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.

Alum purifies muddy water by

Answer»

dialysis
adsorption
COAGULATION
forming a TRUE solution

Solution :coagulation
Discussion
2.

Alum is used by dyers of cloth

Answer»

for FIRE proofing fabrics
as WATER PURIFIER
for SOFTENING hard water
as mordant

Solution :SEE mordant dyes.
Discussion
3.

Alum is used by dyers for cloth

Answer»

for fire- PROOFING fabrics
as FIRST AID for cuts
for softening HARD water
as mordant

Answer :D
Discussion
4.

Alum is used by dyer of cloth :

Answer»

As mordant
For fire-proofing fabrics
As FIRST AID for cuts
For SOFTENING HARD water

Answer :A
Discussion
5.

Alum is used by dyer:

Answer»

for fire-proofing fabrics
as FIRST aid for cuts
for SOFTENING hard water
as mordant

Solution :As a mordant in dye INDUSTRY. The FABRIC which is to be dyed is dipped in a solution of the alum and heated with STEAM.`Al(OH)_(3)`
Discussion
6.

Alum is the name used for all double salts having the composition M_(2)^(-) SO_(4).M_(2)^(III)(SO_(4))_(3).24H_(2)O. Where M^(III) stands for AI^(+3), Cr^(+3), Fe^(+3), while M^(I) stands for

Answer»

`Li^(+), CU^(+), Ag^(+)`
`Li^(+), NH_(4)^(+), Na^(+)`
`Na^(+), K^(+), Rb^(+)`
`Ca^(+2), Mg^(+2), SR^(+2)`

Solution :`M^(-)`-Monovalent cation except `Li^(+)` DUE to small size.
Discussion
7.

Alum is not used :

Answer»

As an insecticide
As a MORDANT in dyeing
In the PURIFICATION of water
In TANNING of leather

Answer :A
Discussion
8.

Alum is found to contain hydrated monovalent cation [M(H_(2)O)_(6)]^(+) trivalent cation [M'(H_(2)O)_(6)]^(+) and SO_(4)^(3-) in the ratio of:

Answer»

`1:1:1`
`1:2:3`
`1:3:2`
`1:1:2`

SOLUTION :Formula of alums contain the ions `[M(H_(2)O)_(6)]^(+),[M'(H_(2)O)_(6)]^(3+)` and `SO_(4)^(2-)` in the ratio `1:1:2` Example : (i)Potash alum `K_(2)SO_(4).Al_(2)(SO_(4))_(3).24H_(2)O` , (ii)Chrome alum `K_(2)SO_(4).Cr_(2)(SO_(4))_(3).24H_(2)O`
Discussion
9.

Alum is an example of simple salt. True or false?

Answer»

SOLUTION :DOUBLE SALT
Discussion
10.

Alum is :

Answer»

NORMAL salt
Basic salt
Double salt
Mixed salt

Answer :C
Discussion
11.

Alum helps purifying water by

Answer»

forming SI complex with CLAY particles
sulphate part which combines with the dirt and removes it
aluminium whichcoagulates the MUD particles
making mud water soluble

Solution :Alum coagaulates mud particles and HELPS in purifying water.
Discussion
12.

Alum in aqueous solution gives positive test for (A)K^(+) ,(B) Al^(3+) , (C)SO_4^(2-)

Answer»

A only
B only
A and B
A, B and C

Answer :D
Discussion
13.

Alum helps in purifying water by:

Answer»

forming SILICON complex with CLAY particles
sulphate part which combines with dirt and removes it
aluminium which coagulates the mud particles
making mud water soluble.

Solution :Alums HELP in purifying water by `Al ^(3+)` ions which COAGULATE the negative mud particles.
Discussion
14.

Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why ?

Answer»

<P>

Solution :This is because DUE to langthanoid CONTRACTION, they have almost the same SIZE `(Zr = 160 p m, Hf = 159 p m )`.
Discussion
15.

Although Zr belongs to 4d and Hf belongs to 5d transition series but it is quite difficult to separate them. Why?

Answer»

Solution :This is because of lanthanoid contraction, ZR and Hf have SIMILAR atomic radii and hence shows similar chemical properties. Thus, they are difficult to SEPARATE.
Discussion
16.

Although Zr belongs to 4d and Hf belongs to 5d transition series, but it is quite difficult to separate them. Why?

Answer»

Solution :The two elements have ALMOST the same SIZE Zr (160 pm) and HF (159 pm). This is DUE to lanthanoid contraction. It becomes difficult to separate them because of almost same size.
Discussion
17.

Although Ziroconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because "…..................".

Answer»

both belong to d-block
both have same number of electrons
both have SIMILAR ATOMIC radius
both belongs to the same GROUP of the periodic table.

Answer :c
Discussion
18.

Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because

Answer»

both BELONG to d-block.
both have same NUMBER of electrons.
both have similar atomic radii.
both belong to the same GROUP of the PERIODIC table

Solution : both have similar atomic radii.
Discussion
19.

Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because…..

Answer»

Both BELONG to d-block
Both have same NUMBER of electrons
Both have similar ATOMIC radius
Both belong to the same group of the periodic table

Solution :This is because of lanthanoid contraction, Zr and HF have similar physical and chemical PROPERTIES.
Discussion
20.

Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because.

Answer»

both belong to d-block
both have same NUMBER of electrons
both have SIMILAR atomic RADIUS
both belong to the same GROUP of the PERIODIC table

Answer :C
Discussion
21.

Although trimethylamine and n-propylamine have the same molecular weight, the former boils at a lower temperature (276 K) than the latter (322 K). Explain.

Answer»

SOLUTION :n-Propylamine has two H-atoms on the N-atom and HENCE undergoes intermolecular H-bonding thereby raising its boiling point. TRIMETHYLAMINE, `(CH_(3))_(3)N` being a `3^(@)` amine does not have a H-atom on the N-atom. As a result, it does not undergo H-bonding and hence its boiling point is LOW.
Discussion
22.

Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why?

Answer»

Solution :Temperatures below the POINT of intersection of `Al_2O_3` and MGO CURVES, magnesium can reduce ALUMINA. But the process will be uneconomical.
Discussion
23.

Although thermodynamically feasible, in practice magnesium metal is not used for the reductionof alumina in the metallurgy of aluminium. Why ?

Answer»

Solution :Inspection of Ellingham diagram shows that `DeltaG` vs T curves for `Al_2O_3 and MgO` intersect at a point corresponding to very high temperaureof the order of 2000 K. This means above this temperature, `DeltaG` for the reaction :
`Al_2 O_3 + 3MG to 2AL + 3MgO`
would BECOME negative and hence reduction will be feasible. However, this temperautre is very high so that the process is uneconomical and technolgically DIFFICULT.
Discussion
24.

Although thermodynamically feasible, in practice alumina is not reduced using magnesium. Why?

Answer»

Solution :Temperature above the point of intersection of `Al_2 O_3` and MgO curves, Mg can REDUCE `Al_2 O_3`. But the temperature required (1600 K) is very high. Hence the reduction process will be uneconomic and difficult.
`Fe_2 O_3+ 2Aloverset( 1000-1500^@ C ) toAl_2 O_3+2Fe `
The ignition helps to OVERCOME the energy of activitation and hence the REACTION is fast.
Discussion
25.

Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium. Why ?

Answer»

SOLUTION :Below the temperature (1623 K), corresponding to the point of intersection of AL,O, and MGO curves inmagnesium can reduce alumina as discussed in above.Butamgnesiumisamuchcostlier metalthan ALUMINIUM and hence the processwill beuneconomical.
Discussion
26.

Although thermodynamically feasible, in practice , magnesium metal is not used for the reduction of alumina in the metallurgy of aluminium .Why?

Answer»

SOLUTION :Temperature abovethe point of intersection of `Al_(2) O_(3)` and MgOcurves, Mg canreduce `Al_(2) O_(3)`. But the temperaturerequired (1600K) is very high. Hence the reductionprocess will be uneconomicand difficult.
`Fe_(2)O_(3) + 2A1 overset (1000-1500^(0)C) to A1_(2) O_(3) + 2FE`
The ignition helps to overcomethe energy of activitation and hence the reaction is fast.
Discussion
27.

Although the chemical formulae of Prussian blue and Turnbull's blue are different yet these are supposed to be identical. How is it justified?

Answer»

Solution :When a ferric salt reacts with POTASSIUM ferrocyanide, Prusian blue is formed.
`4FeCl_(3)+3K_(4)[Fe(CN)_(6)]to Fe_(4)[Fe(CN)_(6)]_(3)+12 KCl`.
Turnbull's blue is formed as a deep blue precipitate when a ferrous salt reacts with potassium ferricyanide.
`3FeCl_(2)+2K_(3)[Fe(CN)_(6)]toFe_(3)[Fe(CN)_(6)]_(2)+6 KCl`.
Other investigations like `X-`ray, have shown that these two compounds Viz.,
Pussian blue and Turnbull's blue are the same. When either of the reactions given above is performed, a redox reaction also TAKES place partially:
`{:(Fe^(3+)+[Fe"(CN)_(6)]^(4-) to Fe^(2+)+[Fe"'(CN)_(6)]^(3-)),(Fe^(2+)+[Fe"'(CN)_(6)]^(3-) to Fe^(3+)+[Fe"(CN)_(6)]^(4-)):}`
So, in both the cases, the compounds have a mixture of `Fe^(2+),Fe^(3+),[Fe(CN)_(6)]^(4-)` and `[Fe(CN)_(6)]^(3-)`. Therefore, we WRITE Prussian and Turnbull's blue as
`Fe[Fe(CN)_(6)]^(-)`. This can be DUE to `Fe^(2+)` and `[Fe'''(CN)_(6)]^(3-)` or due to `Fe^(3+)` and `[Fe"(CN)_(6)]^(4-)`.
Discussion
28.

Although phenoxide ion has more number of resonating structures than carboxylate ion in carboxylic acidis a stronger acid than phenol. Give two reasons.

Answer»

SOLUTION :(i) Pohenoxide ION has non equivalent resonance structures in which the negative charge is at the LESS electtonegative carbon atom, whereas in CASE of carboxylate ion both the resonating structures are equivalent.
Discussion
29.

Although some alkaloids are chiefly used as pain killers, all alkaloids are toxic and cause death if taken in large quantities. Name the alkaloid which was used to kill socrtes, the great Greek philosopher.

Answer»

SOLUTION :Socrates was killed by MAKING him to drink a CUP of the extract of hemlock plant which contains the poisonous alkaloid CONLINE.
Discussion
30.

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?

Answer»

Solution :Let us compare the resonating structures of phenoxide and carboxylate ions :

In structures II, III and IV, the negative charge is on less electonegative carbon. THEREFORE, their contribution to the RESONANCE stabilisation is negligible. Thus, structures I and V contribute MAINLY to the resonance stabilisation. Now consider the resonance structures of carboxylate ION

Negative charge on carboxylate ion is delocalised over two oxygen atoms while in structures and of phenoxide ion the negative charge is delocalised over one oxygen atom. Hence, carboxylate ion is stabilised to a GREATER extent. Thus, carboxylic acid is stronger than phenol.
Discussion
31.

Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol? Why?

Answer»

Solution :In carboxylate ion -ve charge is delocalised over two oxygen atoms WHEREAS in phenoxide ion the -ve charge is delocalised over one oxygen atom. Therefore carboxylate ion is more stable than phenoxide ion and that is why CARBOXYLIC acids are more acidic than PHENOLS.
Discussion
32.

Although phenol is a strong germicide but is not an ideal antiseptic because it causes severe skin burns and kills healthy cells along with harm ful microorganisms. Suggest a better therapeuric agent?

Answer»

SOLUTION :Hexylresorcinol is a more powerful germicide than phenol, and is les damaging to the SKIN. It is used as an anthelmintic annd antiseptic agent for MOUTHWASHES and as skin wounds CLEANSER and anti-browning agent in foods.
Discussion
33.

Although nitrogen does not adsorb on suface at room temperature, it adsorbs on the same surface at 83 K. Which one of the following statement is correct ?

Answer»

At 83 K, there is formation of monomolecular LAYER.
At 83 K, there is formation of multimolecular layer.
At 83 K, nitrogen molecules are held by CHEMICAL BONDS.
At 83 K, nitrogen is adsorbed as ATOMS.

Solution :At 83 K, there is physisorption and HENCE multimolicular.
Discussion
34.

Although hexafluoroethane (C_2F_6,b.p.-79^@C) and ethane (C_2H_6 b.p. – 89^@C) differ very much in their molecular weights, their boiling points differ only by 10^@C. This is due to

Answer»

low polarizability of F
nearly similar SIZE of F and H
both (a) and (b)
Neither of the two

Solution : Small difference in boiling points of `C_2F_6` and `C_2H_6` is DUE to the fact that (i) the F atom is only SLIGHTLY LARGER than H, and (ii) F has low polarizability .
Discussion
35.

Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilize higher oxidation states exceeds that of fluorine. Why?

Answer»

Solution :This is DUE to ability of oxygen to FORM a multiple BONDS with METALS in a covalent compounds.
Discussion
36.

Although fluorine is much more electronegative than hydrogen yet the dipole moment of NF_(3) (0.24 D) is much lower than that of NH_(3) (1.46 D). Explain.

Answer»

Solution :The dipole moment of molecules having lone pairs and polar bonds is the vector sum of the following two MOMENTS.
(i) Dipole moment of the lone pair/s
(ii) RESULTANT of the dipole moments of polar bonds.
Now both `NH_(3) and NF_(3)` have pyramidal structures with FNF and HNH bond angles of `102.4^(@) and 107.8^(@)` respectively. The direction of the dipole moment of the lone pair is in the same direction, i.e., away from the N atom as INDICATED in the Figure

SINE, N is more electronegative the H, therefore, the direction of the N-H bond moments is from H to N. The resultant of these three bonds adds to the moment of lone pair and hence the net dipole moment of `NH_(3)` is 1.46 D.
In contrast, F is more electronegative than N and hence the direction of the N-F bond moments is from N to F/ Tje resultant of the dipole moments of the three N-H bonds in `NH_(3)`. In other words, the only difference is that in cas of `NH_(3)`, the dipole moments of three N-H bonds adds to the dipole moment of lone pair but in case of `NF_(3)`, it OPPOSES the dipole moment of the lone pair. Consequently, `NF_(3)` has a much lower dipole moment (0.24 D) than that if `NH_(3)` (1.46 D).
Discussion
37.

Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why ?

Answer»

Solution :The electronic configuration of fluorine is `1s^(2)2S^(2)2p^(5)`. THUS it can form only one bond as it has only one unpaired electron. Electronic configuration of oxygen is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(4)`.
It may be MENTIONED that oxygen also has vacant d-orbitals along with two 3p orbitals containing single electrons. Thus, oxygen has greater bond formation capacity. In other WORDS, it has greater ability to stabilise higher oxidation states.
Discussion
38.

Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine this is due to

Answer»

Low enthalpy of DISSOCIATION of F- F BOND
High hydration enthalpy of `F^(-)`
Bond 1 & 2
Least ionizationpotential

Answer :C
Discussion
39.

Although electron gain enthalpy of fluorine is less negative as compared to chlorine. Fluorine is a stronger oxidizing agent than chlorine. Why ?

Answer»

Solution :(i) Due to low ENTHALPY of dissociation of F - F bond
(ii) HIGH hydration enthalpy of `F^(-)`.
Discussion
40.

Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidizing agent than chlorine. Why?

Answer»

Solution :The.oxdising capability of a substance can be determined by its electrode potential Higher the electrode potential, stronger is the oxidizing agent. The electrode potential, in TURN, depends upon the following three factors:
(i) Bond dissociation enthalpy (ii) Electron gain enthalpy and (ii) HYDRATION energy. ALTHOUGH electron gain enthalpy of FLUORINE is less negative than that of chlorin e but (a) enthalpy of dissociation of F-F bond is lower than that of CI-Cl bond, and (6) Hydration enthalpy of Fion is higher than that of Crion. Because of these two REASON, electrode potential of `F_(2)` (+ 2.87 V) is much higher than that of `Cl_(2)` (+ 1.36 V) and hence `F_(2)` is a stronger oxidizing agent than `Cl_(2)`.
Discussion
41.

Although electron gain enthalpy of fluorine is less negative as compared to chlorine, fluorine is a stronger oxidising agent than chlorine. Why ?

Answer»

SOLUTION :The OXIDISING capability of a substance can be determined by its electrode potential. Higher the electrode potential, stronger is the oxidising agent. The electrode potential, in turn, depends upon the following three factors : (i) Bond dissociation enthalpy (ii) Electron gain ENETHALPY and, (iii) Hydration energy.
Although electron gain enthalpy of flrorine is less negative than that of chlorine but
(a) enthalpy of disociation of F-F bond (158.8 kJ `mol^(-1)`) is much lower than that of CI-CI bond (242.6 kJ `mol^(-1)`), and
(b) hydration enthalpy of `F^(-)` ion (515 kJ `mol^(-1)`) is much higher than that of `CI^(-)` ion (381 kJ `mol^(-1)`).
Because of these two reasons, electrode potential of `F_(2)`(+ 2.87 V) is much higher than that of `CI_(2)` (+ 1.36 V) and hence `F_(2)` is a stronger oxidising agent than `Cl_(2)`
Discussion
42.

AlthoughDelta H of fluorineis less negative than that of chlorine, but fluorine is a stronger oxidising agent than chlorine, why ?

Answer»


SOLUTION :This is due to,
(i) LOW BOND dissociation enthalpy of F-F bond.
(ii) HIGH HYDRATION enthalpy of `F^(-)`.
Discussion
43.

Although Cr^(3+) and Co^(2+) ions have the same number of unpaired electrons but the magnetic moment of Cr^(3+) is 3.87 BM and that of Co^(2+) is 4.87 BM. Why?

Answer»

Solution :`Cr^(3+)` ion has symmetrical electronic configuration in the OUTERMOST orbit i.e., `3d^(3)`. In such ions there is no ORBITAL configuration to magnetic moment. However, appreciable orbital contribution takes PLACE in `Co^(2+)` with `3d^(7)` configuration.
Discussion
44.

Although C_(P) is usually assumed to be constant, for more accurate calculations we must consider its variation with temperature as well. This relation is given by : C_(P) = a + b T + c T^(2) + d T^(3) mol^(-1) K^(-1) For CO_(2), a = 5.0 , b = 15 xx 10^(-5) .Find the heat required (for one mole) to raise its temperature from 300 K to 500 K in case (i) and case (ii). Case (ii) gives a more accurate value. Find the percent error in case (i)

Answer»

Solution :`a Delta T + (B)/(2) (T_(2)^(2) - T_(1)^(2))` (b) 1000 CAL , 1012 cal , 1.1858 % error
Discussion
45.

Although Cr^(3+) and Co^(2+) ions have same number of unpaired electrons but the magnetic moment of Cr^(3+) is 3.87 B.M. and that ofCo^(2+) is 4.87 B.M. Why ?

Answer»

Solution :`CR^(3+) = [Ar]3d^(3) ` and `Co^(2+) = [Ar] 3d^(7)` . Though bothhavesame number ofunpaired ELECTRONS but total number of electrons present in them are DIFFERENT . Total moment depends upon total spin QUANTUM number (S)as well as the resultant orbital angularmomentum of all the electrons (L) , .e.g. , `mu = sqrt( 4S ( S+1 ) + L ( L + 1))` .Orbital CONTRIBUTION is negligible in case of `Cr^(3+)` but not in case of `Co^(2+)`.
Discussion
46.

Although Cr^(3+) and Co^(2+) ions have same number of unpaired electrons but the magnetic moment of Cr^(3+) is 3.87 B.M and that of Co^(2+) is 4.87 B.M. Why?

Answer»

Solution :In `CR^(3+)` there is no orbital CONTRIBUTION because of SYMMETRICAL configuration while in `CO^(2+)`, the orbital contribution is present SIGNIFICANTLY
Discussion
47.

Although C_(P) is usually assumed to be constant, for more accurate calculations we must consider its variation with temperature as well. This relation is given by : C_(P) = a + b T + c T^(2) + d T^(3) mol^(-1) K^(-1) Find the expression for the amount of heat required to raise the temperature of 1 mole of gas from T_(1) Kto T_(2) K , while (i) keeping the first term [i.e. C_(P) = a] (ii) keeping the first two terms. (ii) keeping all the terms of the above expression. Note that each successive term introduces higher accuracy.

Answer»

SOLUTION :(i) `a DELTA T`
Discussion
48.

Although CO is neutral but it shows acidic nature on reaction at high P and T with

Answer»

`CA(OH)_2`
NaoH
`MG(OH)_2`
LiOH

Answer :B
Discussion
49.

Although chlorine is an electron withdrawing group, yet it is ortho - , paradirecting in electrophilic aromatic substitution reactions. Why ?

Answer»

Solution :

Chlorine withdraws electrons through inductive effect and releases electrons through RESONANCE. Through inductive effect, chlorine destabilies the intermediate carbocation formed during the electrophilic substitution reaction.
Through reasonance, halogen tends to stabilise the carbocation and the effect is more PRONOUNCED at ortho - and para - positions. The inductive effect is stronger than resonance and causes net electron withdrawal and thus cause net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho - and para - positions and HENCE makes the deactivation less for ortho - and para - attack. Reactivity is thus CONTROLLED by the stronger inductive effect and orientation is controlled by resonance effect.
Discussion
50.

Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why?

Answer»

Solution :The LONE pairs of electrons on chlorine take part in REASONANCE with the benzene ring making ortho and para positions negative. Therefore it is ortho and para directing in electrophilic AROMATIC SUBSTITUTION.
Discussion