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 oxidation number of carbon in CH_(2)Cl_(2) is

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0
`+2`
`-2`
`+4`

ANSWER :A
2.

The oxidation number of C in CH_(2)O is

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`-2`
`+2`
0
`+4`

ANSWER :C
3.

The oxidation number of B is Na_2B_4O_7 is +3 . Is the statement correct ?

Answer»


ANSWER :CORRECT
4.

The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following is incorrect in this respect ?

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The OXIDATION number of hydrogen is always +1.
The algebraic sum of the all the oxidation numbers in a COMPOUND is zero.
An ELEMENT in the free or the uncombined state bears oxidation number zero.
In all its compounds, the oxidation number of fluorine is -1.

Solution :Because oxidation number in hydride compound will be (-1) and oxidation number of `H_(2)` will be zero.
5.

The oxidation number of an element in compound is evaluated on the basic of certain rules which of the following rules is not correct in this respect ?

Answer»

Solution :O.N of H is always +1 is a WRONG statement SINCE it is +1 hydrogen HALIDES -1 in hydrdes and zero in the `H_(2)` molecule
6.

The oxidation number of an element in a compound decides its nature to acts as oxidant or reductant. Oxidation number is defined and the residual charge which an atom has or appears to have in a molecule whenal other atoms are removed from the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation number are based on the periodic property-electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can evenn be fractional. Oxidation number of Y in YBa_(2)Cu_(2)O_(7) is +3, then oxidation number of Cu is:

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`+7//3`
`+5//3`
`+2`
`+1`

SOLUTION :`3 +2xx2+3xxa+7xx(-2)=0`
`:.a=+7//3`
7.

The oxidation number of an element in a compound decides its nature to acts as oxidant or reductant. Oxidation number is defined and the residual charge which an atom has or appears to have in a molecule whenal other atoms are removed from the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation number are based on the periodic property-electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can evenn be fractional. Oxidation number of oxygen in K_(2)O, K_(2)O_(2),KO_(2),KO_(3) are in the order:

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`KO_3ltKO_3 ltK_2O_2 LT K_2O`
`KO_2 lt KO_3lt K_2O lt K_2O_2`
`K_2O lt K_2O_2lt KO_2 lt KO_3`
`KO_3 lt K_2Olt KO_2 lt K_2O_2`

SOLUTION :OXIDATION no . if O is –2,–2,–1/2 and –1/3 RESPECTIVELY .
8.

The oxidation number of an element in a compound decides its nature to act as oxidant or reductant. Oxidation number is defined as the residual charge which an atom has or appears to have in a molecule when all other atoms are removed from the molecule as ions. Oxidation number is frequently used interchangebly with oxidation state. The stock notations of oxidation numbers are based on the periodic property - electronegativity. An atom in a molecule can be assigned positive, negative or zero oxidation number by considering its environment. In few cases, oxidation number can even be fractional. Maximum oxidation state shown by Os, Ru and Xe in their compounds is :

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`+8`
`+6`
`+10`
`+4`

SOLUTION :In `OsO_4 and RuO_4 and XeO_6^(4-)`
9.

The oxidation nuber fo an atom in a given species (molecule ion, or free atom ) is the .

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formal charge of the atom
valency of the atom
actual charge of atom
actual charge FO the atom if the atom exists as a mono-atomic ION or the HYPOTHETICAL charge assinged to the atom in the species by simple rules

Solution :In the case of a single-atom ion, OXIDATION number corresponds to the actual charge on the ion. In molecules and `//` or polyatomic ions, oxidation number corresponds to the mumber of charges an atom have in a molecule (or polyatomic ion ) if electrons were transferred completely to more electronegative atom.
10.

The oxidant which is used as an antiseptic is ______

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`KBrO_(3)`
`KMnO_(4)`
`CrO_(3)`
`KNO_(3)`

Answer :B
11.

The oxalic acid is added in test tube and this reaction Fe_((aq))^(3+) + SCN_((aq))^(-) hArr [Fe(SCN)]_((g))^(2+)equilibrium, steered and remain the solution than what will the change in colour of solution ?

Answer»

Solution :The intensity of red colour of solution is DECREASE because the oxalate ion, `C_2O_4^(2-)`of oxalic acid is reacts with `Fe^(3+)` ion and decrease the concentration of `Fe^(3+)`. The reaction should be forward an red colour containing `[Fe(SCN)]_((AQ))^(2+)`is decrease, the red colour decreases.
12.

which is not a moneran

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`s-s`
p-p
`s-p`
both (a) and (B)

ANSWER :B
13.

The overall effect of respiration is similar to that of "_______________".

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photosynthesis
corrosion
combustion
bleaching

Answer :C
14.

The outermost electronic configuration of the element with highest value of electron affinity is ………

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`NS^(2)np^(3)`
`ns^(2)np^(5)`
`ns^(2)np^(4)`
`ns^(2)np^(6)`

ANSWER :b
15.

The outermost electronic configuration of most electropositive element is

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`NS^(s)`
`ns^(2)NP^(2)`
`ns^(2)` `np^(3)`
`ns^(2)np^(5)`

Answer :A
16.

The outermost electronic ccnfiguration of manganese (at.no.=25)is…………….

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SOLUTION :`3D^(5)4S^(2)`
17.

The outer shell configuration of the most electronegative element is

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`NS^(2) np^(3)`
`ns^(2) np^(4)`
`ns^(2) np^(5)`
`ns^(2) np^(6)`

Solution :(`ns^(2) np^(5)` REPRESENTS halogen)
18.

The outer electronic configuration of someelements are given below : (i)3s^(2) 3p^(3) (ii) 3d^(5) 4s^(1) (iii) 3s^(2) 3p^(6) 4s^(2) (iv) 5d^(1) 6s^(2) (v) 4f^(1) 5d^(1) 6s^(2) Stateof which blockof theperiodic tableeach of thetheseelementsbelongs

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<P>

SOLUTION :(i) p, (ii) d, (III) s, (iv) d, (v) F
19.

The outer electronic configuration of chromium atom is 3d^4 4s^2

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ANSWER :F
20.

The outerelectronicconfigurationof Gd (AtomicNo. 64) is

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`4f^(3) 5D^(3)6s^(2)`
`4f^(8)5d^(0)6s^(2)`
` 4f^(4) 5d^(4) 6s^(2)`
` 4f^(7)5d^(1)6s^(2)`

SOLUTION :NA
21.

The outer electronic configuration of alkaline earth metals is

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`ns^2`
`ns^1`
`np^6`
`nd^10`

SOLUTION :`ns^2`
22.

Theouterelectronicconfigurationof alkalineearthmetalsis ______.

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`NS^(2)`
` ns^(1)`
` NP ^(6)`
` nd^(10)`

Answer :A
23.

the outcome of internal liquid of gel on shaking is called :

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SYNERESIS
IMBIBITION
thixotropy
precipitation

Answer :C
24.

The outcome material from Solvay process is

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`Na_(2)CO_(3).10H_(2)O`
`Na_(2)CO_(3)H_(2)O`
`NaHCO_(3)`
`Na_(2)CO`

Solution :The actual manin COMPOUND obtained from sovay process is `NaHCO_(3)` which on ignition gives `Na_(2)CO_(3)`
25.

The ortho/para-directing group among the following is

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COOH
CN
`COCH_3`
`NHCOCH_3`

SOLUTION :DUE to the presence of a LONE PAIR of ELECTRONS on the N-atom, `-NHCOCH_3` is o, p-directing.
26.

The origin of charge on colloidal solution is

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Frictional rubbing
Electron capture during bredig's arc method
Selective adsorption of ION on their surface
It is due to ADDITION of protective colloids

Solution :(a), (b) and (C) are correct, (d) is incorrect, LYOPHOBIC sols are stabilized by addition of protective colloid.
27.

The orientation of hybrid orbitals is tetrahderal in

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`NH_(3)`
`SCl_(4)`
`PCl_(5)`
`XeF_(4)`

Answer :A
28.

The orientation of electron pairsand the shape of molecule are different in

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`BeCl_(2)`
`H_(2)O`
`BCl_(3)`
`"CC"l_(4)`

ANSWER :B
29.

The organic solvent used in differential extraction is…….in water

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ANSWER :IMMISCIBLE
30.

The ore having two different metal atoms is

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haeatite
galena
magnetite
COPPER pyrites.

Solution :Copper pyrites contains CU and Fe as CUFES.
31.

The order of the oxidation state of the phosphorus atom in H_(3)PO_(2),H_(3)PO_(4),H_(3)PO_(3),andH_(4)P_(2)O_(6) is

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<P>`H_(3)PO_(4)gtH_(3)PO_(2)gtH_(3)PO_(3)gtH_(4)O_(2)O_(6)`
`H_(3)PO_(2)gtH_(3)PO_(3)gtH_(4)P_(2)O_(6)gtH_(3)PO_(4)`
`H_(3)PO_(3)gtH_(3)PO_(2)gtH_(3)PO_(4)gtH_(4)P_(2)O_(6)`
`H_(3)PO_(4)gtH_(4)P_(2)O_(6)gtH__(3)PO_(3)gtH_(3)PO_(2)`

SOLUTION :`H_(3)overset(+5)PO_(4)gtH_(4)overset(+4)P_(2)O_(6)gtH_(3)overset(+3)PO_(3)gtH_(3)overset(+1)PO_(2)`
32.

The order of stabiliyt of the following carboncation is…..

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`III gt II gt I`
`II gt III gt I`
`I gt II gt III`
`III gt I gt II`

Solution :
The benzylic cation is EXTRA stabilizaed by resonance.
(I) `CH_(2) = CH- overset(OPLUS)(C )H_(2) harr overset(oplus)(C )H_(2)-CH= CH_(2)` alylic cation is stable by resonance
(II) `CH_(3)-CH_(2) - overset(+)(C )H_(2)` It is least stable as resonance is not possible in it
33.

The order of stability of the following tautomeric compounds is : CH_(2)=overset(OH)overset(|)C-CH_(2)-underset(I)overset(O)overset(||)C-CH_(3) hArr Ch_(3)-overset(O)overset(||)C-CH_(2)-underset(II)overset(O)overset(||)C-CH_(3) hArr CH_(3)-overset(OH)overset(|)C=CH-underset(III)overset(O)overset(||)C-CH_(3)

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`III GT II gt I`
`II gt I gt III`
`II gt III gt I`
`I gt II gt III`

ANSWER :A
34.

The order of stability of the following tautomeric compound is CH_(2) = oversetoverset(C ) - CH_(2) - oversetoverset(O)("||")C - CH_(3) hArr CH_(3) - oversetoverset(O)("||")C-CH_(2) - oversetoverset(O)("||")C - CH_(3)hArr CH_(3) - oversetoverset(OH)(|)C = CH - oversetoverset(O)("||")C - CH_(3)

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III `gt` II `gt` I
II `gt` I `gt` III
II `gt` III `gt` I
O `gt` II `gt` III

SOLUTION :`CH_(3) - oversetoverset(OH)(|)C = CH - oversetoverset(O)(|)C-CH_(3)`
This is more stable because of INTRA molecular hydrogen bonding and conjugation of L.P and `pi` - electrons and between `pi`- electrons
35.

The order of stability of thefollowing cabocations CH_(2) = CH- CH_(2)^(+), CH_(3)CH_(2)CH_(2)^(+),C_(6)H_(5)CH_(2)^(+) is

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`III gt II gt I`
`II gt III gt I`
`I gt II gt III`
`III gt I gt II`

SOLUTION :Order of STABILITY of CATIONS
Benzyl `gt` Allyl `gt I^(@)`
36.

The order of stability of the following carbocations :

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`III GT II gt I`
`II gt III gt I`
`I gt II gt III`
`III gt I gt II`

ANSWER :D
37.

The order of stability of the following carbanions is

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<P>`I gt II gt III gt IV`
`IVgt II gt III gt I`
`I gt III gt II gt IV`
`I gt II gt IV gt III`

Solution :Groups with -I and -M EFFECTS at o- & p positions increase the STABILITY of carbanion
38.

The order of stability of the carbonium ion is-

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`1^(@) gt 2^(@) gt 3^(@)`
`2^(@)GT3^(@)gt1^(@)`
`2^(@)gt1^(@)gt3^(@)`
`3^(@)gt2^(@)gt1^(@)`.

SOLUTION :`3^(@)gt2^(@)gt1^(@)`.
39.

The order of stability of dihalides of Ge, Pb and Sn is

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`GeCl_(2) GT SnCl_(2) gt PbCl_(2)`
`SnCl_(2) gt GeCl_(2)gt PbCl_(2)`
`PbCl_(2)gtGeCl_(2)gt SnCl_(2)`
`PbCl_(2)gtSnCl_(2)gt GeCl_(2)`

Answer :D
40.

The order of solubility of lithium halides in nonpolar solvents follows the order :

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`LiI gt LIBR gt LICL gt LIF`
`LiF gt Lil gt LiBr gt LiCl `
`LiCl gt LiF gt LiI gt LiBr`
`LiBr gt LiClgt LiF gt LiI`

SOLUTION :As the size of the anion increases from `F^(-) ` to `I^(-),` the covalent character increases and hence the solubility in non-polar solvent increases.
`LiI gt liBr gt LiCl gt LiF`
41.

The orderof screening effect of electrons of s, p, d and f orbitals of a givenshell of anatom on itsouter shellelectrons is :

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`s GTP gt d gt f`
`f gt d gt p gt s`
` p lt d lt s gt f`
`f gt p gt s gt d`

SOLUTION :Within the SAMESHELL SCREENINGEFFECT decreases I N the order`: s gt p gt d gt f,` i.e.,option ( a) iscorrect.
42.

The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is :

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`s GT p gt dgt f`
`f gt d gt p gt s`
`p lt d lt s gt f`
`f gt p gt s gt d`

Solution :ORDER of SCREENING effect DECREASES `s gt p gt d gt f` for the same shell.
43.

The order of screening effect of electron of s, p, d and f orbits of a given shell of an atom on its outershell electrons is:

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<P>s>p>d>f
f>d>p>s
p>d>s>f
f>p>s>d

Answer :a
44.

The order of relative ease of formation of various ions is

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`F^(-)GTO^(-2)gtN^(-3)`
`N^(-3)gtO^(-2)GTF^(-)`
`O^(-2)gtN^(-3)gtF^(-)`
`F^(-)gtN^(-3)gtO^(-2)`

ANSWER :A
45.

The order of reactivity of the following compounds in electrophilic monochlorinatic the most favorable position is

Answer»

I lt II lt IV lt III
III lt IV lt I lt II
IV lt III lt II lt I
III lt II lt IV lt I

Answer :B
46.

The order of reactivity of the following alkyl halides for a S_(N)2 reaction is

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`R -F GT R-Cl gt R-Br gt R-I`
`R-F gt R-Br gt R-Cl gt R-I`
`R-Cl gt R-Br gt R-F gt R-I `
`R-I gt RBR gt R-Cl gt R-F`

ANSWER :D
47.

The order of reactivity of the followin alcohols towards conc. HCl is

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

Solution :When alcohols REACT with HX it is the `R-OH` bond which breaks. The more stable the carbocation formed, more is the REACTIVITY of the alcohols. The carbocations formed in case of these alcohols (I, II, III and IV) are
`F -CH_(2)-underset((I))(overset(+)(CH))- CH_(3)`
`underset((II)) (F-CH_(2)-CH_(2)-overset(+)(CH)-CH_(3))`
`underset((III))(CH_(3)-overset(+)(CH)-CH_(3))`
`underset((IV))(C_(6)H_(5)-overset(+)(CH_(2))`
The relative stabilities of carbocations and hence the order of reactivity of corresponding alcohols is `IV gt III gt II gt I`. The greater stability of carbocation (IV) is due to resonance.
Carbocation II is a `2^(@)` carbocation and as such is quite stable. The lesser stability of carbocation I and II is due to the `-I` effect of F. ELECTRON with-drawing effect of F is more in case of carbocation I as in this case F is nearer to the carbon CARRYING +ve charge.
48.

The order of reactivity of halogen acids for reaction with C_(2)H_(5)OH is

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`HCl gt HBr gt HI`
`HI gt HBr gt HCl`
`HBr gt HI gt HCl`
`HBr gt HCl gt HI`

Solution :The reactivity of HX DEPENDS UPON the ease of dissolved of HX, i.e., bond dissociation ENERGY of HX which increases as`Hi lt HBr lt HCl`.
49.

The order of reactivity of differeent hydrogen halides (HCI,HI,HBr)is ……………….

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SOLUTION :`HI GT HBrgt HCI`
50.

The order of reactivity of alkenes,underset"I"((CH_3)_2C=CH_2). underset"II"(CH_3CH=CH_2), underset"III"(CH_2=CH_2) when subjected to acid-catalysed hydration is

Answer»

I gt II gt III
I gt III gt II
III gt II gt I
II gt I gt III

Solution :REACTIVITY increases as the electrondensity in the DOUBLE BOND increases, i.e., I gt II gt III