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.

Fe(OH)_3 and Cr(OH)_3 precipitates are separated by

Answer»

 `Ag.NH_3`
HCI
`NaOH//H_2O_2`
 `H_2SO_4`

ANSWER :C
Discussion
2.

Fe(OH)_(3) and Cr(OH)_(3) ppt are sepurated by

Answer»

AQ `NH_(3)`
`HCI`
`NaOH//H_(2)O_(2)`
`H_(2)SO_(4)`

ANSWER :c
Discussion
3.

Fe(OH)_3 and Cr(OH)_3 ppts. are separated by :

Answer»

AQUEOUS `NH_3`
`H_2SO_4`
`NaOH//H_2O_2`
`HCL`

ANSWER :C
Discussion
4.

Fe(OH)_(2) is precipitated from Fe(II)solutions as a while solid turns dark green and then brown due to the formation of:

Answer»

`Fe(OH)_(2) and Fe(OH)_(3)`
Only `Fe(OH)_(3)`
`Fe_(2)O_(3)*(H_(2)O)_(N)`
`Fe_(2)O_(3)*2H_(2)O`

ANSWER :C
Discussion
5.

Fe(OH)_(3) and Al(OH)_(3) ppt. can be separated by (a) ___when (b)___ becomes soluble due totheformation of (c ) ___ and (d) ____ remaininsoluble.

Answer»


Answer :`NAOH`,b. `AI(OH)_(2)` C. `NaAIO_(2)` d.`FE(OH)_(3)`
Discussion
6.

FeO crystal has a simplecubic structureand each edge of the unitcell is 5 Å . Takingdensity of the oxide as 4g/cc, the number ofFe^(2+) and O^(2-)ionspresentin eachunitcell are

Answer»

`4 Fe^(2+) and 4O^(2-)`
`6FE^(2+) and 6O^(2-)`
`2Fe^(2+) and 2O^(2-)`
`1Fe^(2+) and 1O^(2-)`

Solution :Volume of unit cell = `( 5xx 10^(-8))^(3) cc`
` 1.25xx 10^(22) cc `
Density of FeO = 4 g/cc
Mass of unit cell = ` 1.25 xx 10^(-22) xx 4 = 5 xx 10^(-22) g `
Mass of 1 molecule`= 72/ ( 6.023 xx 10^(23)) = 1.195 xx 10^(22) g `
HENCE,number of FeOmolecules PER UNITCELL
` (5 xx 10^(-22))/( 1.195 xx 10^(-22)) = 4.18 =4 `
Hence, there are four `Fe^(2+)`and four ` O^(2-)`ions in eachunitcell,.
Discussion
7.

FeO is a _________ compound and it is _________ deficient.

Answer»

SOLUTION :NON - STOICHIOMETRIC, METAL
Discussion
8.

Fenton's reagent is

Answer»

`FeSO_(4)+H_(2)O_(2)`
Zn+HCl
Sn+HCl
None of these

Answer :A
Discussion
9.

Fenton's reagent is _______

Answer»

`H_(2)O_(2)//FE^(2+)`
`H_(2)O"//Fe^(2+)`
`H_(2)O_(2)//Fe^(3+)`
`H_(2)O"//Fe^(3+)`

ANSWER :A
Discussion
10.

[Fe(NO_2)_3Cl_3] and [Fe(O-NO)_3Cl_3] shows

Answer»

LINKAGE isomerism
Geometrical isomerism
Optical isomerism
Hydrate isomerism

Answer :A
Discussion
11.

Fenton.s reagent is :

Answer»

`H_2O+FeSO_4`
`H_2O_2+FeSO_4`
`H_2O_2+ZnSO_4`
`NaOH+FeSO_4`

ANSWER :B
Discussion
12.

femous oxide has cubes structure and each edgeof the unit cell is 5.0 Å .Assuming of the oxide as 4.0g//cm^(3) then the number of Fe^(2+) and O^(2) inos present in each unit cell will be

Answer»

`FOUR Fe^(2+) and two O^(2-)`
`Two Fe^(2+) and four O^(2-)`
`Four Fe^(2+) and Four O^(2-)`
`Three Fe^(2+) and three O^(2-)`

Solution :Let the units of femousoxide in a unit cell `= n` molecular weight of femous oxids `(FeO) = 56 + 16 = 73 g mol^(-1)`
Weight of a unit `= (72 xx n)/(6.023 xx 10^(23))`
VOLUME of one cell `= ("length of corner")^(2)`
`= (5 Å)^(2) = 125 xx 10^(-29) cm^(3)`
`"Density" = ("wt. of cell")/("volume")`
`4.09 = (72 xx n)/(6.023 xx 10^(23) xx 125 xx 10^(-24))`
`n = (3079.2 xx 10^(-1))/(72)`
`= 42.7 xx 10 = 4.27 ~~ 4 `
Discussion
13.

Feldspar is :

Answer»

Potassium SODIUM aluminosilicate
A MIXTURE of potassium, aluminium and SILICON oxides
Hydrated CALCIUM silicate
None of the above

Answer :A
Discussion
14.

Fe^(II)Cl(CN)_(4)(O_(2))]^(4-) is named as :

Answer»

Chloridotetracyanidodioxidoferrate (II) ion
chloridotetracyanidoperoxidoferrate (II) ion
chloridotetracyanidosuperoxidoferrate (II) ion
None is correct

Solution :`[overset(+2)(FE)overset(-1)Cloverset(-4)(CN)_(4)overset(-1)(O_(2))]^(4-)`
Discussion
15.

Fehling's solution reduces______aldehydes but not_________alehydes.

Answer»

SOLUTION :ALIPHATIC, AROMATIC
Discussion
16.

Fehling's solution is reduced by :

Answer»

SODIUM FORMATE
Sodium acetate
Sodium chloride
Potassium NITRATE

ANSWER :A
Discussion
17.

Fehling's solution is a mixture of _____ and _____.

Answer»

SOLUTION :ALKALINE solution of `C_USO_4`, solution of SODIUM POTASSIUM TARTRATE
Discussion
18.

Fehling's solution is

Answer»

Alkaline `CuSO_(4)^(+)`Rochelle salt (Sod. pot, tartrate )
Alkaline `CuSO_4`complexed with citrate IONS
Ammoniacal `AgNO_3`SOLUTION
MAGENTA solutions in `H_2SO_3`

ANSWER :A
Discussion
19.

The colour of the precipitate formed when a reducing sugar is heated with Fehling's solution is :

Answer»

Acidified copper SULPHATE solution
AMMONIACAL cuprous CHLORIDE solution
Copper suphate, ROCHELLE salt + NaOH
None of above

ANSWER :C
Discussion
20.

Fehlings solution is

Answer»

`AgNO_(3)` solution `+ NAOH` solution `+ NH_(4)OH`
Alkaline solution of Cupric ion COMPLEXED with citrate ion
Copper sulphate `+` SODIUM potassium tartarate `+ NaOH`
Copper sulphate solution

Solution :Fehlings solution is alkaline solution of `CuSO_(4)` with rochell salt i.e. sodium potassium tartarate.
Discussion
21.

Fehling.s solution is used in the detection of :

Answer»

ACIDIFIED copper sulphate solution
Ammoniacal CUPROUS CHLORIDE solution
Copper suphate, ROCHELLE salt + NaOH
None of above

Answer :C
Discussion
22.

Fehling.s solution consists of two separate alkaline solution. If one is CuSO_4,The other is:

Answer»

`NaHCO_3`
`Na_2SO_4`
`NaKC_4H_6O_8`
`NaKC_2O_4`

ANSWER :C
Discussion
23.

Fehling's solution can make distinction between

Answer»

`CH_(3)CHO` and `C_(6)H_(5)CHO`
`CH_(3)CHO and CH_(3)UNDERSET(O)underset(||)(C)CH_(2)OH`
`CH_(2)underset(OH)underset(|)(C)Hunderset(O)underset(||)(C)CH_(3)` and HCHO
`CH_(3)CHO `and HCHO

Solution :N//A
Discussion
24.

Fehling.s solution and Benedict.s solution are reduced by glucose to form :

Answer»

CUO
`Cu_2O`
`CU(OH)_2`
Cu

Answer :B
Discussion
25.

Fehling's solution 'A' consists of an aqueous solution of copper sulphate, while Fehling's solution 'B' consists of an akaline solution of …………..

Answer»

Solution :SODIUM POTASSIUM tartarate
`NaOOC-CH(OH)CH(OH)COOK`.
Discussion
26.

Fehling solution 'A' consists of an aqueous solution of copper sulphate while fehling solution 'B' consists of an alkaline solution of______.

Answer»

SOLUTION :ROCHELLE SALT.
Discussion
27.

Fehling solution is

Answer»

acidified `CuSO_(4)` solution
ammonical `AgNO_(3)` solution
COPPER sulphate, sodium hydroxide and Rochelle salt
none of these

Answer :C
Discussion
28.

Fehling's reagent contains

Answer»

COPPER sulphate, Rochelle salt and sodium hydroxide
copper sulphate, `Na_(2)CO_(3)` and sodium ACETATE
copper sulphate, sodium CITRATE and sodium acetate
copper sulphate, HCl and Rochelle salt.

Answer :A
Discussion
29.

[Fe(H_(2)O)_(6)]^(3+) is strongly paramagnetic whereas [Fe(CN)_(6)]^(3-) is weakly paramagnetic . Explain.

Answer»

Solution :In both the complexes, `Fe` is in `+3` oxidation state with the CONFIGURATION `3d^(5)`. `CN^(-)` is a STRONG field ligand. In its presence , `3d` electrons pair up leaving only one unpaired electron. The hybridisation is `d^(2)sp^(3)` forming inner orbital COMPLEX. `H_(2)O` is a weak ligand. In its presence `3d` electrons do not pair up. The hybridisation is `sp^(3)d^(2)` forming an outer orbital complex containing five unpaired electrons, hence it is strongly paramagnetic.
Discussion
30.

[Fe(H_(2)O_(6)]^(3+) is strongly paramagnetic whereas [Fe(CN)_(6)]^(3-) is weakly paramagnetic. Explain.

Answer»

Solution :In both the complexes, `Fe is in +3` OXIDATION state with the configuration `3d^(5)` `CN^(-)` is a strong ligand. In its presence, 3d ELECTRONS PAIR up leaving only one unpaired electron. The hybridization is `d^(2)sp^(3-)` forming inner orbital complex. `H_(2)O` is a weak ligand. In its presence, 3d electrons do not pair up. The hybridization is `sp^(3)d^(2)` forming an outer orbital complex containing five unpaired electrons. Hence, it is STRONGLY PARAMAGNETIC
Discussion
31.

[Fe(H_(2)O)_(6)]^(3+) is strongly paramagnetic whereas [Fe(CN)_(6)]^(3-) is weakly paramagnetic. Explain.

Answer»

Solution :Fe is in +3 oxidation STATE in both compounds with the configuration `3d^(5)`. In the PRESENCE of strong `CN^(-)` ligand, 3d electrons pair up leaving only one unpaired electron. The hybridisation is `d^(2)sp^(3)` forming inner ORBITAL complex. In the presence of weak `H_(2)O` ligand, 3d electrons do not pair up. The hybridisation is `sp^(3)d^(2)` forming an outer orbital complex containing five unpaired electrons. HENCE, it is strongly paramagnetic.
Discussion
32.

[Fe(H_(2)O)_(6)]^(3+)is strongly paramagnetic whereas [Fe(CN)_(6)]^(3-) is weakly paramagnetic. Explain.

Answer»

Solution :In both the complexes, Fe is in +3 oxidation state with the configuration `3d^(5)`. `CN^(-)` is a strong ligand. In its presence, 3d electrons pair up leaving only ONE unpaired electron. The hybridisation is `d^(2)SP^(3)` forming inner orbital complex. `H_(2)O` is a WEAK ligand. In its presence, 3d electrons do not pair up. The hybridisation is `sp^(3)d^(2)` forming an OUTER orbital complex CONTAINING five unpaired electrons. Hence, it is stronglyparamagnetic.
Discussion
33.

[Fe(H_(2)O)_(6)]^(2+) is

Answer»

PALE GREEN COMPLEX
Blue COLOURED complex
RED coloured complex
Violet coloured complex

Answer :A
Discussion
34.

[Fe(H_(2)O)_(6)]^(2+) and [Fe(CN)_(6)]^(4-) differ in:

Answer»

GEOMETRY, MAGNETIC MOMENT
geometry, hybridization
magnetic moment, colour
hybridization NUMBER of d-electrons

Solution :
Discussion
35.

[Fe(H_2O)_6 NO] and [Cr(NH_3)_6](NO_2)_3 both are paramagnetic species with 'spin' only magnetic moment of 3.93 B.M. The hybridisation of central metal ions in these species respectively are :

Answer»

Both `sp^3d^2`
Both `d^2sp^3`
`sop^3d^2 and d^2sp^3`
`d^2sp^3 and sp^3d^2`

SOLUTION :`3.93=sqrt(N(n+2))`, so n=3 (here n is NUMBER of UNPAIRED ELECTRONS
Discussion
36.

[Fe(H_(2)O)_(5)NO]^(2+) is a complex formed in the brown ring test for NO_(3)^(-) ion. In this complex.

Answer»

NO TRANSFERS its electronto `Fe^(2+)` so that we have iron as `Fe(I)` and NO as `NO_^(+)`
There are three unpaired electron so that its magnetic moment is 3.87 B.M.
The colour is due to chargetransfer
All the above statements are correct.

Solution :`HgI_(2)overset(Delta)(RARR) Hg_ I_(2) ("VIOLET vapours")`
`NO rarr NO^(+) + E^(-)`,
`{:(Fe^(2+) ,+,e^(-),rarr,Fe^(+)),((3d^(6)),,,,(3d^(7))),("4 unpaired "e^(-),,,,"3 unpaired "e^(-)):}`
`mu` for `Fe^(+) = sqrt(n(n+2))= sqrt(3 xx 5)` BM
`= sqrt(15) BM =3.87 BM`
Discussion
37.

[Fe(H_(2)O)_(5)NO]^(2+) is a complex formed during the brown ring test for NO_(3)^(-) ion. In this complex,

Answer»

there are THREE unpaired ELECTRONS so that its magnetic moment is 3.87 B.M.
NO transfers its electron to `Fe^(2+)` so that IRON exists as Fe (I) and NO as `NO^(+)`
the COLOUR is because of charge transfer
all of the above statements are correct

Solution :`{:(" "":"overset(.)(N)=UNDERSET(..)(O): rarr : overset(+)(N) = underset(..)(O) : + e^(-)),(Fe^(2+)+e^(-) rarr Fe^(+)):}`

`therefore` Magnetic moment of `Fe^(+) = sqrt(n(n+2))` BM
`=sqrt(3xx5)BM = 3.87 BM`
All statements are correct.
Discussion
38.

[Fe(en)_(2)(H_(2)O)_(2)]^(2+)+en to "complex"(X). The correct statement about complex (X) is :

Answer»

It is a low SPIN COMPLEX
It is DIAMAGNETIC
It shows geometrical isomerism
(a) and (b) both

Solution :[FE`(en)_(3)]^(3+), d^(6)` , low spin , diamagnetic
Discussion
39.

[Fe(e n)_(2)(H_(2)O)_(2)]^(2+) +e n to complex (x). The correct statement about the complex (x) it is

Answer»

a LOW spin complex
DIAMAGNETIC
shows GEOMETRICAL ISOMERISM
(1) and (2) both

Solution :`[Fe(e n)_(2)(H_(2)O)_(2)]^(+2) +e n to [Fe(e n)_(3)]^(-2)`
`Fe^(+2) = 3d^(6), t_(2g) ^(6) eg^(0)` low spin, diamagnetic
Discussion
40.

[Fe(en)_(2)(H_(2)O)_(2)]^(2+)+ento complex (x). The correct statement about the complex (x) is

Answer»

it is a LOW spin complex
it is diamagnetic
it shows geometrical isomerism
it does not exhibit any isomerism

Solution :`[Fe(en)_(2)(H_(2)O)_(2)]^(2+)+ENTO[Fe(en)_(3)]^(2+)+2H_(2)O`

(en) is a STRONG field ligand so regrouping takes PLACE as shown below.

So it is a low spin complex and diamagnetic.
Discussion
41.

Fedral regulations set an upper limit of 50 parts per millison (ppm) of NH_(3) in the air in a work environment ( that is , 50 mL NH_(3) per 10^(6) mL of air).The density of NH_(3)(g) at room temperature is 0.771 g/L .Air from a manufacturing operation was drawn through a solution containing 100 mL of 0.0105 M HCl . The NH_(3) reacts with HCl as follows: NH_(3)(aq)+HCl(aq)toNH_(4)Cl(aq) After drawing air through the acid solution for 10 minutes at a rate 10 litres/min the acid was titrated .The remaining acid required13.1 mL of 0.0588 M NaoH to reach the equivalence point . How many ppm of NH_(3) were in the air ?

Answer»

SOLUTION :`61.6 PPM`
Discussion
42.

Fedral regulations set an upper limit of 50 parts per millison (ppm) of NH_(3) in the air in a work environment ( that is , 50 mL NH_(3) per 10^(6) mL of air).The density of NH_(3)(g) at room temperature is 0.771 g/L .Air from a manufacturing operation was drawn through a solution containing 100 mL of 0.0105 M HCl . The NH_(3) reacts with HCl as follows: NH_(3)(aq)+HCl(aq)toNH_(4)Cl(aq) After drawing air through the acid solution for 10 minutes at a rate 10 litres/min the acid was titrated .The remaining acid required13.1 mL of 0.0588 M NaoH to reach the equivalence point . How many grams of NH_(3) were drawn into the acid solution ?

Answer»

SOLUTION :`0.00475g`
Discussion
43.

FeCr_(2)O_(4)+NaOH+"air" to (A)+Fe_(2)O_(3)(A)+(B) to Na_(2)Cr_(2)O_(7) Na_(2)Cr_(2)O_(7)+X overset(Delta) to Cr_(3),O_(3) Cr_(3)+Y(Delta) to Cr Na_(2)CrO_(4) and Fe_(2)O_(3) are separated by :

Answer»

DISSOLVING in conc `H_(2)SO_(4)`
dissolving in `NH_(3)`
dissolving in `H_(2)O`
dissolving in dil. HCl

Solution :`Na_(2)Cr_(2)O_(7)` solubel in `H_(2)O` but `Fe_(2)O_(3)` is insoluble in `H_(2)O`
Discussion
44.

FeCr_(2)O_(4)+NaOH+"air" to (A)+Fe_(2)O_(3)(A)+(B) to Na_(2)Cr_(2)O_(7) Na_(2)Cr_(2)O_(7)+X overset(Delta) to Cr_(3),O_(3) Cr_(3)+Y(Delta) to Cr (X) and (Y) are :

Answer»

C and Al
AI and C
C in both
Al in both

Solution :`X=C,y=Al`
Discussion
45.

FeCr_(2)O_(4)+NaOH+"air" to (A)+Fe_(2)O_(3)(A)+(B) to Na_(2)Cr_(2)O_(7) Na_(2)Cr_(2)O_(7)+X overset(Delta) to Cr_(3),O_(3) Cr_(3)+Y(Delta) to Cr Compounds (A) and (B) are :

Answer»

`Na_(2)CrO_(4). H_(2)SO_(4)`
`Na_(2)Cr_(2)O_(7),HCl`
`Na_(2)CrO_(5),H_(2)SO_(4)`
`Na_(4)[Fe(OH)_(6)].H_(2)SO_(4)`

Solution :`4FeCr_(2)O_(4)+8NaOH+O_(2)OVERSET(Delta) to underset((A)) (4Na_(2)CrO_(4)) +2Fe_(2)O_(3)+4H_(2)O`
`underset((A))(2Na_(2)CrO_(4))+underset((B))(H_(2)SO_(4))to Na_(2)SO_(4) to Na_(2)Cr_(2)O_(7)+Na_(2)SO_(4)+H_(2)O`
`{:(Na_(2)Cr_(2)O_(7)+3C to 2Na_(2)CrO_(2)+3CO),(2NaCrO_(2)+H_(2)O to Cr_(2)O_(3)+2NaOH), (Cr_(2)O_(3)+2Al to Al_(2) O_(3)+2Cr):}`
Discussion
46.

FeCr_(2)O_(4)+NaOH+"air" to (A)+Fe_(2)O_(3) (A)+(B) to Na_(2)Cr_(2)O_(7) Na_(2)Cr_(2)O_(7)+X overset(Delta)to Cr_(2)O_(3) Cr_(2)O_(3)+Y overset(Delta)to Cr Na_(2)CrO_(4) and Fe_(2)O_(3) are separated by :

Answer»

dissolving in CONC. `H_(2)SO_(4)`
dissolving in `NH_(3)`
dissolving in `H_(2)O`
dissolving in DIL. HCl

Answer :C
Discussion
47.

FeCr_2O_4+Na_2CO_3+O_2overset("Fusion")to[X]underset(H_2O)overset(H^+)to[Y]underset(H_2O_2)overset(H^+)to[Z] Which of the following statements is true for the compounds [X],[Y] and [Z] ?

Answer»

In all three COMPOUNDS , the CHROMIUM is in +6 oxidation STATE
[Z] is a deep blue -VIOLET coloured compound which decomposes rapidly in aqueous solution into `Cr^(3+)` and dioxygen
Saturated solution of [Y] gives bright organge compound , chromic anhydride, with concentrated `H_2SO_4`
All of these

Solution :`4FeCr_2O_4 8Na_2CO_3+7O_2overset("Fusion")tooverset(+VI)(Na_2CrO_4)underset(H_2O)overset(H^+)tooverset(+VI)(Na_2Cr_2O_7)overset(H^+ // H_2O_2)tooverset(+VI)(CrO(O_2)_2)`("deep blue violet")overset(H_2O)toO_2+H_2O+Cr^(3+)`
`Na_2Cr_2O_7+H_2SO_4to2CrO_3("chromic anhydride")+Na_2SO_4+H_2O`
Discussion
48.

FeCr_(2)(chromite ) is a good soure of chromium and its compounds like Na_(2)CrO_(4),Na_(2)Cr_(2)O_(7) P:FeCr_(2)O_(4)+NaOH+airto A+Fe_(2)O_(3)Q: A+Bto Na_(2)Cr_(2)O_(7)R:Na_(2)Cr_(2)O_(7)+Xoverset(Delta)toCr_(2)O_(3)S:Cr_(@)O_(3)+YtoCr Na_(2)CrO_(4)and Fe_(2)O_(3)are separated by:

Answer»

Dissolving in CONC ,`H_(2)SO_(4)`
Dissolving in `NH_(3)`
Dissolving in `H_(2)O`
Dissolving in DIL,HCl

Answer :a
Discussion
49.

FeCr_(2)O_(4)+Na_(2)CO_(3)+O_(2) overset("Fusion")(to) [X] overset(H^(+))underset(H_(2)O)(to) [Y] overset(H_(2)O"/"H^(+))(to) [Z] Which of the following statement is true for the compounds [X], [Y] and [Z]?

Answer»

In all THREE COMPOUNDS, the chromium is in +6 oxidation state
[Z] is a deep blue-voilet coloured compound which decomposes rapidly in aqueous solution into `Cr^(3+)` and dioxygen
Saturated solution of [Y] GIVES BRIGHT orange compound, chromic anhydride, with concentrated `H_(2)SO_(4)`
All of these

Answer :D
Discussion
50.

[Fe(CN)_(6)]^(4-) is

Answer»

INNER complex
outer complex
square planar
trigonal bipyramidal

Answer :A::C
Discussion