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.

Hydro-metallurgical process of extraction of metals is based on

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COMPLEX formation
Hydrolysis
Dehydration
Dehydrogenation

Solution :For EXAMPLE, `Ag_(2)S`is converted into `NA[Ag(CN)_(2)]` when ZN is ADDED. Ag is displaced.
Discussion
2.

Hydroazobenzene on treatement with H_(2)SO_(4) forms

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Azobenzene
Azobenzene sulphonic acid
BENZIDINE
None of the above.

Solution :
This is KNOWN as Benzidine REARRANGEMENT.
Discussion
3.

Hydro chloric acid is used in manufacture of {:(A,"Chlorine",B.NH_(4)Cl),(C.,C_(6)H_(12)O_(6),D."Bromine"),(E,C_(12)H_(22)O_(11),):}

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All are correct
A,B,D
A,B,E
A,B,C

Answer :D
Discussion
4.

Hydro chloric acid decomposes salts of which compounds

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STRONG acid
Strong bases
Weak acid
2 & 3

Answer :D
Discussion
5.

Hydrides of oxygen and sulphur differ in physical state due to

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presence of intermolecular HYDROGEN bonding in the hydrides of OXYGEN
more electronegativity of oxygen
stronger S- S bonds compared to O-O bonds
repulsion of lone pair of electrons on oxygen ATOMS

SOLUTION :`H_2O` has intermolecular H-bonding , whereas `H_2S` does not.
Discussion
6.

Hydrides of group 16 are weakly acidic in nature. The correct order of acidity is

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`H_2O gt H_(2)S gt H_(2)Se gt H_(2)TE`
`H_(2)Te gt H_(2)O gt H_(2)S gt H_(2)Se`
`H_2Te gt H_(2)Se gt H_(2)S gt H_(2)O`
`H_(2)Te gt H_(2)Se gt H_(2)O gt H_(2)S`

SOLUTION :Owing to increase in the size of central atom, ACIDITY of hydrides increases on moving down the group, The distance between the central atom and hydrogen increases as we move from 'O' to Te', which favors the release of hydrogen as proton.
Discussion
7.

......hydrides are non inflammable.

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`NH_(3)`
`PH_(3)`
`AsH_(3)`
`SbH_(3)`

ANSWER :A
Discussion
8.

Hydride of sulfur is more stable than that of oxygen. The given statement is correct or not?

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SOLUTION :HYDRIDE of sulfur is LESS stable than of oxygen
Discussion
9.

Hydride of sulfur is more stable than that of oxygen.

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SOLUTION :HYDRIDE of sulfur is less STABLE than of OXYGEN
Discussion
10.

Hydride of ether is

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aldehyde
alcohol
ketone
carboxylic acid

Answer :B
Discussion
11.

In an aqueous solution, hydrogen (H_2) will not reduce:

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HEATED CUPRIC oxide
Heated FERRIC oxide
Heated STANNIC oxide
Heated ALUMINIUM oxide

Answer :D
Discussion
12.

Hydrogen cannot reduce :

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HEATED CUPRIC oxide
Heated FERRIC oxide
Heated STANNIC oxide
Heated ALUMINIUM oxide

Answer :D
Discussion
13.

Hydrazones of aldehydes and ketones are not prepared in highly acidic medium. Explain.

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Solution :In highly ACIDIC medium, the `NH_(2)` group of hydrazine gets protonated.

Due to strong -I-effect of the `OVERSET(+)(N)H_(3)` group, the lone pair of ELECTRONS on the `-NH_(2)` group of protonated hydrazine is not available for nucleophilic attack on the C=O group and hence hydrazone formation does not occur.
Discussion
14.

Hydrazoic acid is

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`H_(2)NOH`
`N_(3)H`
`NH_(2)NH_(2)`
`C_(6)H_(5)-N=N-OH`

ANSWER :B
Discussion
15.

Hydrazobenzene reacts with cold conc. HCl to form :

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Azobenzene
Benzidine
Azoxybenzene
Azobenzene-4-sulphonic acid.

Answer :B
Discussion
16.

Hydrazobenzene can be obtained by reducing nitrobenzene with

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`SN + HCI`
`Fe+HCI`
`ZN + NH_(4)CL`
`Zn+NaOH`

Solution :
Discussion
17.

Hydrazinium ion is chemically

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`N_2H_3^-`
`N_2H_3^+`
`N_2H_5^+`
`N_2H_5^-`

ANSWER :C
Discussion
18.

Hydrazine is formed when cyanamide gives

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`N_2`
`NH_3`
`CaOCl_2`
`N_2O`

ANSWER :B
Discussion
19.

Hydrazine as a drug is also used in the treatment of

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TUBERCULOSIS
CHOLERA
TYPHOID
MALARIA

SOLUTION :Hydrazineused to trate T-B
Discussion
20.

Hydrazine as a drug is also used in the treatment of :

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TYPHOID
CHOLERA
MALARIA
TUBERCULOSIS

ANSWER :D
Discussion
21.

Hydration reaction of alkene is catalyzed by dilute acid. Selection of acid is important . Conjugate base of the acid should not interfere in the reaction. There are other means by which alkenes can be converted to alcohols. Oxymercuration demercuration gives Markovnikoff's alcohols while hydroboration oxidation give Anti Markovnikoff's alcohol. When subjected to acid catalyzed hydration which of the following alkene will give rearrangement alcohol as the predominant product?

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ANSWER :A
Discussion
22.

Hydraulic classifier is __________.

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CONICAL in shape
triangular in shape
cyclindrical in shape
rectangular in shape

Answer :A
Discussion
23.

Hydration of which one of the following yields a ketone

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Propyne
Ethene
Propene
Ethyne

Solution :
Discussion
24.

Hydration of which of the given compounds leads to the formation of 2-methylpropan-2-ol?

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2-methylpropane
1-bromopropane
2-methylpropene
Propene

Answer :C
Discussion
25.

Hydration of which of the following yields ketone ?

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Propyne
Ethene
Propene
Ethyne

Solution :`CH_(3) - C -= CH overset(H_(2)O)rarr CH_(3) - underset(OH)underset(|)(C) = CH_(2) harr CH_(3) - underset(O)underset(||)(C) - CH_(3)`
Discussion
26.

Hydration of an alkyne to get an aldehyde or ketone is possible in the presence of Hg^(2+) ions and H_2SO_4

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ANSWER :T
Discussion
27.

Hydration of alkynes catalysed by H_2 SO_4 in presence of Hg^(+2) give carbonyl compounds. Which of the following is correct.

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Acetylene gives acetaldehyde
Terminal alkynes (R-C`-=`CH) GIVE aldehydes
Non-terminal alkynes give ketones
All the above

Solution :`(A) `Acetylene givesacetaldehyde`HC -= CHunderset(HG^(+))overset(H_2 SO_4) to [H_2 C= CH -OH] toCH_3 -CHO`
`(C ) ` Terminalalkynes`R-C-= CH underset(H_2SO_4) overset(Hg^(+2))to[R -underset(OH) underset(|)C=CH_2] to R- underset(O) underset(||)C-CH_3`
(D) Non-terminalalkynesgive ketons
`R-C-=C -Runderset(H_2SO_4)overset(Hg^(+2))to[R-CH= underset(OH) underset(|)C-R ] toR- CH_2 - underset(O) underset(||)C-R ` ( ketones)
Discussion
28.

Hydration of 3-phenyl-1.butene with dil. H_(2)SO_(4), is not a satisfactory method for preparing 3-phenyl-2-butanol because 2-phenyl-2-butanol is obtained instead. Explain.

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Solution :Hydration form `2^(@)` carbocation which undergoes a HYDRIDE SHIFT to STABLE `3^(@)` benzylic carbocation.
Discussion
29.

Hydration of alkenes (except ethylene ) in presence of acid produce-

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`1^@` ALCOHOLS
`2^@ or 3^@` alcohols
MIXTURE of `1^@ and 2^@` alcohols
mixture of `2^@ and 3^@`alcohols

Answer :B
Discussion
30.

Hydration of alkene produces

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`1^@`ALCOHOLS or `2^@` alcohols
`1^@`alcohols or `3^@` alcohols
1, `2^@` and `3^@` alcohols
`1^@` or `2^@` or `3^@` alcohols

ANSWER :D
Discussion
31.

Hydration of 3- phenylbut -1- ene with dil , H_2SO_4 mainly gives

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3- Phenlbutan -1-OL
3- Phenylbutan -2-ol
2- Phenylbutan -1- ol
2- Phenybutan -2-ol

ANSWER :D
Discussion
32.

Hydrogenation of 2-butyne yields :

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Butanane
Butanol
BUTANONE
3- butanone

ANSWER :C
Discussion
33.

Hydration energy of Mg^(+2) is more than

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`Al^(+3)`
`Na^(+)`
`Be^(+2)`
`MG^(3+)`

Answer :B
Discussion
34.

Hydrated sulphate of a divalent metal of atomic weight 65.4 loses 43.85% of its weight on dehydration. Find the number of molecules of water of crystallisation in the formula of hydrated salt.

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SOLUTION :Formulae of divaelnt hydrated metal SULPHATE will be
`MSO_(4)xH_(2)O`
MOLECULAR MASS of salt`=65.4+96+18x`
`=(161.4+18x)`
% of water`=(18x)/(161.4+18x)xx100=43.85`
On solvingx=7
`therefore` Molecular FORMULA of hydrated salt `=MSO_(4).7H_(2)O`
Discussion
35.

Hydration :

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TAKES place through CARBOCATION
is a reversible PROCESS
FOLLOWS electrophilic ADDITION
follows all of the above

Answer :D
Discussion
36.

Hydrated salts are solution of …………………in…………………..

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SOLUTION :LIQUID in SOLID
Discussion
37.

Hydrated magnesium chloride, on strong heating gives finally

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`MgCl_(2)`
`MgO`
`MG(OH)CL`
`Mg_(2)OCl_(2)`

Answer :B
Discussion
38.

Hydrated oxalic acid contains :

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5 WATER MOLECULES
1 water molecules
2 water molecules
4 water molecules

Answer :C
Discussion
39.

Hydrated Cu^(2+) ions absorb light of ..... colour and transmit light of ....... colour

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RED and BLUE
GREEN and PURPLE
purple and red
blue and red

Solution :red and blue
Discussion
40.

Hydrate isomerism.

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Solution :The phenomenon of isomerism in the coordination compounds arising due to the EXCHANGES of `H_(2)O` molecules inside the coordination sphere and OUTER sphere of the COMPLEX is KNOWN as hydrate isomerism.
Discussion
41.

Hybridization, shape and magnetic moment of K_(3)[Co(CO_(3))_(3)] is

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`d^(2)sp^(3)`, octahedral, 4.9 B.M.
`sp^(3)d^(2)`, octahedral, 4.9 B. M.
`DSP^(2)`, square planar, 4.9 B.M.
`sp^(3)`, tetrahedral, 4.9 B.M.

Solution :In each case, magnetic moment is 4.9 B.M. which shows that each of them has 4 unpaired electrons. Further, in the given complex, oxidation state of cobalt is +3. Hence, it will have `3d^(6)` configuration, i.e.,

Thus, hybridisation possible is `sp^(3)d^(2)`, i.e., outer ORBITAL (HIGH spin) octahedral complex.
Discussion
42.

Hybridized state of bromine in bromine penta fluoride is :

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`sp^(3)d`
`d^(2)sp^(3)`
`dsp^(3)`
`sp^(3)d^(2)`.

ANSWER :B
Discussion
43.

Hybridization of .S. in SF_6 is:

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`sp^3 d^3`
`sp^3 d^2`
`d^2 sp^3`
`sp^3 d`

ANSWER :B
Discussion
44.

Hybridization is a concept of mixing or merging of orbitals of same atom with slight differences in energies to redistribute their energies and give new orbitals of equivalent energy called 'Hybrid Orbitals'. Hybridisation is a hypothetical concept and never actually exists. One should not be confused by a common misconception that hybridization is responsible for particular geometry. Geometry of a molecule is decided by energy factor not by hybridization. It is the orbital (which may be half filled, completely filled or empty) that undergoes hybridization and not the electron. The bond angles in hybridised orbitals are influenced by presence of lone pair, presence ofmultiple bonds, presence of one electron and electronegativity of atom. An increase in s-character of hybridised orbitals results in decrease in size of orbitals. This results in decrease in bond length and increase in energy. Which of the following statement is not true?

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<P>`O-F` bond length in `OF_(2)` is less than `O-F` bond length in `O_(2)F_(2)`
In `HCO_(3)^(-)`, all `C-O` bond lengths are not identical
In diborane, two different `B-H` bond lengths are observed ALTHOUGH the hybridization of both boron atoms are same.
In hydrazine, the `N-N` bond length is larger than NORMAL `N-N` bond length.

Solution :(a) `{:(""O-H),("|"),(H-O-underset(sp^(2))B-underset(sp^(2))O-H):}`
(b) `BF_(4)^(-)` is tetrahedral with B.A. = `109^(@)28'`
`P_(4)rarrsp^(3),SO_(3)rarrsp^(2)`
Discussion
45.

Hybridization of Fe in [Fe(H_2O)_5NO]SO_4(brown ring complex) is

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`dsp^2`
`sp^3d`
`sp^3d^2`
`d^2sp^3`

ANSWER :C
Discussion
46.

The percentage of p-character of the hybrid orbitals in graphite and diamond are respectively.

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`SP^(3), sp^(3)`
`sp^(3), sp^(2)`
`sp^(2), sp^(2)`
`sp^(2), sp^(3)`

ANSWER :D
Discussion
47.

Hybridisedstateof N-atomsin trimethyl amineis

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`SP^(3)`
`sp^(2)`
`sp`
`sp^(3) -d`

ANSWER :A
Discussion
48.

Hybridised states of carbon in graphite and diamond are respectively

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`SP^(3),sp^(3)`
`sp^(3),sp^(2)`
`sp^(2),sp^(2)`
`sp^(2),sp^(3)`

ANSWER :D
Discussion
49.

Hybridisation of Xe atom in XeF_6 is:

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`sp^3`
`sp^3d`
`sp^3d^2`
`sp^3d^3`

ANSWER :D
Discussion
50.

Hybridisation of Xenon in Xenon tetrafluoride is ____.

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SOLUTION :`sp^3d^2`
Discussion