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(CN)_6]^(4-) ion is:

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Octahedral
Square planar
Bipyramidal
Tetrahedron

Answer :A
2.

[Fe(CN)_(6)]^(4-) and [Fe(H_(2)O)_(8)]^(2+) are of different colours in dilute solutions. Why

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Solution :In both the complexes, `Fe is in +2` state with the configuration `3d^(6)`, i.e., it has FOUR unpaired electrons. As the ligands `H_(2)O and.`CN^(-)` posses, different CRYSTAL FIELD splitting energy(triangle_(o))`, they absorb different components of the visible light (VIBGYOR) for d-d TRANSITION. HENCE, the transmitted colours are different
3.

[Fe(CN)_(6)]^(4-) and [Fe(H_(2)O)_(6)]^(2+) are of different colours in dilute solutions. Why ?

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SOLUTION :In both the complexes, Fe is in +2 state with the CONFIGURATION `3d^(6)`, having four unpaired electrons. As the ligands `H_(2)O` and `CN^(-)` POSSESS different crystal field splitting energy (`Delta_(0)`), they absorb different wavelengths of the visible light for d - d transition. HENCE, the transmitted colours are different.
4.

[Fe(CN)_(6)]^(4-) and [Fe(CN)_(6)]^(3-) have different _________ but same _________.

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SOLUTION :MAGNETIC MOMENT, STRUCTURE
5.

[Fe(CN)_6]^(4-) and Fe(H_2 O)_6]^(2+) are different colours in dilute solutions . Why ?

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Solution :In both the COMPLEXES, Fe is in +2 state with the configuration `3d^6`, i.e., it has FOUR unpaired electrons. In the presence of WEAK `H_2 O` ligand, they do not pair up. In the presence of strong CN ligand, they pair up leaving no unpaired electron. DUE to the difference in the number of unpaired electrons, they have different colours.
6.

[Fe(CN)_5NO]^(2-)+S^(2-)toProduct[X] Answer the following questions for the product [X]. (i)Number of negative monodenate ligand(s). (ii)Effective atomic number of iron in the complex. (iii)Number of unpaired electrons

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Solution :(i)five `CN^(-)` and ONE `NOS^(-)`=6
(ii)Iron is in +2 OXIDATION state and atomic number of iron is 26.
Therefore, EAN=24+12=36
(iii)`Fe^(2+)=3d^(5)4s^(0)`

`CN^-` is strong FIELD ligand which compels for paring of electron. Therefore , there is one unpaired electron.
7.

FeCL_(3)+KI to Fe^(2+)(aq.)+KI_(3)

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For COLOURED ppt./Black ppt
For coloured solution.
for clear/colourless solution
For WHITE ppt.

Answer :B
8.

FeCl_3 on reaction with K_4[Fe(CN)_6] in aq solution gives blue colour and these are separated by a semi permeable membrane Due to osmosis there is

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blue COLOUR FORMATION in the SOLUTION of`FeCl_3`
blue colour formation in the solution of`K_4[Fe(CN)_6]`
blue colour formation in both the solutions
no blue colour formation

Answer :D
9.

FeCl_(3) on reactiSon with K_(4)[Fe(CN)_(6)] in aq solution gives blue colour of these are separated by a semi permeable memrane AB as shown, Due to osmosis there is

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BLUE colour formation in side X
blue colour formation in side Y
blue colour formation in both SIDES X & Y
no blue colour formation

ANSWER :D
10.

FeCl_3 solution on reaction with SO_2 changes to:

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`FeCl_2`
`Fe_2(SO_4)_3`
`Fe_2(SO_3)_3`
`FeSO_4`

ANSWER :A
11.

FeCl_(3) and CoCl_(2) have primary valencies of

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2 and 2
2 and 1
3 and 2
2 and 3

Answer :C
12.

FeCl_(3) is known but not Fel_(3) Why?

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Solution :`I^(-)` is a strong REDUCING agent, it reduces `Fe^(3+)` to `Fe^(2+) (or) Fe^(3+)`can oxidise `I^(-)` to `I_(2)`but not `CL^(-)` to `Cl_(2).2Fe^(3+)+2I^(-)rarr 2Fe^(2+)+I_(2)`. THUS `FeCl_(3)` is KNOWN but not `Fel_(2)`.
13.

FeCl_(3) has each of the following characteristics except :

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it exists as dimer
its aqueous solution is basic in nature
its is used for stopping from a FRESH cut.
on heating above 973 K it dissociates to ferrous CHLORIDE and GIVES `Cl_(2)`

ANSWER :B
14.

FeCl_2 reacts with SO_2

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to GIVE FeS
to give FeO
Fe will be oxidized
Fe will be reduced

Answer :C
15.

FeC_3 is added separately to excess of boiling water and excess of NaOH solution, the sols obtained contain respectively the disperse particles to be :

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`F_3e_3 C`
`F_3e_3^(2) C`
`F_3e_3^(3) C I`
`F_3 e_3^(3) C I C`

Solution :A red SOL of FERRIC hydroxide is OBTAINED by the hydrolysis of ferric chloride with BOILING WATER.
16.

FeC_(2)O_(4) can decolorise acidified KMnO_(4) dueto the oxidation of (a) _____and (b)____.

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Answer :a. `Fe^(2+)` B. `C_(2)O_(4)^(2-)`
17.

Fear or excitement generally cause one to breathe rapidly and it results in the decreases of concentration of CO_2 in blood. In what way it will change pHof blood :

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PH will increase
pH will decrease
No change
pH will be 7

Answer :C
18.

Fe_3O_4 is known ashaematite. True or False

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SOLUTION :MAGNETITE
19.

Fe_2O_3 is known as___-.

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SOLUTION :MAGNETITE
20.

Fe_3O_4 is known as _____ .

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SOLUTION :MAGNETITE
21.

Fe_(3)O_(4) is ferrimagnetic at room temperature but at 850 K it becomes :

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DIAMAGNETIC
Ferromagnetic
Non-magnetic
PARAMAGNETIC

ANSWER :D
22.

Fe_3O_4 is ferrimagnetic at room temperature but becomes paramagnetic at 850K. Explain.

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Solution :DUE to RANDOMISATION of SPINS at high TEMPERATURE.
23.

Fe_(3)O_(4) is:

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FERROMAGNETIC
DIAMAGNETIC
PARAMAGNETIC
FERRIMAGNETIC

ANSWER :D
24.

Fe_3O_4 has inverse spinel structure. What is not true about this solid ?

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`FE^(3+)` ions are EQUALLY distributed between octahedral and tetrahedral voids
Tetrahedral voids are equally distributed between `Fe^(2+)` and `Fe^(3+)` ions
No. of `O^(-2) gt Fe^(+3) gt Fe^(+2)`
Coordination no. of `Fe^(3+)` =8 through out the unit cell

ANSWER :B::D
25.

Fe^(3+)overset(SNC^(Θ)("Excess"))rarrunderset("Bloodre"d)(A)overset(F^(ΘExcess))rarrunderset("Colourless")(B) Identify A and B (a) Write the IUPAC name of A and B (b) Find out the spin only magnetic moment of B .

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Solution :`NaS_(2)O_(3)+2HCI rarr 2NaCI +H_(2)O+SO_(2) +S`
A is `[Fe(SCN)(H_(2)O)_(5)]^(2+)`
The `IUPAC` NAME is pentaaquathiocyanatoferrate(III) ion `B` is `[FeF_(6)]^(3-)`
The `IUPAC` mane is hexafluridoferrate(III)
`Fe^(3+) + SCNbar rarr [Fe(SCN)(H_(2)O)_(5)]^(2+)`
`Fe[SCN(H_(2)O)_(5)]^(2+)+6F rarr [FeF_(6)]^(3-) + SCN^(Θ) +5H_(2)O`
`B[FeF_(6)]^(3-)`
The oxidation number of ion is + 3 Atomic number = 26
`Fe = [Ar]3d^(6) 4s^(2)`
`Fe =[Ar]3d^(5)`
Which means five UNPAIRED electrons
Spin magnetic moment` sqrt(N(+2))`
`=sqrt(5(5+2)) =sqrt35BM`
.
26.

Fe^(3+)overset(SCN^(-)(excess))to "blood red"(A)overset(F^(-)(excess))to"colourless"(B)Identify A and B (a) Write IUPAC name of A and B (b) Find out spin only magnetic moment of B.

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Solution :`Fe^(3+)unerset(excess)(SCN)^(-)overset(aqueous)tounderset((A)"(blood RED))[Fe(SCN)(H_(2)O)_(5)]]^(2+)``unerset((A)("blood red"))([Fe(SCN)(H_(2)O)_(5)]^(2+))
`unerset(excess)(6F^(-))tounerset((B)("colour LESS")([FeF_(6)]^(3-))+SCN^(-)+5H_(2)O`
(a) Pentaaquathiocyanatoiron (III)ion
hexafluoroferrate (III) (b) Magnetic moment = SQRT(n(n+2)) = sqrt35 = 5.92 B.M., where n = number of UNPAIRED electrons which is equal to 5 here.
27.

Fe^(3+) is reduced to Fe^(2+) by using

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`H_(2)O_(2)` in PRESENCE of `NaOH`
`Na_(2)O_(2)` in WATER
`H_(2)O_(2)` in presence of `H_(2)SO_(4)`
`Na_(2)O_(2)` in presence of `H_(2)SO_(4)`

SOLUTION :In Basic MEDIUM, `Fe^(3+)` will be precipitated as `Fe(OH)_(3)`.
28.

Fe^(3+) is reduced of Fe^(2+) by using

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`H_(2)O_(2)` is PRESENCE of NAOH
`Na_(2)O_(2)` in water
`H_(2)O_(2)` in presence of `H_(2)SO_(4)`
`Na_(2)O_(2)` in presence of `H_(2)SO_(4)`

Answer :A::B
29.

Fe^(3+) does not give prussian blue colour with K_(4)[Fe(CN)_(6)] but on its reaction with (X), prussian blue colour appears (X) can be:

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`MnO_(4)^(-)//H^(+)`
`Zn//NaOH`
`NH_(3)(AQ)`
all true

Solution :N//A
30.

Fe^(3+)ions react with potassium thiocyanide to give :

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yellow PPT
BLUE colouration
blood red colouration
black ppt.

Solution :`FeCl_3 + underset("POT. sulphocyanide")(3KCNS) to underset("Blood red colouration")(FE(CNS)_3) + 3KCI`
31.

[Fe^(3+)(biphy)_(X)] , (bipy = 2,2' bipyridine)'X' value is

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SOLUTION :`[FE(bipy)_(3)]^(3+)` 6 COORDINATION
32.

Fe_(2)(SO_(4))_(3) is used in water and sewage treatment to aid the removal of suspended impurities. Calculate the mass percentage of iron, sulphur and oxygen in this compound.

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Solution :`"MOL. MASS of "Fe_(2)(SO_(4))_(3)=2xx56+(32+64)xx3=400`
`%" of FE"=(2xx56)/(400)xx100=28%","%" of S"=(32xx3)/(400)xx100=24%,""%" of O"=(4xx16xx3)/(400)xx100=48%`
33.

Fe_(2)O_(3) is reduced to spongy iron near the top of blast furnace by

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CO
`CO_(2)`
C
`H_(2)`

SOLUTION :C is used as reducing agent
34.

Fe_(2)O_(3) is reduced to spongy iron near near the top of blast furnace by

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`H_(2)`
CAO
`SiO_(2)`
CO

Answer :D
35.

Fe_(2)O_(3) is converted to FeO in the presence of CO at ________

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473-573 K
573-673 K
673-773 K
773-873 K

Answer :D
36.

Fe_2(CO)_9is diamagnetic. Which of the followingreasons is correct:

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PRESENCE of one CO as BRIDGE group
Presence ofmonodentate ligand
Metal-metal (Fe-Fe) BOND in molecule
Resonance hybridisation of CO

Answer :C
37.

Fe_(2)(CO)_(9) is diamagnetic. Which of the following is the correct reason ?

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One CO is present as BRIDGE group
CO is a `pi`-ACCEPTOR ligand
CO can form `pi`-bond with Fe by BACK bonding
Metal-metal (Fe-Fe) bonding TAKES place.

Solution :Metal-metal bonding pairs up the UNPAIRED electron.
38.

Fe_(2)(CO)_(9) is diamagnetic. Which of the following reasons is correct :

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Presence of one CO as bridge group
Presence of MONODENTATE ligand
Metal - metal (Fe-Fe) bond in molecule
Resonance HYBRIDISATION of CO

Solution :All electrons are PAIRD up and M-M bond is absent.
39.

Fe_2(CO)_9 is diamagnetic. Which of the following reasons is correct:

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PRESENCE of one CO as BRIDGE group
Presence of MONODENTATE ligand
Metal-Metal (Fe-Fe) bond in molecule
Resonance HYBRIDIZATION of CO

Answer :C
40.

Fe^(2+)+NO_(3)^(-) +H_(2)SO_(4) ("conc.") to "X" (Brown ring complex) The magnetic moment of complex 'X' to its nearest integer is :

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ANSWER :4
41.

Fe^(2+) ions react with nitric oxide formed from reduction of nitrate and yields a brown coloured complex-

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`[Fe(CO)_(5)NO]^(2+)`
`[Fe(NH_(3))_(5)NO]^(2+)`
`[Fe(CH_(3)NH_(2))_(5)NO]2^(+)`
`[Fe(H_(2)O)_(5)NO]^(2+)`

Answer :B::D
42.

Fe^(2+) gives blue coloue, called(a)____ with (b)___white Fe^(2+) gives blue coloue, called(c)____ with (d)____.

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ANSWER :a.Turabull's BLUE b. `K_(2)[Fe(CN)_(6)]`C. Prussium blue d. `K_(6)[Fe(CN)_(6)]`
43.

Fe^2+ ion can be distinguished by Fe^3+ ion by:

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ANSWER :C
44.

Fe^(2+) does not give prussian blue colour with K_4[Fe((CN)_6], but on its reaction with (X), prussian blue colour appears with the reagent (X) is :

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`MnO_4^(-)//H^+`
`H_2SO_4`
`NH_3`
All true

Solution :`MnO_4 H^(+) to[O] : FE^(2+)OVERSET([O])toFe^(3+)+[Fe(CN)_6]^(4-)to Fe_4[Fe(CN)_6]_3`
45.

Fe^(2+) does not givebluecolour with K_(4)[Fe(CN)_(6)] but onitsreaction with (X) ,bluecolour oppears (X) can be

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`MnO_(4)^(THETA) //H^(o+)`
`H_(2)SO_(4)`
`NH_(3)`
`HCI`

Answer :a
46.

Fe^(2+)._((aq))+NO_(3)^(Θ)._((aq))+H_(2)SO_(4)(conc.)rarr Brown ring .The oxidationnumber of iron in brownring complex is

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Solution :`[overset(+1)(Fe)(H_(2)O)_(5)overset(+1)(NO)]^(2+)SO_(4)^(2-)`
47.

Fe^(2+) (aq)+NO_(3)^(-) (aq)+H_(2)SO_(4) (conc.) to Brown ring The brown ring is due to the formation of complex, [Fe(H_(2)O)_(5)NO]SO_(4). What is the oxidation state of iron in the complex ?

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Solution :Here `NO` is `NO^(+)`, so in `[Fe(H_(2)O)_(5)NO^(+)]^(2+)`, the oxidation state of IRON is `+1`
48.

Fe^(2+) and Fe^(3+) can be distinguished by :

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`NH_(4)SCN`
`CaCl_(2)`
`AgNO_(3)`
`H_(2)SO_(4)`.

ANSWER :A
49.

Fe ore is concentrated by :

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MAGNETIC TREATMENT
FROTH FLOATATION
Electrolysis
Roasting

ANSWER :A
50.

Fe is made passive by :

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DIL.`H_(2)SO_(4)`
dil. HCl
conc.`HNO_(3)`
conc.`H_(2)SO_(4)`

SOLUTION :Chloroplatinic ACID is `H_(3)[PtCl_(6)],` which is DIBASIC