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.

Moderator is a material which is used to slow down the neutrons produced during nuclear fission. The neutrons from the source are of high speed and energy. Heavy water or graphite moderators slow down the speed of the neutrons. The energy of fast moving neutrons decreases from 2MeV to 0.02535 eV, it corresponds to the velocity of 220 m "sec"^(-1). At this velocity, the neutrons are in thermal equilibrium with the moderator. such neutrons are called thermal neutrons. Thermal neutrons cause further fission reaction. The essential characterstices of moderators are: (i) Its molar mass must be low, (ii) It should not absorb neutrons. (iii) It should undergo elastic collisions with neutrons. Moderator in the reactor yields:

Answer»

FAST MOVING neutrons
thermal neutrons
magnetic neutrons
electric neutrons

Answer :B
Discussion
2.

Moderator is a material which is used to slow down the neutrons produced during nuclear fission. The neutrons from the source are of high speed and energy. Heavy water or graphite moderators slow down the speed of the neutrons. The energy of fast moving neutrons decreases from 2MeV to 0.02535 eV, it corresponds to the velocity of 220 m "sec"^(-1). At this velocity, the neutrons are in thermal equilibrium with the moderator. such neutrons are called thermal neutrons. Thermal neutrons cause further fission reaction. The essential characterstices of moderators are: (i) Its molar mass must be low, (ii) It should not absorb neutrons. (iii) It should undergo elastic collisions with neutrons. Which of the following is not used as a moderator?

Answer»

HEAVY water
Graphite
Beryllium
Sodium

Answer :d
Discussion
3.

Moderator is a material which is used to slow down the neutrons produced during nuclear fission. The neutrons from the source are of high speed and energy. Heavy water or graphite moderators slow down the speed of the neutrons. The energy of fast moving neutrons decreases from 2MeV to 0.02535 eV, it corresponds to the velocity of 220 m "sec"^(-1). At this velocity, the neutrons are in thermal equilibrium with the moderator. such neutrons are called thermal neutrons. Thermal neutrons cause further fission reaction. The essential characterstices of moderators are: (i) Its molar mass must be low, (ii) It should not absorb neutrons. (iii) It should undergo elastic collisions with neutrons. The moderator in a reactor:

Answer»

ABSORBS neutrons
acclerates neutrons
slows down neutrons
absorbs THERMAL ENERGY. PRODUCED in the reactors

Answer :C,d
Discussion
4.

Moderator is a material which is used to slow down the neutrons produced during nuclear fission. The neutrons from the source are of high speed and energy. Heavy water or graphite moderators slow down the speed of the neutrons. The energy of fast moving neutrons decreases from 2MeV to 0.02535 eV, it corresponds to the velocity of 220 m "sec"^(-1). At this velocity, the neutrons are in thermal equilibrium with the moderator. such neutrons are called thermal neutrons. Thermal neutrons cause further fission reaction. The essential characterstices of moderators are: (i) Its molar mass must be low, (ii) It should not absorb neutrons. (iii) It should undergo elastic collisions with neutrons. A good moderator should:

Answer»

not be a GAS only
not have appertite for neutrons only
be light in MASS NUMBER only
be all the above three

Answer :d
Discussion
5.

Moderate acts as a bleaching agent only in presence of

Answer»

Silica
Graphite
Diamond
Carborundum

SOLUTION :Graphite show moderate conductivity due to the presence of unpaired or FREE fourth valence electron on each CARBON ATOM.
Discussion
6.

Mobility of H^+(in aq medium) is high because:

Answer»

Of the SMALL size of `H^+`
Of the high hydration ENERGY of `H^+`
It exhibits a Grotthus TYPE of conduction
Hydrogen is the lightest element

Answer :C
Discussion
7.

Mn^(x+) has a magnetic moment of 4.9 BM. Calculate the value of x in the given ion.

Answer»


ANSWER :3
Discussion
8.

MnSO_(4)+Sr(NO_(3))_(2) to SrSO_(4) darr

Answer»

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

Answer :d
Discussion
9.

MnO_(4)^(2-) in acidic solution undergoes disportionation to give "…....................." and "…...................".

Answer»


ANSWER :`MnO_(4)^(-), MnO_(2)`
Discussion
10.

MnO_(4)^(2-) can be coverted to MnO_(4)^(-) 1) by oxidation of Cl_(2) 2) by oxidation of CO_(2) 3) by oxidation of H_(2)SO_(4) 4) by oxidations of O_(3) 5) by electrochemical oxidation at anode

Answer»

`1,2,3`
`1,3,5`
`1,2,3,5`
`1,2,3,4,5`

ANSWER :D
Discussion
11.

MnO_(4)^(2-) is

Answer»

Diamagnetic
Ferromagnetic
PARAMAGNETIC
NONMAGNETIC

ANSWER :C
Discussion
12.

MnO_(4)^(-) reacts with Br^(-) in alkaline pH to give

Answer»

`BrO_(3)^(-), MnO_(2)`<BR>`Br_(2),MnO_(4)^(2-)`
`Br_(2), MnO_(2)`
`BRO^(-), MnO_(4)^(2-)`

Solution :`2KMnO_(4)+ KBr+H_(2)O rarr 2KOH+ 2MnO_(2) + KBrO_(3)`
Discussion
13.

MnO_(4)^(-) react with Br in alkaline pH to give

Answer»

`BrO_(3)^(-), MnO_(2)`
`Br_(2), MnO_(4)^(2-)`
`Br_(2), MnO_(2)`
`BRO^(-), MnO_(4)^(2-)`

SOLUTION :`BrO_(3)^(-), MnO_(2)`
Discussion
14.

MnO_(4)^(-) is off intense pink colour, though Mn is in (+7) oxidation state, it is due to:

Answer»

oxygen gives colour to it
Charge TRANSFER when Mn(+7) gives its electron to oxygenn and oxidise to Mn (+8) temporarily
Charge transfer when oxygen gives its electron to Mn(+7) CHANGING in Mn(+6) NONE is CORRECT explanation.

Answer :C
Discussion
15.

MnO_(4)^(-) is of intense pink colour, though Mn is in (7+) oxidation state, it is due to :

Answer»

OXYGEN gives colour to it
Charge transfer when MN `(7+)` gives its ELECTRON to oxygen and oxidise to Mn `(8+)` temporaily
Charge transfer when oxygen gives its electron to Mn `(7+)` CHANGING inMn `(6+)` temporaily
none is correct explanation

Answer :C
Discussion
16.

MnO_(4)^(-) react with Br^(-) in alkaline P^(H) to give

Answer»

`BrO_(3)^(-)MnO_(2)`<BR>`Br_(2) MnO_(4)^(2-)`
`Br_(2) MnO_(2)`
`BrO_(4)^(-) MnO_(4)^(2-)`

Answer :A
Discussion
17.

MnO_(4)^(-) react with Br in alkaline pH to give ………………..

Answer»

`BrO_(3)^(-), MnO_(2)`<BR>`Br_(2), MnO_(4)^(2-)`
`Br_(2). MnO_(2)`
`BrO^(-), MnO_(4)^(2-)`

Solution :Hint : `2MnO_(4)^(-)+Br^(-)+H_(2)Orarr 2OH^(-)+2MnO_(2)+BrO_(3)^(-)`
Discussion
18.

MnO_4 ions are reduced in acidic condition to Mn^(2+)ions whereas they are reduced in neutral condition to MnO_2. The oxidation of 25 mL of a solution X containing Fe^(2+) ions required in acidic condition 20 mL of a solution Y containing MnO_4 ions. What volume of solution Y would be required to oxidise 25 mL of solution X containing Fe^(2+) ions in neutral condition?

Answer»

11.4 mL
12.0 mL
33.3 mL
35.0 mL

Solution :

INACIDICMEDIUM
5 volof `Fe^(2+)` requires1 volof `MnO_(4)^(-)` inacidicmedium
` therefore25Vol` of ` Fe^(2+)` requires` 1/5xx 25vol ` of` MnO_4^(- ) `in acidicmedium
`1/5xx 25vol` of `MnO_4 ^(- )~=20volor20 mL`
In neutralmedium
3 molof ` Fe^(2+)` requires1volof `MnO_(4)^(-)`in neutralmediumthen25volof ` Fe^(2+)` requires`1/3xx 25vol` of ` MnO_4`inneutralmedium
`1/5 XX 25vol` of `MnO_(4)^(- )~=20vol`
`therefore1/3x 25vol` of ` MnO_4^(-)=(20) /(5) xx(25)/( 3 )vol`of `MnO_(4)^(-)`
`= 33.3volor mL`
Discussion
19.

MnO_(4)^(-) ions can be reduced in strong alkaline medium to give :

Answer»

`Mn^(2+)`
`MnO_(2)`
`MnO_(4)^(2-)`
`MnO_(3)^(-)`

Answer :C
Discussion
20.

MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_(2). The oxidataion of 25 mL of a solution X containing Fe^(2+) ions required in acidic medium 20 mL of a solutionY containing MnO_(4)^(-) ions. What volume of solution Y would be required to oxidise 25 mL of solution Y containing Fe^(2+) ions in neutral condition ?

Answer»

11.4 mL
12.0 mL
33.3 mL
25.0 mL

Solution :In acidic medium,
`overset(+7)MnO_(4)^(-) to overset(+2)Mn^(2+)`
Change in oxidation number = 5
`underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))`
`N xx 25 = 5 M xx 20 "" …(i)`
25 N = 100 M
IN neutral medium,
`overset(+7)MnO_(4)^(-) to overset(+4)MnO_(2)`
Change in oxidation number = 3
`underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))`
`25 xxN = 3 M xx V`
25N = 3MV
Equating (i) and (ii)
100 M = 3 MV
or `V = (100)/(3) = 33.3` mL
Discussion
21.

MnO_(4)^(-) + 8H^(+) + 5e to Mn^(2+) + 4H_(2)O Fe^(3+) + e to Fe^(2+) , E^(@) = +1.51 V Cl_(2) + 2e to 2Cl^(-), E^(@) = +0.77 V Cl_(2) + 2e to 2Cl^(-), E^(@) =+1.36 V Ce^(4+) + e to Ce^(3+), E^(@) = 1.76 Choose the incorrect statement among the following:

Answer»

`MnO_(4)^(-)` is a sufficiently strong oxidant in acidic solution (pH=0) to oxidise `FE^(2+)` ion.
`Fe^(2+)` ion cannot be titrated against STANDARD `KMnO_(4)` solution is the medium is MADE acidic (pH=0) by adding HCl.
`MnO_(4)` ion cannot oxidise `Ca^(3+)` in acidic medium (pH=0)
`Fe^(2+)` cannot be titrated against standard `KMnO_(4)` solution in ACID medium (pH=0 in presence of `Ce^(3+)` ion.

ANSWER :D
Discussion
22.

MnO_(4)^(-)+8H^(+)+5e to Mn^(2+)+4H_(2)O""E^(0)=+1.51V Fe^(3+)+e to Fe^(2+)""E^(0)=+0.77V Cl_(2)+2e to 2Cl^(-) ""E^(0)=+1.36V Ce^(4+)+e to Ce^(3+) ""E^(0)=1.76 Choose the incorrect statement among the following:

Answer»

`MnO_(4)^(+)` is a sufficlently strong oxidant In acidic solution (pH =0) to oxidize `FE^(2+)` ion.
`Fe^(2+)` ion cannot be titrated against standard `KMnO_4` solution if the, medium is made acidic (pH =0) by adding HCl.
`MnO_(4)^(-),` ion cannot oxidize Cet in acidic medium (pH =0)
`Fe^(2+)` cannot be titrated against standard `KMnO_4` solution in acid medium (pH =0) in presence of `Ce^(3+)` ion.

Answer :D
Discussion
23.

MnO_(4)^(-) +8H^++5e^-to Mn^(2+) +4H_2 O ,E^@ = 1.51 V MnO_(4)^(-) +4H^++2e^-to Mn^(2+) +2H_2 O ,E^@ = 1.23 VE_(MnO_(4)^(-) "|" MnO_(2))is

Answer»

`1.70 v`
`0.91 V`
`1.37 V`
`0.548 V`

Solution :`MnO_(4)^(- )+ 8H^( +) + 5e^(- )to MN^(2+)+ 4 H_2O ,E^@= 1.51V `
` DeltaG_(1)^(@)=- 5 ( 1.51 )F=- 7.55F `
`MnO_2+4H^( +) +2e^(-) toMn^(2+) +2 H_2 O ,E^@= 1.23V `
`DeltaG_(2)^(@)=- 2(1.23 )F= - 2.46F `
onsubtracting
`MnO_(4)^@=- 5 ( 1.51) F = - 7.55F `
`E_("MnO_(4)^(-) |MnO_2 ) ^@ = (DeltaG_(3) ^(@) )/(-nF) =(-5.09 F)/(-3F) = 1.70V `
Discussion
24.

MnO_(3) in an acidic medium dissociates into

Answer»

`MnO_(2) and MnO_(4)^(-)`
`MnO and MnO_(4)^(-)`
`MnO_(2) and MnO`
`MnO_(2) and MnO_(3)`

SOLUTION :`5MnO_(3) + 2H^(+) + 4e^(-) rarr 3MnO_(3) + 2MnO_(4)^(-) + H_(2)O`
Discussion
25.

MnO_2 liberates oxygen from a solution of H_2O_2 (the action being catalytic) only if the solution is:

Answer»

Basic
Acidic
Neutral
None

Answer :B
Discussion
26.

MnO_2is the most important oxide of manganese. MnO_2 occurs naturally as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_3O_4 on heating to 530^(0) C. It is used in the preparation of potassium permanaga-nate and in the production of Cl_2 gas. Over half a million tonnes per year of MnO_2 is used in dry batteriesWhen MnO_2 is fused with KOH in the presence of air, the product formed is :

Answer»

purple COLOUR `KMnO_4`
green colour `K_2MnO_4`
COLOURLESS `MnO_4`
purple colour `K_2MnO_4`

Solution :`MnO_2 + KOH + AIR RARR K_2MnO_4`
Discussion
27.

MnO_2is the most important oxide of manganese. MnO_2 occurs naturally as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_3O_4 on heating to 530^(0) C. It is used in the preparation of potassium permanaga-nate and in the production of Cl_2 gas. Over half a million tonnes per year of MnO_2 is used in dry batteriesMnO_2 dissolve in concentrated HCl to form :

Answer»

`Mn^(4+)` ion and `Cl_2`
`Mn^(2+)` ion and `Cl_2`
`[MnCl_2]^(-2) and Cl_2`
only `[MnCl_4]^(-2)`

SOLUTION :`MnO_2 +HCL RARR MnCl_2 +Cl_2`
Discussion
28.

MnO_(2) is the most important oxide of manganese , MnO_(2) occurs natually as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_(3)O_(4) on heating to 530^(@)C. It is used in the preparation of potassium permanaganate and in the productioon of Cl_(2) gas. Over half million tonnes per year of MnO_(2) is used in dry batteries. Q.In the laboratory, MnO_(2) is made by:

Answer»

heating MN in `O_(2)`
oxidising `Mn^(2+)` in air
electrolytic oxidation of `MnSO_(4)`
PRECIPITATING `MnO_(2)` from SOLUTION when PERFORMING titration of `KMnO_(4)` in alkaline medium

Answer :D
Discussion
29.

MnO_(2) is the most important oxide of manganese , MnO_(2) occurs natually as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_(3)O_(4) on heating to 530^(@)C. It is used in the preparation of potassium permanaganate and in the productioon of Cl_(2) gas. Over half million tonnes per year of MnO_(2) is used in dry batteries. Q.MnO_(2) dissolves in concentrated HCl to form:

Answer»

`Mn^(4+)` ion and `Cl_(2)`
`Mn^(2+)` ion and `Cl_(2)`
`[MnCl_(4)]^(2-) and Cl_(2)`
only `[MnCl_(4)]^(2-)`

Answer :B
Discussion
30.

MnO_(2) is the most important oxide of manganese , MnO_(2) occurs natually as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_(3)O_(4) on heating to 530^(@)C. It is used in the preparation of potassium permanaganate and in the productioon of Cl_(2) gas. Over half million tonnes per year of MnO_(2) is used in dry batteries. Q.When MnO_(2) is fused with KOH in the presence of air, the product formed is:

Answer»

PURPLE colour `KMnO_(4)`
GREEN colour `K_(2)MnO_(4)`
Colourless `MnO_(4)^(-)`
Purple colour `K_()MnO_(4)`

ANSWER :B
Discussion
31.

MnO_2is the most important oxide of manganese. MnO_2 occurs naturally as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_3O_4 on heating to 530^(0) C. It is used in the preparation of potassium permanaga-nate and in the production of Cl_2 gas. Over half a million tonnes per year of MnO_2 is used in dry batteries In the laboratory, MnO_2 is made by :

Answer»

heating Mn in `O_2`
OXIDISING `Mn^(2+)` in air
electrolytic oxidation of `MnO_4`
Precipitating `MnO_2` from SOLUTION when perfoming TITRATION of `KMnO_(4)`

Solution :`MnO_4^- overset(OH^-) rarr MnO_2`
Discussion
32.

MnO_(2) is the most important oxide of manganese , MnO_(2) occurs natually as the black coloured mineral pyrolusite. It is an oxidising agent, and decomposes to Mn_(3)O_(4) on heating to 530^(@)C. It is used in the preparation of potassium permanaganate and in the productioon of Cl_(2) gas. Over half million tonnes per year of MnO_(2) is used in dry batteries. Q. In which of the following species, the colour is due to charge transfer? (I)[Mn(OH)_(4)]^(2-)(II)MnO_(4)^(2-)(III)MnO_(4)(IV)KMnO_(4)

Answer»

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

Answer :D
Discussion
33.

MnO_(2) is fused with an alkali. We obtain a green mass which on acidic hydrolysis produces a solution which has the colour

Answer»

DARK PURPLE.
yellow.
colourless.
BLUE.

ANSWER :A
Discussion
34.

MnO_2 is antiferromagnetic/ferromagnetic substance.

Answer»

SOLUTION :ANTIFERROMAGNETIC
Discussion
35.

MnO_(2) + HCl overset(Delta)to A_((g)) A_((g)) + F_(2("excess")) overset(573K)to B_((g)) B_((l)) + U_((s)) to C_((g))+ D_((g))The gases A,B, C and D are respectively ________.

Answer»

`Cl_(2), CLF, UF_(6), ClF_(3)`
`Cl_(2), ClF_(3), UF_(6) , ClF`
`O_(2), OF_(2), U_(2)O_(3), O_(2)F_(2)`
`O_(2), O_(2)F_(2), U_(2)O_(3), OF_(2)`

Answer :B
Discussion
36.

{:(MnO_(2) + HCl overset(Delta)rarr A(g)),(A(g) + F_(2)(excess)overset(573 K)rarr B(g)),(B(l) + U(s) rarr C(g) + D(g)):} The gases A, B, C and D are respectively

Answer»

`Cl_(2), ClF, UF_(6), ClF_(3)`
`Cl_(2), ClF_(3), UF_(6), ClF`
`O_(2), OF_(2), U_(2)O_(3), O_(2)F_(2)`
`O_(2), O_(2)F_(2), U_(2)O_(3), OF_(2)`

Solution :`{:(MnO_(2) + 4 HCl rarr MnCl_(2) + underset((A))(Cl_(2))(g) + 2H_(2)O),(Cl_(2)(g) + 3F_(2)(EXCESS) overset(573 K)rarr underset((B))(2ClF_(3))(g)),(3ClF_(3)(l) + U(s) overset(Delta)rarr underset((C))(UF_(6))(g) + 3 underset((D))(ClF)(g)):}`
The gases (A), (B), (C) and (D) are `Cl_(2), ClF_(3), UF_(6) and ClF` respectively, i.e., option (b) is correct.
Discussion
37.

MnO_2 + HCl overset(Delta)to A_((s)) A_((g)) + F_(2"(excess)") overset(573K)to B_((s)) B_((l)) + U_((s)) to C_((g)) + D_((g)) The gases A,B,C and D are respectively

Answer»

`Cl_(2), ClF, UF_(6), ClF_(3)`
`Cl_(2), ClF_(3), UF_(6), ClF`
`O_(2), OF_(2), U_(2)O_(3), O_(2)F_(2)`
`O_(2), O_(2)F_(2), U_(2)O_(3), OF_(2)`

Solution :`MnO_(2)+4HCl overset(DELTA)rarr underset((A))(Cl_(2(g)))+MnCl_(2)+2H_(2)O`
`underset((A))(Cl_(2(g)))+3F_(2("excess")) overset(573K)rarr underset((B))(2ClF_(3(g))`
`underset((B)"")(3ClF_(3(l))+U_((s))) overset(Delta)rarr underset((C)"")(UF_(6(g)))+underset((D)"")(3ClF_(3(g))`
Discussion
38.

MnO_(2) + HCl overset(Delta) to A_((g)) + MnCl_(2) + 2H_(2)O A_((g)) + F_(2) ("excess") overset (573 K) to B_((g)) B_((g)) + U((s)) to C_((g)) + D_((g)) The gases A, B, C and D are respectively

Answer»

`Cl_(2), ClF, UF_(6),ClF_(3)`
`Cl_(2),ClF_(3), UF_(6),ClF`
`O_(2),OF_(2), U_(2)O_(3), O_(2)F_(2)`
`O_(2),O_(2)F_(2),U_(2)O_(3),OF_(2)`

Solution :
Discussion
39.

MnO_(2) and Ca_(3)(PO_(4))_(2) present in iron ore get reduced to Mn and P in the zone of-

Answer»

combustion
reduction
FUSION
SLAG FORMATION

ANSWER :C
Discussion
40.

MnO_(2) and Ca_(3)(PO_(4))_(2) present in iron are get reduced to Mn and P in the zone of :

Answer»

COMBUSTION `
Reduction
FUSION
SLAG formation

ANSWER :B
Discussion
41.

MnO_(2) and H_(2)SO_(4) added to NaCl, the greenish yellow gas liberated is

Answer»

`Cl_(2)`
`NH_(3)`
`N_(2)`
`H_(2)`

Solution :Yellowish-green gas (chlorine) with suffocating odour is evolved when the sodium chloride mixed with manganese dioxide is HEATED with concentrated `H_(2)SO_(4)`.
`NaCl+H_(2)SO_(4)toNaHSO_(4)+HCl`
`MnO_(2)+4HCL toMnCl_(2)+2H_(2)O+Cl_(2)`
Discussion
42.

[Mn(CO)_(4)NO] is diamagnetic because:

Answer»

Mn metal is diamagnetiic in free state
Mn is in +1 oxidation state in this complex
No is present as positive ligand
All of these

Solution :`[OVERSET(+1)(Mn)(CO)_(4)(NO^(-))]`: Paramagnetic due to presence of UNPAIRED `e^(-)` s in `NO^(-)`
`[overset(-1)(Mn)(CO)_(4)(NO^(+))]:`No unpaired `e^(-)` EITHER on LIGANDS or on `overset(-1)(Mn)` hence it is diamagnetic.
Discussion
43.

The two electrons in helium atom:

Answer»

3D, 4s
4s, 3d
4f, 3d
3d, 4f

Answer :B
Discussion
44.

An atom has its K and L shells completely filled and five electrons in M shell. Find out the total number of p electrons in the atom ?

Answer»

3D, 4s
4s, 3d
4f, 3d
3d, 4f

Answer :B
Discussion
45.

Mn_(3)O_(4) is a mixed oxide of :

Answer»

`MnO, Mn_(2)O_(3)`
`MnO, MnO_(2)`
`MnO, MnO_(3)`
`MnO_(2), Mn_(2)O_(3)`

Answer :B
Discussion
46.

Mn_(2)O_(7) is

Answer»

ACIDIC oxide
basic oxide
neutral oxide
amphoteric oxide

Answer :A
Discussion
47.

Mn_(2)O_(7) dissolves in water to give ....

Answer»

SOLUTION :permanganic ACID` (HMnO_(4))`
Discussion
48.

Mn_2O_7, CrO_3, V_2O_5 are____oxides

Answer»

Basic
Acidic
Neutral
Amphoteric

Answer :B
Discussion
49.

Mn^(2+), Fe^(3+)have high magnetic moment. Prove it.

Answer»

Solution :(i) `Mn^(2+), Fe^(3+)` configuration is `d^(5)`
(ii) `MU= sqrt(5(5+2))= sqrt(35)= 5.916 mu_(B)`
AMONG 3d series, `Mn^(2+), Fe^(3+)` have HIGH magnetic moment as 5.916 Hg.
Discussion
50.

Mn^(2+) is more stable than Mn^(4+). Why?

Answer»

SOLUTION :(i) The relative STABILITY of different oxidation states of 3d metals is correlated with the extra stability of half-filled and FULLY filled ELECTRONIC configurations.
(ii) Example: `Mn^(2+)(3d^(5))` is more stable than `Mn^(4+)(3d)`
Discussion