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 pair of compounds in which both the metals are in the highest possibel oxidation state is

Answer»

`[Fe(CN)_(6)]^(4-), [Co(CN)_(6)]^(3-)`
`CrO_(2)Cl_(2).MnO_(4)^(-)`
`TiO_(2),MnO_(2)`
`[Co(CN)_(6)]^(3-),MnO_(3)`

Solution :`[Fe(CN)_(6)]^(4-)` i.e. `Fe^(2+)` (Highest for Fe is +3)
`[Co(CN)_(6)]^(3-)` i.e. `Co^(3+)` (Highest for Co)
`CrO_(2)Cl_(2)` i.e. `Cr(+6)` Highest for Cr
`MnO_(4)^(-)` i.e. `Mn(+7)` Highest for Mn
`TiO_(2)` i.e. `TI(+4)` Highest for Ti
`MnO_(2)` i.e `Mn(+4)`
`MnO_(3)` i.e. Mn(+6)
Discussion
2.

The pair of compounds in which both the compounds give positive test with Tollen's reagent is

Answer»

Glucose and Sucrose
Fructose and Sucrose
Acetophenone and Hexanal
Glucose and Fructose

Solution :Glucose being an aldose responds to TOLLEN's test while fructose, ALTHOUGH a KETOSE, undergoes rearrangement in presence of basic medium (PROVIDED by Tollen's reagent) to form glucose, which then responds to Tollen's test.
Discussion
3.

The pair of compounds in which both the compounds give positive test Tollens reagent is :

Answer»

GLUCOSE and sucrose
fructose and sucrose
acetophenone and hexanal
glucose and fructose

Solution :N//A
Discussion
4.

The pair of compounds having identical shapes for their molecules is

Answer»

`CH_(4), SF_(4)`
`BCl_(2), ClF_(3)`
`XeF_(2), ZnCl_(2)`
`SO_(2), CO_(2)`

Solution :`XeF_(2)` is linear in shape with 2 BOND PAIRS and 3 lone pairs of electrons.
`ZnCl_(2)` is also linear in shape with 2 bond pairs of electrons.
Discussion
5.

The pair of compound which cannot exist together in solution is

Answer»

`NaHCO_(3) and H_(2)O`
`Na_(2)CO_(3) and NaOH`
`NaHCO_(3) and NaOH`
`NaHCO_(3) and Na_(2)CO_(3)`

Solution :`NaHCO_(3)` being an acidicsalt will REACT with NaOH .
`NaHCO_(3) +NaOH to Na_(2)CO_(3) +H_(2)O`
Discussion
6.

The pair ofcompound whichcannotexist togatherinsolution is

Answer»

`NaHCO_(3) and NaOH`
`Na_(2)CO_(3) and NaHCO_(3)`
`Na_(2)CO_(3) and NaOH`
`NaHCO_(3)` and `NaCI`

Solution :`NaHCO_(3)` and `NaOH`cannot coexist insolution because `NaHCO_(3)` is an acid sqalt it REACTS withthe base `NaOH` as follows
`NaHCO_(3) + NaOH rarr Na_(2)CO_(3) +H_(2)O`
Discussion
7.

The pair of coloured ions among the following is

Answer»

`Cu^(+2),Zn^(+2)`
`SC^(+3),TI^(+2)`
`Ni^(2+),FE^(2+)`
`Ti^(4+),V^(+3)`

Answer :C
Discussion
8.

The pair of amphoteric oxides is…..

Answer»

VO, `Cr_(2)O_(3)`
`V_(2)O_(5), CrO_(3)`
`V_(2)O_(3), Cr_(2)O_(3)`
`VO_(2), Cr_(2)O_(3)`

Answer :D
Discussion
9.

The pair in which species have the same magnetic moment is:

Answer»

`"["Cr(H_(2)O)^(6)"]"^(2+)[CoCl_(4)]^(2-)`
`"["Cr(H_(2)O)^(6)"]"^(2+)[FE(H_(2)O_(6))]^(2+)`
`"["Mn(H_(2)O)^(6)"]"^(2+)[Cr(H_(2)O)_(6)]^(2+)`
`"["CoCl_(4)"]"^(2-)[Cr(H_(2)O)_(6)]^(2+)`

Solution :The magnetic moment depend UPON the no. of unparied ELECTRONS i.e, `sqrt(n(n+2))`
No. of UNPAIRED electrons in :
`Cr^(2+),[Ar]3d^(4)`( four)
`Co^(2+),[Ar]3d^(7)` (three)
`Fe^(2+),[Ar]3d^(6)` (four)
`Mn^(2+),[Ar]3d^(5)` (five)
Both `Cr^(2+)` and `Fe^(2+)` have four electrons.
Discussion
10.

The pair in which phosphorus atoms have a formal oxidation state of +3 is-

Answer»

Pyrophosphorus and pyrophosphoric acids
Orthophosphorous and pyrophosphorous acids
Pyrophosphorous and HYPOPHOSPHORIC acids
Orthophosphorous and hypophosphoric acids

Answer :B
Discussion
11.

The pair in which phosphorous atoms have a formal oxidation state of +3 is :

Answer»

Pyrophosphorous and pyrophosphoric acids.
Orthophosphorous and pyrophosphorous acids.
Pyrophosphorous and hypophosphoric acids.
Orthophosphorous and hypophosphoric acids.

Solution :Orthophosphorous ACID `to H_(3)PO_(3)` OS value=3
Orthophosphoric acid `to H_(3)PO_(4)` OS value= +5
Pyrophosphorous acid `to H_(4)P_(2)O_(5)` OS value = +3
Pyrophosphoric acid `to H_(4)P_(2)O_(7)` OS value = + 5
Hypophosphoric acid `to H_(4)P_(2)O_(6)`, OS value = +4
Discussion
12.

The pair in which both the species have the same hybridization is

Answer»

`[NI(NH_(3))_(6)]^(2+)` and `[MN(CN)_(6)]^(4-)`
`[Co(NH_(3))_(6)]^(2+)` and `[Mn(CN)_(6)]^(4-)`
`[Cr(NH_(3))_(6)]^(3+)` and `[Co(H_(2)O)_(6)]^(2+)`
`[Co(H_(2)O)_(6)]^(2+)` and `[Ni(NH_(3))_(6)]^(2+)`

Solution :
In both COMPLEX, Ni and Co have same hybridization.
Discussion
13.

The pair in which both the species are used in the preparation of antacid is

Answer»

`NaHCO_(3)` and `Mg(OH)_(2)`
`Na_(2)CO_(3)` and `Ca(HCO_(3))_(2)`
`Ca(HCO_(3))_(2)` and `Mg(OH)_(2)`
`Ca(OH)_(2)` And `NaHCO_(3)`

ANSWER :A
Discussion
14.

The pair in which both species have the same magnetic moment (spin only value) is

Answer»

`[CR(H_2O)_6]^(2+) , [CoCl_4]^(2-)`
`[Cr(H_2O)_6]^(2+) , [FE(H_2O)_6]^(2+)`
`[Mn(H_2O)_6]^(2+) , [Cr(H_2O)_6]^(2+)`
`[CoCl_4]^(2-) , [Fe(H_2O)_6]^(2+)`

Solution :`[Cr(H_2O)_6]^(2+)=Cr^(2+) , 3d^4` = FOUR unpaired electrons
`[Fe(H_2O)_6]^(2+)= Fe^(2+) =3d^6` , Four unpaired electrons.
Hence, both the species have same magnetic MOMENT.
Discussion
15.

The pair in which both species have same magnetic moment (spin only value ) is :

Answer»

`[Cr(H_(2)O)_(6)]^(2+), [CoCl_(4)]^(2-)`
`[Cr(H_(2)O)_(6)]^(2+), [Fe(H_(2)O)_(6)]^(2+)`
`[Mn(H_(2)O)_(6)]^(2+), [Cr(H_(2)O)_(6)]^(2+)`
`[CoCl_(4)]^(2-),[Fe(H_(2)O)_(6)]^(2+)`

Solution :`(b) [Cr(H_(2)O)_(6)]^(2+) IMPLIES d^(4),[FeH_(2)O)_(6)]^(2+) implies d^(6).` Both have 4 UNPAIRED ELECTRONS.
Discussion
16.

The pair in which both species have iron is

Answer»

Nitrogenase, CYTOCHROMES
Carboxypeptidase, haemoglobin
haemoglobin, nitrogenase
haemoglobin, cytochromes

Solution :The proteins, HEAMOGLOBIN and cytochromes CONTAIN iron as CENTRAL atom.
Discussion
17.

The pair having the same magnetic moment is_________. [ At. No. :Cr = 24, Mn =25 , Fe= 26 , Co =27]

Answer»

`[Cr(H_(2)O)_(6)]^(2+) and [CoCl_(4)]^(2-)`
`[Cr(H_(2)O)_(6)]^(2+) and [FE(H_(2)O)_(6)]^(2+)`
`[MN(H_(2)O)_(6)]^(2+) and [Cr(H_(2)O)_(6)]^(2+)`
`[CoCl_(4)]^(2-) and [Fe(H_(2)O)_(6)]^(2+)`

Answer :B
Discussion
18.

The pair having the same magnetic moment is [At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27]

Answer»

`[Cr(H_(2)O)_(6)]^(2+)`and `[CoCl_(4)]^(2-)`
`[Cr(H_(2)O)_(6)]^(2+) and [Fe(H_(2)O)_(6)]^(2+)`
`[Mn(H_(2)O)_(6)]^(2+) and [Cr(H_(2)O)_(6)]^(2+)`
`[CoCl_(4)]^(2-) and [Fe(H_(2)O)_(6)]^(2+)`

Solution :`[Ci(H_(2)O)_(6)]^(2+):Cr^(2+)=[Ar]3D^(4)4s^(0)`

Thus, `[Cr(H_(2)O)_(6)]^(2+)` and `[Fe(H_(2)O)_(6)]^(2+)` have the same number of UNPAIRED electrons and HENCE have same magnetic MOMENT.
Discussion
19.

The pair having the same magnetic moment is [at. No.Cr = 24, Mn = 25, Fe = 26 " and " Co = 27]

Answer»

`[CoCl_(4)]^(2-) and [FeH_(2)O)_(6)]^(2+)`
`[CR(H_(2)O)_(6)]^(2+) and [CoCl_(4)]^(2-)`
`[Cr(H_(2)O)_(6)]^(2+) and [Fe(H_(2)O)_(6)]^(2+)`
`[Mn(H_(2)O)_(6)]^(2+) and [Cr(H_(2)O)_(6)]^(2+)`

Answer :C
Discussion
20.

The pair [Co (NH_(3))_(4) CI_(2) ] Br_(2) and [Co(NH_(3))_(4)]CI_(2) will show

Answer»

LINKAGE isomerism.
Hydrate isomerism.
Ionisation isomerism.
Coordinate isomerism.

Answer :C
Discussion
21.

The pair of compounds in which both the metals are in the highest possible oxidation state is :

Answer»

`TiO_3 and MnO_2`
`[Fe(CN)_6]^(3-) and [CO(CN)_6]^(3-)`
`CrO_2Cl_2 and MnO_4^(-)`
`[Co(CN)_6]^(3),MnO_3`

ANSWER :C
Discussion
22.

The packing fraction of a simple cubic structure is pi//6. Prove.

Answer»

Solution :In a simple cube, the FACE edge length, a = 2R Number of atoms present in one unit cell = 1 Packing fraction =
`("volume OCCUPIED by atoms" )/("VOLUMEOF theunitcell" ) =(4/3pir^(3))/(a^(3))=4/3(pir^(3))/(2r)^(3)=(pi)/(6)`
Discussion
23.

The packing fraction of a simple cubic structure is (pi)/(6). Prove.

Answer»

Solution :In a SIMPLE CUBE, the face edge length, `a= 2r`. Number of atoms present in one unit cell `= 1`
Packing Fraction `=("VOLUME occupied by atoms")/("Volume of the unit cell")=((4)/(3) pi r^(3))/( a^(3) ) = (4)/( 3) (pir^(3) )/( (2r)^(3))= (pi)/( 6) =0.524`
Discussion
24.

The packing fraction in bcc arrangement is …………………………………..

Answer»

`52.31%`
`68%`
`100%`
`80%`

ANSWER :B
Discussion
25.

The packing fraction for a body-centred cubic is

Answer»

0.42
0.53
0.68
0.82

Solution :The p.f. for BODY CENTRED CUBE= 0.68
Discussion
26.

The packing fraction for a body centred cube

Answer»

0.74
0.76
0.68
0.86

Answer :C
Discussion
27.

The packing efficiencyof the two dimensional square unit cell shown below is

Answer»

0.5025
0.6802
0.7405
0.785

Solution :`4r=Lsqrt(2)` so `L=2sqrt(2)r`
Area of square unit cell `=(2sqrt(2)r)^(2)=8r^(2)`
Area of atoms present in one unit cell

`=PIR^(2)+r((pir^(2))/4)=2pir^(2)`
so packing EFFICIENCY `=(2pir^(2))/(8r^(2))XX100`
`=(pi)/4xx100=78.5%`
Discussion
28.

The packing efficiency in hcp and ccp structures is

Answer»

`68%`
`74%`
`78%`
`64%`

Answer :B
Discussion
29.

The packing efficiency in a simple cubic cell system of crystals is

Answer»

`52%`
`68%`
`74%`
`92%`

ANSWER :C
Discussion
30.

Calculate the efficiency of the packaing in case of face - centered cubic crystal .

Answer»

0.52
0.68
0.74
0.92

Answer :C
Discussion
31.

The packaging fration for a body centred cubic structure is

Answer»

0.42
0.53
0.68
0.82

Answer :C
Discussion
32.

The P-P-P bond angle in white phosphorus is:

Answer»

`120^@`
`90^@`
`60^@`
`109^@ 28' `

Answer :C
Discussion
33.

The P-P-P angle in P_(4) molecule is .......... Degree while S-S-S angle in S_(8) is ........... Degree.

Answer»

SOLUTION :`60^(@), 107^(@)`
Discussion
34.

The p-p bond energy is 'x' KJ/mole. Then the energy needed for the dissociation of 124 g of white phosphorous is

Answer»

X KJ
4x KJ
6x KJ
8x KJ

Answer :C
Discussion
35.

pH of humen blood is

Answer»

2 to 4
5 to 6.5
7.2 to 7.5
10 to 12.

Answer :C
Discussion
36.

The ozonolysis of gt C= C ltproduces

Answer»

`gt C=O+O =C lt`
`gt C= O =O`

Solution :Ozonolysis of C=C GIVES ALDEHYDES and ketones
Discussion
37.

The ozonolysis of (CH_(3))_(2) C = C (CH_(3))_(2) followed by treatment with zinc and water will give :

Answer»

ACETONE
ACETALDEHYDE and acetone
formaldehyde and acetone
ACETIC acid

Solution :
Discussion
38.

The ozonolysis of an olefin gives only propanone. The olefin is

Answer»

but-1-ene
but-2-ene
2, 3-dimethylbut-2-ene
PROPENE

SOLUTION :
Discussion
39.

The ozonolysis of an alkene X followed by hydrolysis gives ethanal and propanone . X is

Answer»

2 - BUTENE
2-Methyl- 2- butene
2- Methyl- 3- butene
2- Pentene

ANSWER :B
Discussion
40.

The ozonlysis of isobutene gives:

Answer»

`CH_3CHO`
`CH_3COCH_3 and HCHO`
`CH_2CH_2OH`
`CH_3OH`

ANSWER :B
Discussion
41.

The ozone layer forms naturally by

Answer»

the interaction of CFCs with OXYGEN
the interaction of UV radiation with oxygen
the interaction of IR radiation with oxygen
the interaction of oxygen and water vapour.

Solution :Ozone is formed in the ATMOSPHERE by the DECOMPOSITION of oxygen by ultra-violet radiation from the sun.
`O_(2(g)) + hv OVERSET(UV)to O_((g)) + O_((g))`
` O_((g)) + O_(2(g))overset(UV) to O_(3(g))`
Discussion
42.

The oxygen of H_2O_2 used for oxidation is bound by :

Answer»

ELECTROVALENT bond
Co-ordinate bond
Covalent bond
None of the above

Answer :C
Discussion
43.

The ozonelayerformsnaturallyby

Answer»

theinteraction of CFCwith OXYGEN
the interaction of UVradiationwith oxygen
theintercationof IRRADIATION withoxygen
the interaction of OXYGENAND watervapour.

Answer :B
Discussion
44.

The oxygen present today in atmosphere :

Answer»

Is a PLANT product
Came from ozone
Was PRESENT in the beginning
Produced by CARBON dioxide

Answer :A
Discussion
45.

The oxygen and hydrogen formed during electrolysis of water are in the weight ratio of

Answer»

`8:1`
`16:1`
`1:4`
`2:1`

ANSWER :A
Discussion
46.

The oxygen atom in a either molecule is

Answer»

Very active
Replaceable
Comparatively inert
Active

Answer :C
Discussion
47.

The oxyacid of SO_2 is :

Answer»

`H_2SO_3`
`H_2SO_4`
`H_2S_2O_8`
None

Answer :A
Discussion
48.

The oxyacid which acts both as oxidising and reducing agent is :

Answer»

`H_2SO_4`
`H_3PO_4`
`HNO_2`
`HCIO_4`

ANSWER :C
Discussion
49.

The oxyacid of phosphorous in which phosphorus has lowest oxidation state is.

Answer»

<P>Hypophosphorus ACID
Orthophosphoric acid
Pyrophosphoric acid
Metaphosphoric acid.

Solution :`O.S` of `P` is HYPOPHOSPHOROUS acid, `H_3 PO_2 = + 1`.
Discussion
50.

The oxoanion which contains all equivalent M-O bondis: (I)Cr_(4)^(2-)(II)MnO_(4)^(-)(III)Cr_(2)O_(7)^(2-)

Answer»

III only
I, II, III
I, II
I only

Answer :C
Discussion