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.

Among ortho, meta and para substituted diahalobenzenes which has high melting point? Give reasonw ith example.

Answer»

Solution :The boiling POINTS of isomeric dihalobenzene are nearly the same. The melting point of para of para ISOMER is generally higher than the melting points of ortho and meta isomers. The higher melting point of p-isomer is due to its symmetry which leands to more close packing of its molecules in the crystal lattice and consequently strong INTERMOLECULAR attractive FORCE which requires more energy for melting.
p-Dihalobenzene`GT`o-Dichlorobenzene`gt`m-Dichloro benzene.
Discussion
2.

Among OH^(-), CH_(3)O^(oplus), CH_(3)-CH_(2)-OH, CH_(3)-OH which solution in water has maximum pH value

Answer»

OH-

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

ANSWER :B
Discussion
3.

Among NH_3, H_2O and HF which would you expect to have highest magnitude of hydrogen bonding? Why?

Answer»

SOLUTION :HF.Due to high ELECTRONEGATIVITY of F,H - BONDING will be the strongest in HF.
Discussion
4.

Among NH_3,H_2Oand HF which would you expect to have highest-magnitude of hydrogen bonding and why ?

Answer»

Solution :On the basis of molecular masses of `NH_3, H_2O` and HF, their boiling points are expected to be lower than those of the subsequent group MEMBER hydrides.
However, DUE to higher electronegativity of N, O and F, the MAGNITUDE of HYDROGEN bonding in their hydrides will be quite appreciable. Hence,the boiling points `NH_3, H_2O`and HF will be higher than the hydrides of their subsequent group MEMBERS.
Discussion
5.

Among NH_(3),H_(2)O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?

Answer»


ANSWER :HF, because FLOURINE is more ELECTRONEGATIVE.
Discussion
6.

Among NH_(3), H_(2)O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?

Answer»

Solution :Hydrogen bondling represents dipolar attraction. Its strength DEPENDS UPON the magnitude of the polarity of the bond. The electronegativities of threee non metals F, O and N involved in these COMPOUNDS are 4,3.5and 3respectively. This means that H-F bond is MAXIMUM POLAR and as a result, highest magnitude of hydrogen bonding is HF molecules
Discussion
7.

Among NH_(3), H_(2)O and HF , which would you expect to have highest magnitude of hydrogen bonding and why ?

Answer»

Solution :Due to GREATER electronegativity of N, O and F over, H, all these molecules undergo intermolecular H-bonding

However,since electronegativity of F is the highest, THEREFORE, magnitude of the +ve charge on HYDORGEN and -ve charge of F is the highest and HENCE electrostatic attraction or the H-bonding is the strongest in H-F
Discussion
8.

Among NaF, RbF, CsF more ionic and strong ionic are

Answer»

CSF, CsF
CsF, NaF
NaF, NaF
KF, CsF

Answer :B
Discussion
9.

Arrange NaCl, MgCl_(2) and AlCl_(3) in the increasing order of covalent character.

Answer»

Solution :Among the ionic compounds `NACL, MgCl_(2) and AlCl_(3)` the charge of the cation INCREASES in the order `Na^(+)lt Mg^(2+)ltAl^(3+),`thus the covalent character also FOLLOWS the same order `NaCl lt MgCl_(2) lt AlCl_(3).` shows more covalent character.
Discussion
10.

Among N, Cu, Rn and U, identify the element that (i) belongs to d-block, (ii) is an actionid.

Answer»


ANSWER :(i) CU (II) U
Discussion
11.

Among N^(-3), O^(-2) , Mg^(+2) and Al^(+3) (a) What is common in them. (b) Arrange them in the increasing ionic radii

Answer»


Answer :ISO electronic species `Al^(3+) lt MG^(2+) lt O^(2-) lt N^(3-)`
Discussion
12.

Among MgSO_4, CaSO_4, SrSO_4 and BaSO_4, which one is most soluble in water ?

Answer»

SOLUTION :`MgSO_4`
Discussion
13.

AmongMg, Mg^(2+) ,A1 and A1^(3+) which willhas the largestand the smallestsizerespectively ?

Answer»

`Mg, A1^(3+)`
`A1^(3+) ,Mg`
`Mg^(2+) , A1`
`A1,Mg^(2+)`

SOLUTION :Alonga periodatomicradiidecrease .ThereforeMg hassmalleratomicradiusthant A1.
Amongstisoelectronicions (i.e., `Mg^(2+) =12-2=10` electronicand `A1^(3+)=13- 3 =10 ` electrons ) , higherthe + Vechargesmallerthe ionicradiusThus `A1^(3+)` has smallerionicradiusthan `Mg^(2+)`
Combining thetwo trendsMg hasthe largestadn `A1^(3+)` has thesmallestsize i.e.,OPTION(a) ISCORRECT.
Discussion
14.

AmongMe_(3)N, C_(5)H_(5)N andMe CN (Me= methylgroup) , theelectronegativityof N is theorder

Answer»

`MeCN GT C_(5) H_(5) N gt Me_(3)N`
`C_(5) H_(5) N gt Me_(3)N gt MeCN`
`Me_(3) NGT MeCN gt C_(5) N_(5) N`
electronegativity is samein all

Solution :Theelectronegativitydecreasesas s- characterof the hybridorbitalof Ndecreases , i.e.,`sp gt sp^(2) gt sp^(3)`. Thuselectronegativityof N in the followingcompoundsdecreasesin theorder :

Thusoption (a) is correct.
Discussion
15.

Among LiCl, RbCl, BeCl_2, MgCl_2, the compounds with greatest and least ionic character respectively are

Answer»

`LICL, RBCL`
`RbCl, BeCl_2`
`RbCl, MgCl_2`
`MgCl_2, BeCl_2`

SOLUTION :`RbCl, BeCl_2`
Discussion
16.

Among Licl, BeCl_(2), BCl_(3) and "CC"l_(4) the covalent bond character follows the order

Answer»

`LiCl LT BeCl_(2)gt BCl_(3)gt"CC"l_(4)`
`LiClgtBeCl_(2)ltBCl_(3)lt"CC"l_(4)`
`LiCl lt BeCl_(2)lt BCl_(3)lt"CC"l_(4)`
`LiClgtBeCl_(2)gtBCl_(3)gt"CC"l_(4)`

Answer :C
Discussion
17.

Among KO_(2), AlO_(2)^(-), BeO_(2) and NO_(2)^(+), unpaired electron is present in

Answer»

`NO_(2)^(+) and BaO_(2)`
`KO_(2) and AlO_(2)^(-)`
`KO_(2)` only
`BaO_(2)`only

Solution :ELECTRONS present :
`NO_(2)^(+) = 7 + 16 - 1 = 22 , BaO_(2) = 56 + 16 = 72 `
`AlO_(2)^(-) = 13 + 16 + 1 = 30 `
and `KO_(2) = 19 + 16 = 35`
Thus , OLNY `KO_(2)` is odd electron species.
Discussion
18.

AmongKO_(2), AlO_(2)^(-), BaO_(2) and NO_(2)^(+), unpaired electrons is present in

Answer»

`NO_(2)^(+)` and `BaO_(2)`
`KO_(2)` and `AlO_(2)^(-)`
`KO_(2)` only
`BaO_(2)` only.

Solution :A superoxide ion `O_(2)^(-)` (in `KO_(2)`) contains UNPAIRED electron.
Discussion
19.

Among KO_(2), AlO_(2)^(-), BaO_(2) and NO_(2)^(+), unpaired electron is present in

Answer»

`NA^(+) and BaO_(2)`
`KO_(2) and AlO_(2)^(-)`
`KO_(2)` only
`BaO_(2)` only

Answer :C
Discussion
20.

Among KO_(2), AlO_(2)^(-), BaO_(2) and NO_(2)^(+), unpaired electron is present in …………

Answer»

`NO_(2)^(+) and BaO_(2)`
`KO_(2) and AlO_(2)^(-)`
`KO_(2)` only
`BaO_(2)` only

ANSWER :C
Discussion
21.

Among KO_2, AlO_2 , BaO_2 and NO_2^+ unpaired electron is present in …..

Answer»

`NO_2^+` and `BaO_2`
KO and `AlO_2^-`
`KO_2` only
`BaO_2` only

ANSWER :C
Discussion
22.

Among isomeric butenes, .................is the most stable.

Answer»


ANSWER :trans-2-butene
Discussion
23.

Among : I.18 g water at 298 K II . 18 g ice at 263 K III . 18 g steam at 373 K IV. 9 g ice at 263 K , correct order of "U"

Answer»

`IV gt II gt I gt III`
`IV lt II lt I lt III`
`II lt IV lt I lt III`
`III gt I gt IV gt II`

ANSWER :B
Discussion
24.

Among HCl, HBr and HI which reactant gives Kharash reaction with alkene ?

Answer»

SOLUTION :Only HBR
Discussion
25.

Among halogens, the correct order of amount of energy released in electron gain enthalpy is:

Answer»

`FgtClgtBrgtI`
`FgtClltBrltI`
`FltClgtBrgtI`
`FltClgtBrltI`

ANSWER :C
Discussion
26.

Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is :

Answer»

`F gt Ci gt Br gt I`
`F LT CI lt Br lt I`
`F lt CI gt Br gt I`
`F lt CI lt Br lt I`

SOLUTION :Moving in table Cl to I, electron gain enthalpy becomes less negative due to the corresponding increase in the atomic size.
(Electron gain ecthalphy = ENERGY RELEASED when electron is GAINED).
Electron enthalpy of F is less negative as of CI, due to its small size.
Order for electron gain enthalpy is as follows `F lt CI gt Br gt I`
Discussion
27.

Amonghalogens the correctorder of amountof energyreleasedin electron gain (electron gainenthalpy ) is :

Answer»

`F gt C1 gt BR gt I`
`F gt C1 LT Br gt I`
`F lt C1 gt Br I`
` F lt C1 lt Be lt I`

Solution :As we move from C1 to I theelectrongainenthalpy(i.e.,energy in electron gain )becomes less and lessnegative due to a corresponding increases in theatomic size. Howeverthe electron gainenthalpy of Fis less NEGATIVE than that of C1due to its small size.Thus , the negativeelectron gain ENTHALPY follows the ORDER`: F lt C1 gt Br gt I` i.e.,option( c)iscorrect.
Discussion
28.

Among halides of alkaline earth metals fluorides are less soluble, but BeF_(2) is more soluble. Why?

Answer»

Solution :Lattice energies of ionicfluridesare HIGH .
HENCE , floridesof ALKALINE earth metals are less SOLUBLE .
`BeF_(2)`is covalentand has the tendency toformhydorgenbondingwith water .
`BeF_(2)`is more solublein water due to high hydration energy of `Be^(2+)` ion .
Discussion
29.

Among H_(2),He_(2)^(+) ,Li_(2),Be_(2)_,C_(2),N_(2),O_(2)^(-) and F_(2) , the number of diamagnetic species is (Atomic numbers : H=1, He=2, Li=3, Be=4, B=5, C=6, N=7, O=8, F=9)

Answer»


Solution :M.O. Configuration are
`H_(2)(2) = sigma (1s)^(2)` (Diamagnetic)
`He_(2)^(+) = sigma (1s)^(2), sigma^(**)(1s)^(1)` (Paramagnetic)
`LI (2) (6) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2S)^(2) ` (Diamgnetic)
` Be_(2) (8) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2)` (Diamagnetic)
` Be_(2) (10) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(1) = pi(2p_(y))^(1)`(Paramagnetic)
` C_(2) (12) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(2) = pi(2p_(y))^(2)`(Diamagnetic)
` N_(2) (14) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`pi(2p_(x))^(2) = pi(2p_(y))^(2),sigma(2p_(z))^(2)`(Diamagnetic)
` O_(2)^(-) (17) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`sigma(2p_(z))^(2) = pi(2p_(y))^(2),sigma(2p_(z))^(2)`
` pi^(**) (2p_(x))^(2) = pi^(**) (2p_(y))^(1) ` (paramagnetic)
` F_(2) (18) = sigma (1s)^(2) , sigma^(**) (1s)^(2) , sigma(2s)^(2) , sigma^(**) (2s)^(2), `
`sigma(2p_(z))^(2) = pi(2p_(x))^(2),sigma(2p_(y))^(2)`
`pi^(**) (2p_(x))^(2) = pi^(**) (2p_(y))^(2)`(Diamagnetic)
THEREFORE , the TOTAL NUMBER of diamagnetic species
is six .
Discussion
30.

Among group 14 elements, most acidic oxide is formed by

Answer»

`PB`
C
`Si`
`Ge`

ANSWER :B
Discussion
31.

Among groups 1 and 2, the elements of which group have higher ionisation enthalpies ?

Answer»

SOLUTION :GROUP 14
Discussion
32.

Among ethanol (I), acetic acid (II), phenol (III) and benzoic acid (IV), the correct order of increasing acid strength is

Answer»

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

SOLUTION :Benzoic ACID is stronger than acetic acid due to `-I` EFFECT of `-C_(6)H_(5)` group. Acetic acid is stronger acid than phenol which is stronger acid than ethanol
Discussion
33.

Among elements with the following electronic configurations, the one with the largest radius is

Answer»

`1s^2 2s^2 2p^6 3s^2`
`1s^2 2s^2 2p^6 3s^2 3p^1`
`1s^2 2s^2 2p^6 3s^2 3p^2`
`1s^2 2s^2 2p^6 3s^2 3p^5`

ANSWER :A
Discussion
34.

Among elements of carbon family, which act as 7 semi conductors?

Answer»

SOLUTION :SILICON and GERMANIUM ACT as SEMICONDUCTORS.
Discussion
35.

Among diamond and graphite, which is a good conductor of electricity ?

Answer»

SOLUTION :GRAPHITE
Discussion
36.

Among chalcogens electron affinity is highest for

Answer»

O
S
Se
Te

Answer :B
Discussion
37.

Among CH_(3)F, CH_(3)Cl, CH_(3)Br and CH_(3)I which does not give methane on reduction ?

Answer»

SOLUTION :`CH_(3)F`
Discussion
38.

Among CH_3CH_2CH_2CH_2Cl, CH_3 -underset(Cl)underset(|)CH - CH_2 - CH_3, (CH_3)_3 C - Cl. Which one has low boling point ? Give reason.

Answer»

Solution :For a compound with MOLECULAR formula `C_5H_9Cl_3` isomers `(1^@, 2^@, 3^@)` are possible.
AMONG the isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl chain. This is because with increase in branching, the MOLECULE attains spherical shape with less surface area. As a result, the intermolecular forces become weak, resulting in LOWER boiling point. Therefore the boiling point decreases in the order:
`underset(351.5 K)(CH_3-CH_2 - CH_2)-CH_2Cl gt underset((341 K))(CH_3-underset(Cl)underset(|)CH-CH_3) - CH_3 gt underset((324K))(CH_3 - underset(Cl)underset(|)overset(CH_3)overset(|)C-CH_3)`
Hence, `(CH_3)_3C -Cl`has the lowest boiling point.
Discussion
39.

Among the hydrides given below which are reasonable good acids

Answer»

`NH_3 and B_2H_6`
`CaH_2 and B_2H_6`
`NAH and CaH_2`
`NaH and NH_3`

SOLUTION :CONCEPTUAL
Discussion
40.

Among CaH_(2) , BeH_(2) , BaH_(2) , the order of ionic character is

Answer»

`BeH_(2) lt CaH_(2) lt BaH_(2)`
`CaH_(2) lt BaH_(2) lt CaH_(2)`
`BeH_(2) lt BaH_(2) lt CaH_(2)`
`BaH_(2) lt BeH_(2) lt CaH_(2)`

Solution :As the electronegativity of the ELEMENT DECREASES down the group , the ionic character INCREASES , i.e., option (a) is correct .
Discussion
41.

Among CaH_2, BeH_2, BaH_2the order of ionic character is :

Answer»

`BaH_2 lt BeH_2 lt CaH_2`
`BeH_2 lt CaH_2 lt BaH_2`
`BeH_2 ltBaH_2 lt CaH_2`
`CaH_2 lt BeH_2 lt BaH_2`

Solution :Be, Ca and BA all are belongs to second GROUP. So their ELECTRONEGATIVITY decreases, Atomic radius INCREASES and therefore ionic character increases.
Discussion
42.

Among CaH_(2),BeH_(2),BaH_(2), the order of ionic character is-

Answer»

`BeH_(2) lt CaH_(2) lt BaH_(2)`
`CaH_(2) lt BeH_(2) lt BaH_(2)`
`BeH_(2) lt HaH_(2) lt CaH_(2)`
`BaH_(2) lt BeH_(2) lt CaH_(2)`.

SOLUTION :Down a group METALLIC character INCREASES. Thus IONIC character of the metal hydride increases down the group. Hence least ionic compound is `BeH_(2)`.
Discussion
43.

Among CaH_2, BeH_2 , BaH_2, the order of ionic character is

Answer»

`BeH_2 LT CaH_2 lt BaH_2`
`CaH_2lt BeH_2 lt BaH_2`
`BeH_2 lt BaH_2 lt CaH_2`
`BaH_2 lt BeH_2 lt CaH_2`

SOLUTION :`BeH_2 lt CaH_2 lt BaH_2`
Discussion
44.

Among CaH_(2), BeH_(2), BaH_(2) the order of ionic character is :

Answer»

`BaH_(2) LT BeH_(2) lt CaH_(2)`
`BeH_(2) lt CaH_(2) lt BaH_(2)`
`BeH_(2) lt BaH_(2) lt CaH_(2)`
`CaH_(2) lt BeH_(2) lt BaH_(2)`

Solution :Because Be, Ca and Ba all three elements are of group two. For these three, as their electronegativity decreases, their atomic radius INCREASES and hence their ionic characteristic (ionic PROPERTIES) increases.
Discussion
45.

Among [BeCl_(2)]_(n), [BeH_(2)]_(n), B_(2)H_(6), Al_(2)(CH_(3))_(6), Al_(2)Cl_(6), Be(BH_(4))_(2), Al(BH_(4))_(3), the number of compounds having 3c-2e bonds are :

Answer»


ANSWER :5
Discussion
46.

Among Be ,B,Mg andA1 thesecondionizationpotential ismaximumfor

Answer»

B
Be
Mg
A1

Solution :The E.Csofthe ionsobtainedafterremovalof firstelectronfrom theseelementsare `: Be^(+)(1s^(2)2s^(1))`
`B^(+) (1s^(2)2s^(2) ) ,Mg^(+)(1s^(2)2s^(2)3s^(1))` and `A1^(+)(1s^(2) 2p^(2)2p^(6)3s^(2))`
Since`Mg^(+)` and `A1^(+)` belongto 3rdperiodwhile `Be^(+)` and `B^(+)`belongto 2ndperiod , thereforedue tobiggersize the`Delta_(i)H_(2) `of `Mg^(+)` and `A1^(+)` are lowerthan thoseof `Be^(+)`and `B^(+)`
Since `B^(+)`the nextelectronhas to beremovedfrom amorestabletally filled 2s-orbital whilein`Be^(+)` it has to belost fromthe lessstable hafl- filled2s- orbital adn furthermore LOSS of anelectronfrom `Be^(+)` givesmorestable `Be^(2+)` ion withinet gasconfigurationtherefore`Delta_(i)H_(2)` ofB ishigherthanthat of Be , i.e.,option( a) ISCORRECT.
Discussion
47.

Write the general valence shell electronic configuration of group 13 elements.

Answer»

SOLUTION :`B_2 O_3`
Discussion
48.

Between atoms of a molecule, there exists

Answer»

only attractive forces
only REPULSIVE forces
both attractive and repulsive forces
neither attractive nor repulsive forces

ANSWER :C
Discussion
49.

Among alpha-nitro acetic acid (I), alpha- fluoroacetic (II), alpha-bromoacetic acid (III) and alpha-cyanoacetic acid (IV), the order of increasing acid strength is

Answer»

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

Solution :Electron withdrawing EFFECT decreases as
`-NO_(2) gt -CN gt - F gt -Br`
The ACID strength is
`III lt II lt IV lt I`
Discussion
50.

Among all the isomers of pentane which has the lowest boiling point ?

Answer»

SOLUTION :neo-pentane
Discussion