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 increasingorder of atomicradii of the followinggroup 13 element is

Answer»

`Al lt GA lt In lt Tl`
`Ga lt Al lt In lt Tl`
`Al lt In lt Ga lt Tl`
`Al lt Ga ltTl lt In`

Solution :Due to ineffectiveshielding of 3d-orbitals , theeffective nuclearcharge of GAIS greater in magnitudethanof Al. As a result,ATOMIC radii ofFa is lower thanthat of Al.The atomic radiiof remaining elementincreases regularly as we movedown the group from LN to Tl. Thus,atomic radii of the four elements from Al to Tl increase in the order : `Ga lt Al lt In lt Tl`, i.e.,OPTION(b) is correct.
Discussion
2.

The increasing order of basicity of the following compounds is :

Answer»

`(B) LT(a) lt (d) lt (C)`
`(d) lt (b) lt (a) lt (c)`
`(a) lt (b) lt (c) lt (d)`
`(b) lt (a) lt (c) lt (d)`

ANSWER :A
Discussion
3.

The increasing order of acidic nature among ZnO, Na_(2) O_(2) , P_(2) O_(5) and MgO is

Answer»

<P>`ZnO LT P_(2) O_(5) lt Na_2 O_(2) lt MgO`
`MgO lt Na_(2) O_(2) lt ZnO lt P_(2) O_(5)`
`Na_(2) O_(2) lt MgO lt ZnO lt P_2 O_5`
`P_2 O_5 lt ZnO lt Na_(2) O_(2) lt MgO`

ANSWER :C
Discussion
4.

The increasing basicity order of the following compounds is : (A) " " CH_(3)CH_(2)NH_(2)""(B) {:(""CH_(2)CH_(3)),(""|),(CH_(3)CH_(2)NH):} ""(C) H_(3)C-overset(CH_(3))overset(|)N-CH_(3)""(D) Ph-overset(CH_(3))overset(|)N-H

Answer»

`(A) LT (B) lt (C) lt (D)`
`(D) lt (C) lt (B) lt(A)`
`(A) lt (B) lt (D) lt (C)`
`(D) lt (C) lt (A) lt (B)`

ANSWER :D
Discussion
5.

The increase in the temperature of the aqueous solution placed in a closed vassel will result in its

Answer»

MOLALITY to increase
molarity to DECREASE
mole FRACTION to increase
MASS % increase

ANSWER :B
Discussion
6.

The incorrectly metched pair among the following is :

Answer»

`{:(" MoleculeShape "),(BrF_(5)"Trigonal bipyramidal "):}`
`{:(" MoleculeShape "),(SF_(4)"See saw "):}`
`{:(" MoleculeShape "),(CIF_(3)"T-shape"):}`
`{:(" MoleculeShape "),(NH_(4)^(+) "Tetrahedral"):}`

Solution :`BrF_(5)` has SQUARE PYRAMIDAL shape.
Discussion
7.

The incorrect statements (S) is/are

Answer»

`BeCl_(2)` is an IONIC compound
`BeCl_(2)` can FORM dimer
`BeCl_(2)` is an electron deficient compound
`BeCl_(2)` has seesaw SHAPE

Solution :`BeCl_(2)` is covalent compound `BeCl_(2)` has linear shape.
Discussion
8.

The incorrect statement regarding the formation of ionic bond

Answer»

It INVOLVES electrostatic attraction
It is a redox process
It is an exothermic process
It involves the ABSORPTION of ENERGY

ANSWER :D
Discussion
9.

The INCORRECT statement is

Answer»

LITHIUM is least reactive with water AMONG the alkali metals
LiCl crystallises from AQUEOUS solution as `LiCl.2H_2O`
Lithium is the STRONGEST reducing agent among the alkali metals
`LiNO_3` DECOMPOSES on heating to give `LiNO_2 and O_2`.

Solution :`LiNO_3` decomposes on heating to give `LiNO_2 and O_2`.
Discussion
10.

The incorrect statement is :

Answer»

Lithium is least REACTIVE with WATER AMONG the alkali metals.
LiCl crystallises from aqueous solution as `LiCl.2H_2O`.
Lithium is the STRONGEST reducing agent among the alkali metals.
`LiNO_3` decomposes on heating to give `LiNO_2 and O_2`

Solution :(d): is INCORRECT
`4LiNO_3 overset(Delta) to 2Li_2O + 4NO_2 + O_2 `
Discussion
11.

The incorrect statement in respect of chromyl chloride test is

Answer»

formation of red vapour
formation of LEAD chromate
formation of chromyl chloride
liberation of chlorine

Solution :`Cl_(2)` GAS is not LIBERATED in the chromyl chloride test.
Discussion
12.

The incorrect statement are

Answer»

`CH_3 COOH ` is a weak acid
`NH_4 Cl `GIVES an alkaline solution in WATER
`CH_3 COONa ` gives an acidic solution in water
`NH_4 OH` is a strong base

Solution :`NH_4 Cl ` is acidic due to cation HYDROLYSIS `CH_3COONa ` is basic due to anionhydrolysis`NH_4OH ` is a weakbase
Discussion
13.

Select the incorrect statement

Answer»

`CH_3 COOH ` is a weak acid
`NH_4 Cl `gives an alkaline SOLUTION in WATER
`CH_3 COONa ` gives an acidic solution in water
`NH_4 OH` is a STRONG base

Solution :`NH_4 Cl ` is acidic due to cation hydrolysis `CH_3COONa ` is BASIC due to anionhydrolysis`NH_4OH ` is a weakbase
Discussion
14.

The incorrect statement among the following is

Answer»

The first ionisation potential of AL is less than the first ionisation potential of Mg
The second ionisation potential of Mg is greater than the second ionisation potential of NA
The first ionisation potential of Na is less than the first ionisation potential of Mg
The third ionisation potential of Mg is greater than the third ionisation potential of Al

ANSWER :B
Discussion
15.

The incorrect statement among the following is :

Answer»

CRYSTALLINE boron is obtained by the reduction of boric anhydride with AL powder
Amorphous boron is obtained by the reduction of boric anhydride by FUSION with `NA, K` or `Mg`.
Amorphous boron is CHEMICALLY inert in nature.
Crystalline boron is black and chemically inert.

Answer :C
Discussion
16.

The incorrect statement among the following are

Answer»

`AlCl_(3)` exists as `Al_(2)Cl_(6)` in vapour state
All the `Al - Cl` bonds in `Al_(2)Cl_(6)` are equivalent below `473 K`
Borax when STRONGLY heated with `NH_(4)Cl` FORMS Boron nitride as one of the PRODUCT
`AlF_(3)` has a higher melting point while `AlCl_(3)` has lower melting point

Answer :B
Discussion
17.

The incorrect statement according to second law of themodynamics is

Answer»

HEAT cannot flow from colder body to a hotter body of its own
All spontaneous processes are themody-namically irreversible
Heat can be converted into workcompletely without causing some PERMANENT change in the system (or) surroundings
Perpetual motion machine of second KIND is not possible

Solution :`ETA ne 1`
Discussion
18.

The incorrect statement about step-1 is :

Answer»

It is `S_N2` REACTION
Only one transition STATE is formed in this reaction
Walden inversion has occurred at reactant 'X'
The reaction has MOLECULARITY two

Answer :C
Discussion
19.

The incorrect relations (s) about the velocity of photon is (are)

Answer»

Dependent on its WAVELENGTH
DEPEND on its intensity
equal to cube of its AMPLITUDE
Independent of its wavelength

SOLUTION :The velocity of PHOTON is independent of its wavelength
Discussion
20.

The incorrect order of second ionization energies in the following is

Answer»

RB gtK 
NA gt Mg
Cr gt MN 
S gt P 

Answer :A
Discussion
21.

The incorrect option regarding the chemical rection is

Answer»

Process is called Lane's process
Water gas at as reducintg agent for `Fe_(3)O_(4)`
`Fe_(3)O_(4)` showsmagnetic PROPERTY
`Fe_(3)O_(4)` CANT be concentrate by magnetic separation method.

SOLUTION :`Fe_(3)O_(4)` is magnetic oxide of iron and it has a NON magnetic impurities so it is concentrate by magnetic separation method.
Discussion
22.

The incorrect match is

Answer»




ANSWER :D
Discussion
23.

The incorrect IUPAC name is

Answer»

`{:(""O),("||"),(CH_(3)-C-CH-CH_(3):" 2-Methyl-3-butanone"),("|"),(""CH_(3)):}`
`{:(CH_(3)-CH-CH-CH_(3):" 2, 3-Dimethylpentane"),("|""|"),(""CH_(3)" "CH_(2)-CH_(3)):}`
`CH_(3)-C equiv C-CH (CH_(3))_(2)`: 4-Methylpent-2-yne
`{:(CH_(3)-CH-CH-CH_(3):" 2-Bromo-3-chlorobutane."),("|""|"),(""Cl""Br):}`

Solution :(a) is the INCORRECT name. CORRECT name is 3- Methylbutan-2-one.
Discussion
24.

The incorrect expression among the following is

Answer»

in isothermal process, `W_("reversible") = - nRT"In (V_(f))/(V_(i))`
In `K = (DeltaH^(@) - T DELTA S^(@))/(RT)`
`K = e^(-DeltaG^(@)\RT)`
`(DeltaG_("SYSTEM"))/(DeltaS_("total")) = -T`

Solution :is incorrect because
`DeltaG^(@) = DeltaH^(@) -TDeltaS^(@)`
`RT In K= DeltaH^(@) -TDeltaS^(@)`
`therefore In K = -(DeltaH^(@) -TDeltaS)/(RT)`
Discussion
25.

The incorrect expression among the following is :

Answer»

`K= e^(- DeltaG^(@) //RT)`
`(DeltaG_("system"))/( DeltaS_("total"))=- T`
In ISOTHERMAL process,
`W_("reversible") = -nRT ln.((V_(f))/(V_i))`
`ln K =( DELTAH^(@) - T DeltaS^(@))/( RT)`

SOLUTION :`DeltaG^(@) =- RTlnK`
or `LNK = - (DeltaG^(@))/( RT) - ( DeltaH^(@) - T DeltaS^(@))/( RT)`
Hence, (d) is incorrect.
Discussion
26.

The inceasing order of the acidity of the following carboxylic acids is :

Answer»

`I LT III ltII lt IV`
`IV lt II lt III lt I`
`II lt IV lt III lt I`
`III lt II lt IV lt I`

ANSWER :D
Discussion
27.

The inceasing order of reactivity among group 1 elements is Li lt Na lt K lt Rb lt CS whereas that among group 17 is F gt Cl gt Br gt I. Explain .

Answer»

Solution :GROUP I elements are electropositive in which IE DECRASES down the group.Therefore reactivity increases down the group. Group 17 elements are ELECTRONEGATIVE which show strong electron accepting capacity. This property depends on their elctrode potentials which very as `F gt CL gt Br gt I`. So the reactivity follows the same order.
Discussion
28.

The impossible set of quantum number is

Answer»

n=2, l=0, m=0, s = +1/2
n=2, l=1, m=0, s = +1/2
n=2, l=0, m=1, s = -1/2
n=3, l=1, m=-1, s = -1/2

Answer :C
Discussion
29.

The impossible set of quantum numbers is

Answer»

N = 2 , L= 0 , m =0 , s =+1/2
n=2 , l=1 , m=0 , s =+1/2
n=2 , l=0 , m=1 , s=-1/2
n=3 , l=1 , m=-1 , s=-1/2

Solution :`l=0 , m=1 ` (IMPOSSIBLE)
Discussion
30.

The important oxide ore of iron is

Answer»

Sidherite
HAEMATITE
Pyrites
Bauxite

Solution :Haematite is an IMPORTANT OXIDE ORE of IRON .
Discussion
31.

The III A group element that does not displace hydrogen from hydrochloric acid is

Answer»

B
Al
both (1) and (2)
Tl

Solution :BORON remains uneffected by non-oxidising ACIDS LIKE HCl.
`therefore` Boron does not displace HYDROGEN from HCl.
Discussion
32.

The IIIA group element that does not displace hydrogen from hydrochloric acid is

Answer»

B
AL
both B and Al
TI

Solution :Boron REMAINS uneffected by non-oxidising ACIDS like HCI 
`:.` Boron does not dispalce hydrogen from НСІ
Discussion
33.

The II A carbonate that decomposes at room temperature of 25^(@) C is

Answer»

`CaCO_(3)`
`BeCO_(3)`
`BaCO_(3)`
`MgCO_(3)`

Solution :`BeCO_(3)` can DECOMPOSE even at room tempera ture DUE to low lattice energy
Discussion
34.

The ignition mixture used in alumino-thermy is a mixture of

Answer»

Mg powder `+Na_(2)O_(2)`
Al powder `+ BaO_(2)`
Al powder `+ Na_(2)O_(2)`
Mg powder + `BaO_(2)`

SOLUTION :Ignition MIXTURE used in ALUMINOTHERMY is a mixture of aluminium powder and barium peroxide `(BaO_(2))`.
Discussion
35.

The idea of stationary orbits was first given by………….

Answer»

SOLUTION :NIELS BOHR
Discussion
36.

The +I (Inductive effect) is show by

Answer»

`CH_(3)-`
`HO-`
`-F`
`C_(6)H_(5)-`

ANSWER :A
Discussion
37.

The -I effect is shown by ?

Answer»

`-COOH`
`-OCH_(3)`
`-CH_(3)CH_(2)`
`-F`

Solution :E.N`ALPHA`-I
Discussion
38.

The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to

Answer»

`SIGMA-p` (empty) and `sigma-pi^(**)` electron delocalisations
`sigma-sigma^(**)` and `sigma-pi^(**)` electron delocalisations
`sigma-p` (filled) and `sigma-pi` electron delocalisations
p (filled) `rarr sigma^(**)` and `sigma-pi^(**)` electron delocalisations

Solution :In tert-butyl CATION, the CARBON bearing positive charge has an empty p-orbital and hence hyperconjugation here INVOLVES `sigma`-p (empty) electron delocalisation.

In 2-butene, `pi`-bond contains two electrons, one electron from each of the two carbon atoms, i.e., `pi`-MO contains a PAIR of electrons, but `pi^(**)` MO is empty and hence electrons of the `sigma` C-H bond can move only into the `pi^(**)` orbital. Thus, option (a) is correct.
Discussion
39.

The hyparconjugative stabilities of tert-butyl cation and 2-butene, respectively ,are due to

Answer»

<P>`sigma to p("sigma") ("empty ") and sigma to pi^("*")`ELECTRON delocalization.
`sigma to sigma^("*") and sigma to pi` electron delocalization.
`sigma to `p (FILLED ) and ` sigma to pi` electron delocalization.
p (filled ) `to sigma^("*") and sigma to pi^("*")` electron delocalization .

ANSWER :A
Discussion
40.

The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water . Explain .

Answer»

Solution :Due to larger size of Na and K as compared to that of Mg and Ca , the lattice ENTHALPIES of hydroxides and CARBONATES of sodium and potassium are MUCH lower than those of the hydroxides and carbonates of magnesium and calcium . As a RESULT , the hydroxides and carbonates of Na and K are EASILY soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water .
Discussion
41.

The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.

Answer»

Solution :SODIUM and potassium ions are larger in size than magnesium and calcium ions respectively. DUE to larger size, the lattice energies of hydroxides and CARBONATES of sodium and potassium are much lower than those of magnesium and calcium. This is why the hydroxides and carbonates of sodium and potassium are easily soluble in water, whereas the CORRESPONDING salts of magnesium and calcium are sparingly soluble in water.
Discussion
42.

NaOH, KOH, Na_2CO_3, K_2CO_3 are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain

Answer»

Solution :Alkali metals form MONOVALENT cations (such as `Na^(+), K^(+)`) while alkaline earth metals form DIVALENT cations (such as `Mg^(2+), Ca^(2+)`). Due to increase in charge of the cations, the LATTICE energy of the corresponding salt increases. For this reason, hydroxides and carbonates of sodium and potassium have LOWER lattice enthalpy values than the hydroxides and carbonates of magnesium and calcium. As hydration enthalpies of the hydroxides and carbonates of sodium and potassium are greater than their lattice enthalpies, these salts readily dissolve in water. HOWEVER, in case of the hydroxides and carbonates of calcium and magnesium, the lattice enthalpy values is greater than that of hydration enthalpy and consequently these salts are less soluble in water.
Discussion
43.

The hydroxides and carbonates of sodium and potassium are easily solub while the corresponding salts of magnesium and calcium are sparingly soluble in waterExplain

Answer»

Solution :Since group 1 hydroxides and CARBONATES DUE to large size CONTAIN higher hydration en the lattice energy so, they are easily soluble in water. Whereas, in magnesium and calci to small size their lattice energy dominates over hydration energy they are sparingly sol water.
Discussion
44.

The hydroxides and carbonates of Na and K are solube in the water while those of Mg and Ca are spairingly soluble .Explain.

Answer»

SOLUTION :Lattice energies of the hydroxides and CARBONATES of Na and K are much smaller than those compounds of and MG.(This is due to the LARGER size of Na and K compared to Mg and Ca.)
Discussion
45.

The hydroxide which sublime on heating are i)LiOH ii)KOH iii)RbOHiv) mg(OH)_(2)

Answer»

All undergoes sublmation
I,ii,&ivOonly
I,iii&IV only
I,ii&iiiOnly

Answer :1
Discussion
46.

Give reasons for the following. Sodium and potassium are stored in kerosene.

Answer»

SOLUTION :Since group 1 hydroxides and carbonates due to LARGE size contain HIGHER hydration EN the lattice energy so, they are easily soluble in water. Whereas, in magnesium and calci to small size their lattice energy dominates over hydration energy they are SPARINGLY sol water.
Discussion
47.

The hydroxide of which metal ion(s) is/are soluble in excess of NaOH solution

Answer»

`Al^(3+)`
`ZN^(2+)`
`FE^(3+)`
`Cu^(2+)`

Solution :`3AI+2NaOHrarr2NaAIO_(2)+H_(2)`
`Zn+2NaOHrarrNa_(2)Zn_(2)_+H_(2)`
Discussion
48.

The hydrolysis reaction of ethyl acetate CH_3COOC_2H_(5(l)) + H_2O_((l)) hArr CH_3COOH_((l)) + C_2H_5OH_((l)) Is this reaction get equilibrium in open vessel ?

Answer»

SOLUTION :YES, because the GAS is not in this.
Discussion
49.

The hydrolysis of NCl_(3)" by "H_(2)O produces

Answer»

`NH_(2)OHandHOCl`
`NH_(2)NH_(2)andHCl`
`NH_(4)OH+HOCl`
`NH_(2)ClandHOCl`

Solution :Refer to Ans. Q 4, PAGE 11/168. `NH_(3)` THUS produced combines with `H_(2)O` to form `NH_(4)OH`. Thus, the PRODUCT of hydrolysis are `NH_(4)OHandHOCl`.
Discussion
50.

The hydrolysis constant ofNaX (K_a of HX is 2xx 10 ^(-6))is

Answer»

` 5 xx 10^(-9) `
` 2XX10^(10) `
` 2xx10 ^(-6) `
` 10^(-7) `

Solution :` K_h =(K_W)/(K_a)=(10 ^(-14))/(2 xx 10 ^(-6))= 5 xx 10 ^(-9) `
Discussion