquid by cooling under high pressure.
Discussion
27.

Basic principle of hydrogen economy is transportation and storage of energy in the form of liquid or gaseous hydrogen . Which property of hydrogen may be useful for this purpose ? Suppose your answer with the chemical equation if required.

Answer»

Solution :Hydrogen is a gas at room temperature . Because of its BULK, it is DIFFICULT to transport it either by rail or by road. However, by cooling and applying high pressure , gaseous `H_(2)` can be converted into liquid `H_(2)` which has much smaller volume and HENCE can be transporated EASILY. Thus, the baseic property of hydrogen which is useful for hydrogen economy is that if can be converted into a liquid by cooling under high pressure.
Discussion
28.

Basic nature of the oxides of a period from left to right

Answer»

INCREASES
decreases
REMAINS CONSTANT
FIRST increases and then decreases

ANSWER :B
Discussion
29.

Explain the structure of water molecule ?

Answer»

Solution :ELECTRONIC configuration of `O:1s^(2)2s^(2)2p^(4)`
Valence orbital representation.

1. When `2s^(2)` orbital hybridised with `2p_(x),2p_(y)` and `2p_(Z)` forms the following `sp^(3)` hybridisation structure.
2. In `2s^(2)` and `2p_(x)^(2)`, the ELECTRONS are paired up hence they arenot involved in bond formation called loan paired of electron (L.P.)
3. When `sp^(3)` hybridised O combines with 2 hydrogen atom forms Bond STRUCTRUE water.
Discussion
30.

Based on VSERP theor explain the structure of water.

Answer»

Solution :Electronic CONFIGURATION of `O, 1s^(2) 2s^(2) 2p^(4)`
Valence ORBITAL representation.

1 When `2s^(2)` orbital hybridised with `2p_(x), 2p_(y) and 2p_(z)` forms the following `sp^(3)` hybridisation structuree.
2In `21s^(2) and 2p_(x)^(2)`, the electrons are paired up hence they are not involved in bond formation called loan paired of electron (L.P)
3. When `sp^(3)` hybridised O combines with 2 hydrogen atom forms Bond structure water.
Discussion
31.

Based on VSEPR theory , the number of 90 degree F-B angles in BrF_(5) is

Answer»


Solution :
DUE to repulsion by the lone pair on the bond PAIRS
all the four planar bond anlges DECREASE from ` 90^(@)`
to ` 84.8^(@)` . The axial bond ALSO no LONGER remains
`90^(@)` with the planar bonda . In other words , there
are no `90^(@)FBrF` bonds in `BrF_(5)`
Discussion
32.

Based on VSEPR theory explain the structure of methane.

Answer»

Solution :In the formation of methane, CARBON utilizes `sp^(3)`hybrid orbitals for BONDING. The four `sp^(3)` hybrid orbitals of carbon atom overlaps axially with a orbitals of four hydrogen atoms to GIVE four `C-H` s bond. The C-H s bond is `sp^(3)`-s bond. H-C-H bond angle is `109^(@)28`. the SHAPE of the molecule is tetrahedral.
Discussion
33.

Based on the values of B.E. given, Delta_(f)H^(0)" of " N_(2)H_(4(g)) is : Given BE of : N- N is 159 kJ mol^(-1), H-H is 436 kJ mol^(-1), N =- N is 941 kJ mol^(-1), N-H is 398 kJ mol^(-1)

Answer»

`711 kJ MOL^(-1)`
`62 kJ mol^(-1)`
`-98 kJ mol^(-1)`
`-711 kJ mol^(-1)`

Solution :`N_(2) + 2H_(2) rarr N_(2)H_(4)`
`Delta H_(f) = ((N -= N ) + 2 XX (H - H))- ((N - N) + 4(N - H))`
`=((941) + 2 xx (436)) - (159 + 4 xx (398))` = 62 kJ
Discussion
34.

Based on VSEPR theory explain the structure of ammonia.

Answer»

Solution :Ammonia `(NH_(3))`
Electronic configuration of `N: 1s^(2)2s^(2)2p^(3)`
Valance orbital REPRESENTATION

1 When `2s^(2)` orbital hybridised with `2p_(x), 2p_(y) and 2p_(Z)` forms the following `sp^(3)` hybridised structure.
2 in `2s^(2)` orbital, the ELECTRONS are paired up hence they are not involved in bond formation CALLED loan paired of electron (L.P)
3 When `sp^(3)` hybridized N combines with 3 hydrogen atom, forms pyramidal structure of ammonia. Pyramidal structure angle `=107^(@)`.
Discussion
35.

Based on the third law of themodynamics, the entropy can be obtained using the equation.

Answer»

`Delta S = (Delta H)/(T)`
`underset(0)overset(T)INT T.C_(P)^(-1) DT = S`
`Delta G = T Delta S`
`int_(0)^(T) C_(P) T^(-1) dT= S`

Solution :From third law, `S = underset(0)overset(T)int (C_(P))/(T) dT`
Discussion
36.

Based on the lattice energy and other consideraction which one of the following alkali metal chloride is expected to have the highest melting point.

Answer»

LiCl
NaCl
KCl
RbCl

Answer :B
Discussion
37.

Based on the following equilibrium reactions HNO_(2)+HF rarr H_(2)F^(+) +NO_(2)^(-) CH_(3)COOH +HF rarrF + CH_(3)COOH_(2)^(+) H_(2)O+CH_(3)COOH rarr H_(3)O^(+) +CH_(3)COO^(-) Which are the correct order (s) regarding acid strength

Answer»

`H_2O LT CH_3 COOH`
` HF lt HNO_2`
` H_2O lt HF`
` CH_3 COOH lt HNO_2`

SOLUTION :`H_2O lt CH_3COOH lt HF lt HNO_3`
Discussion
38.

Based on the equation E = - 2.178 xx 10^(-18) J ((Z^(2))/(n^(2))) certain conclusions are written. Which of them is not correct ?

Answer»

The NEGATIVE sign in equation simply means that the energy of electron bound to the NUCLEUS is lower than it would be if the electrons were at infinite distance from the nucleus.
Lower the value of n, the larger is the orbit radius
Equation can be used to claculate the change in energy when the electron changes orbit
For `n = 1`, the electron has a more negative energy that it does for `n = 6` which means that the electron is more lossely bound in the smallest allowed orbit.

Solution :Statement (d) is not correct because in the smallest allowed orbit, the electron is more tightly bound and not loosely bound
Discussion
39.

Based on the above answer the followingoverset(PCl_(5))toIf H is replaced withCH_(3)COO^(-) , ClCH_(2)COO^(-),C_(2)H_(5)COO,PhSO_(3)-then rate of reaction is

Answer»

`CH_(3)COO^(-)ltClCH_(2)COO^(-)ltC_(2)H_(5)ltPhSO_(3)^(-)`
`PhSO_(3)^(-)ltClCH_(2)COO^(-)ltC_(2)H_(5)COOltCH_(3)COO^(-)`
`C_(2)H_(5)COOltClCH_(2)COO^(-)ltCH_(3)COO^(-)ltPhSO_(3)^(-)`
`C_(2)H_(5)COOltCH_(3)COO^(-)ltClCH_(2)COO^(-)ltPhSO_(3)^(-)`

Solution :
Discussion
40.

Based on the above answer the followingWhich of the following statements regarding above sequence is incorrect ?

Answer»

The reaction involves nucleophillic addition reaction.
Reaction involvesa unimolecular STEP.
Rate of FORMATION of white precipitate changes as`F^(-)GTCL^(-)gtBr^(-)gtI^(-)`
B on reaction with silver nitrate GIVES an AROMATIC compound.

Solution :
Discussion
41.

Based on the above answer the following overset((1)Sn//HCl(2)KOHBr_(2))underset((3)Br_(2)//C Cl_(4))tomajor product ?

Answer»




SOLUTION :
Discussion
42.

Based on the compounds of group 15 elements, the correct statement(s) is (are)

Answer»

`Bi_(2)O_(5)` is more basic than `N_(2)O_(5)`
`NF_(3)` is more covalent than `BiF_(3)`
`PH_(3)` boils at lower temperature then `NH_(3)`
the N-N single bond is STRONGER than the P-P single bond

Solution :Due to small size, lone PAIRS on N ATOMS repel the bond PAIR between two N-atoms, and hence N-N single bond is weaker than P-P single bond. All other statements are correct.
Discussion
43.

Based on reductive ozonolysis reaction, how do you distinguish between an alkene and alkyne?

Answer»

SOLUTION :An alkene, on reductive ozonolysis TWO carbonyl COMPOUNDS are formed.
`R-CH=CH-R.underset((ii)H_(2)OZn)overset((i)O_(3))toR-CHO+R.CHO`
An alkyne, on reductive ozonolysis, a dicarbonyl COMPOUND is formed.
`R-C-=C-R.underset((ii)H_(2)O,Zn)overset((i)O_(3))toR-underset(O)underset(||)C-underset(O)underset(||)C-R.`
Discussion
44.

Based on lattics energy and the other considerations which one of the following alkali metal chlorides is expected to have the highest melting point ?

Answer»

LICL
NaCl
KCl
RbCl

SOLUTION :ALTHOUGH lattice energy of LiCl is higher than that
of NaCl but LiCl is covalent in NATURE and NaCl is
ionic. After NaCl , lattice energy decreases as the
size of the alkali metal ion increases . As MELTING
point depends upon lattice energy and ionic/
covalent nature of the compound , NaCl has the
highest melting point .
Discussion
45.

Based on kinetic theory of gases following laws can be proved

Answer»

BOYLE's law
Charles' law
Avogadro's law
all of the above.

Solution :all of the above.
Discussion
46.

Based on first law of thermodynamics , which one of the following is correct ?

Answer»

For an isochoric proces`,DeltaU =-q`
For an ADIABATIC PROCESS, `DeltaU = -W`
For an isothermal process`,q= +w`
For a cyclic process,`q= -w`

Solution :(a) `DeltaU = q+w`. For an isochoric process,`w = - P DeltaV =0`. Hence,`DeltaU = q_(v)`
(b) For an adiabatic process , `q=0`. Hence, `DeltaU =w`
(C ) For an isothermal process, `DeltaU =0` Hence, `q=-w`
(d) For a cyclic process, `DeltaU =0`. Hence, `q= - w`
Discussion
47.

Based on his alpha- ray scattering experiment, Rutherford proposed the nuclear model of an atom. The threshold frequency v0 for a metal is 6.2xx10^14s^(-1). Calculate the K.E. emitted for an electron when the radiation of frequency v=8.7xx10^14s^(-1) strikes the metal.

Answer»

SOLUTION :`K.E=1/2mv^2=hgamma-hgamma_0=h(gamma-gamma_0)=6.626xx10^(-34)(8.7xx10^14-6.2xx10^14)6.626xx10^(-34)xx2.5xx10^14=1.656xx10^(-19)J`
Discussion
48.

Based on equation E=-3.178xx10^(-18)(Z^(2)/(n^(2)))J certain conclusions are written. Which of them is not correct?

Answer»

Equation can be used to calculate the energy change when the electron changes orbit.
For n = 3 , then electron has more negative energy than it does for n = 5 which means that the electron is more tightly bound in the smallest allowed orbit.
The negative SIGN in the equation simply means that the energy of electron bound to the nucleus is lower it would be if ELECTRONS were at the infinite DISTANCE from nucleus.
SMALLER the value of n , the larger is the orbit RADIUS.

Answer :D
Discussion
49.

Based on equation E=-22.178xx10^(-18)J(z^(2))/(n^(2)) certain conclusions are written.

Answer»

Equation can be used to calculate the change in ENERGY when the electron changes orbit. <BR>For n=1, The electron has a more negtative then it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit.
the negative sign in equation simply means that the energy of electron bound to the NUCLEUS is lower than it would br if the electrons were at the infinite distance from the nucleus.
Larger the value of n, the larger is the orbit radious

Solution :Correct STATEMENT: For n=1, The electron has a more negtative then it does for n=6 which means that the electron is more STRONGLY bound in the smallest allowed orbit.
Discussion
50.

Based on equationE=-2.178xx10^(1-18)J((z^2)/(n^2))certain concluusions are written.Which of them is nnot correct ?(NEET)

Answer»

Equation can be used to calculate the change in energy when the ELECTRON CHANGES orbit
For n= 1. the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the SMALLEST allowed orbit
The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus.
Larger the value of n. the larger is the orbit radius

Solution : For n = 1। the electron has a more negative energy than it does for n= 6 which means that the electron is more loosely bound in the smallest allowed orbit
Correct STATEMENT: For n = 1। the electron has more negative energy than it does for n = 6 which means that the electron is strongly bound in the smallest allowed orbit.
Discussion