MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 7

A transition element has an electronic configuration such that 4s orbital has only one electron. Which of the following is correct?

0%
The element is $Mg$ and contains one electron in each of its 3d orbital
0%
The element is $Mg$ and contains an empty 3d orbital
0%
The element if $Cr$ and contains one electron in each of its 3d orbital
0%
None of these

Explanation

The electron configuration of Chromium

$(Cr)$ is

$[Ar] {3d}^{5}4s^{1}$

Electron configuration of Magnesium

$(Mg)$ is

$[Ne] 3s^{2}$

What is geometry of chromate ion?

0%
Tetrahedral
0%
Octahedral
0%
Trigonal planer
0%
Linear

Explanation

The chromate ion

$(CrO^{2-}_4)$ has a tetrahedral shape, whereas dichromate ion has two tetrahedral units with one common oxygen having

$Cr-O-Cr$ bond angle of

$126^o$ .

In which pair highest oxidation states of transition metals are found?

0%
Nitrides and Chlorides
0%
Fluorides and Chlorides
0%
Fluorides and Oxides
0%
Nitrides and Oxides

____________ is used in gas lamp material.

0%
$MnO_2$
0%
$CeO_2$
0%
$N_2O_5$
0%
$Fe_2O_3$

Explanation

Cerium dioxide

$\displaystyle CeO_2$ is used in gas lamp material. The oxides of lanthanum, cerium, yttrium and magnesium are used for gas lamp materials.

Silver salt used in photography is:

0%
$AgCl$
0%
$AgNO_3$
0%
$AgF$
0%
$AgBr$

Explanation

Silver salt used in photography is silver bromide

$AgBr$ . On exposure to light,

$AgBr$ is reduced to black silver.

$\displaystyle 2AgBr (s) \xrightarrow {light} 2Ag(s) +Br_2(l)$

Bromide ion transfers an electron to silver ion forming silver metal.

Hence, option

$D$ is correct.

The metals present in Nichrome alloy is:

0%
$Cr,Ni,Fe$
0%
$Cr,Co,Ni$
0%
$Cr,Fe$
0%
$Cr,Fe,Cu$

Explanation

Nichrome alloy is an alloy of nickel, chromium and iron. It is used in electric heating elements in the application such as hairdryer, ironing, heat guns, etc.

Hence, option

$A$ is correct.

Fuel used in nuclear power plants is:

0%
Uranium - 235
0%
Plutonium - 235
0%
Plutonium - 238
0%
Uranium - 238

Which of the following compounds will not give

$+ve$ chromyl chloride test?

0%
$CuCl_2$
0%
$HgCl_2$
0%
$ZnCl_2$
0%
$C_6H_5NH_{2}HCl$

Which of these ions is expected to be coloured in aqueous solution?
I.

${Fe}^{3+}$
II.

${Ni}^{2+}$
III.

${Al}^{3+}$

0%
I and II
0%
II and III
0%
I and III
0%
I, II and III

Explanation

$Fe^{2+}\ : \ [Ar] 3d^6$ , i.e., 4 unpaired electrons

$\therefore$ Coloured ion, [ref. image I]

II.

$Ni^{2+} \ : \ [Ar]3d^8$ , i.e., 2 unpaired electrons

$\therefore$ Coloured ion, [ref. image II]

III.

$Al^{3+}=[Ne]$ , i.e., no unpaired electron in 3d, colourless ion.

Which metal oxide is used to obtain blue coloured glass?

0%
Ferric oxide
0%
Chromium oxide
0%
Manganese oxide
0%
Cobalt oxide

Which of the following is true regarding derivation of the name of californium?

0%
The name of californium was derived from the name of the place.
0%
The name of californium was derived from its color.
0%
The name of californium was derived from the name of the scientist who discovered it.
0%
The name of californium was derived from mythological character name.

The only radioactive element among the lanthanoids is:

0%
gadolinium
0%
holmium
0%
promethium
0%
neodynium

Which of the following weights less when weighted in magnetic field?

0%
$Sr{ Cl }_{ 3 }$
0%
$Fe{ Cl }_{ 3 }$
0%
$Ti{ Cl }_{ 3 }$
0%
$V{ Cl }_{ 3 }$

Explanation

$Sn{ Cl }_{ 3 }\longrightarrow { Sc }^{ 3+ }+3{ Cl }^{ - }\quad$
No unpaired electron so, will show diamagnetic character so will weights less when, weighted in magnetic field.

Which of following ion has the maximum theoretical magnetic moment?

0%
$Cr^{3+}$
0%
$Fe^{3+}$
0%
$Ti^{3+}$
0%
$Co^{3+}$

Explanation

$Fe^{3+}$
n=5 (unparied electrons)

$\therefore \mu =\sqrt{n(n+2)}BM$

$=\sqrt {5(5+2)}$

$=\sqrt{35}= 5.91 \, BM$

Which lanthanide compound is used as a pigment?

0%
$CeO_{2}$
0%
$Ce(OH)_{3}$
0%
$Lu(OH)_{3}$ $
0%
$Tb(OH)_{3}$

Explanation

Ceria or cerium oxide,

$CeO_{2}$ , a lanthanide compound is used as a pigment and as a polishing agent for glass.

Which one of the following forms a colourless solution in an aqueous medium?

[Atomic number

$Sc=21, Ti=22, V=23, Cr=24$ ]

0%
${ V }^{ 3+ }$
0%
${ Cr }^{ 3+ }$
0%
${ Ti }^{ 3+ }$
0%
${ Sc }^{ 3+ }$

Explanation

The electronic configuration of

$Sc$ is:

$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 2 },3{ p }^{ 6 },3{ d }^{ 1 },4{ s }^{ 2 }$

Electronic configuration of

${ Sc }^{ 3+ }$ is:

$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 6 },3{ s }^{ 2 },3{ p }^{ 6 }$

So,

${ Sc }^{ 3+ }$ is colourless due to the absence of unpaired electrons in

$d$ - orbital.

Hence, option

$D$ is correct.

The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition elements which shows highest magnetic moment?

0%
$3d^2$
0%
$3d^8$
0%
$3d^7$
0%
$3d^5$

Explanation

The magnetic nature of elements depends on the presence of unpaired electrons.

$\displaystyle 3d^5$ configuration of transition elements which shows highest magnetic moment as it has maximum number of unpaired electrons (5 unpaired electrons).

How is sodium chromate converted into sodium dichromate in the manufacture of potassium dichromate from chromite ore?

0%
By the action of concentrated sulphuric acid
0%
By roasting with soda ash
0%
By the action of sodium hydroxide
0%
By the action of lime stone

Explanation

In the manufacture of potassium dichromate from chromite ore, sodium chromate is converted into sodium dichromate by the action of concentrated sulphuric acid.

$\displaystyle 2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O$

The basic character of the transition metal monoxides follows the
(At. no. of

$Ti=22,V=23,Cr=24,Fe=26$ )

0%
$TiO> VO> CrO> FeO$
0%
$VO> CrO> TiO> FeO$
0%
$CrO> VO> FeO> TiO$
0%
$TiO> FeO> VO> CrO$

Explanation

The basic character of oxide decreases from left to right in a period of Periodic Table. Thus, the correct order is:

$TiO> VO> CrO> FeO$ .

Metals are usually used as catalysts belong to:

0%
s-block
0%
p-block
0%
d-block
0%
f-block

Identify the metal that forms colourless compounds.

0%
Iron (Z = 26)
0%
Chromium (Z = 24)
0%
Vanadium (Z = 23)
0%
Scandium (Z = 21)

Explanation

$\displaystyle Scandium (Z = 21)$ forms colourless compounds.

$\displaystyle Scandium (Z = 21)$ has valence shell electron configuration of

$\displaystyle 3d^1 \: 4s^2$

In its

$\displaystyle +3$ oxidation state, it looses 3 electrons and has valence shell electron configuration of

$\displaystyle 3d^0 \: 4s^0$ . Since no unpaired electron is present, it forms colourless compounds.

Select the coloured compound amongst the following:

(Atomic no.

$Ti = 22, Cr = 24, Cu = 29, Zn = 30$ )

0%
$TiCl_4$
0%
$CrCl_3$
0%
$ZnCl_2$
0%
$CuCl$

Explanation

TiCl

$_4$

Electronic configuration of Ti is

$[Ar]3d^2 4s^2$ .

For Ti

$^{4+}$ -

$[Ar]3d^04s^0$

Since no electron is present, so it is colourless.

CrCl

$_3$

Electronic configuration of Cr is

$[Ar]3d^54s^1$

For Cr

$^{3+}$ -

$[Ar]3d^34s^0$ .

Since there is a presence of d-electrons, so it is coloured. There is partially filled d-orbital so due to d-d transition colour is imparted to this compound.

ZnCl

$_2$

Electronic configuration of Zn -

$[Ar]3d^{10}4s^2$

For Zn

$^{2+}$ -

$[Ar]3d^{10}4s^0$

Because of fully filled d orbital, the compound is colourless.

CuCl

Electronic configuration of Cu

$^+$ is

$3d^{10}4s^0$

Due to fully filled d-orbital, it is colourless.

Identify a 'Chemical twin' among the followings.

0%
$Zr-Ta$
0%
$Nb-Tc$
0%
$Hf-Re$
0%
$Nb-Ta$

Spin only magnetic moment value of

${ Cr }^{ 3+ }$ ion is:

0%
$2.87$ BM
0%
$3.87$ BM
0%
$3.47$ BM
0%
$3.67$ BM

Explanation

${ Cr }^{ 3+ }=3{ d }^{ 3 }$ ;

Hence,

$\mu =\sqrt { n\left( n+2 \right) }$

$=\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3.87$ BM

Hence, option

$B$ is correct.

What is the atomic number of the element with

$M^{2+}$ ion having electronic configuration

$[Ar]3d^{8}$ ?

0%
$25$
0%
$28$
0%
$27$
0%
$26$

Explanation

$E.C$ of

$M : [Ar]4s^{2}3d^{8}$

$E.C$ of

$M^{2+} : [Ar]4s^{0}3d^{8}$

Total electrons

$= 28 =$ atomic number.

Select the coloured and paramagnetic ions.

0%
$Cu^+, Zn^{2+}, Ca^{2+}$
0%
$Sc^{3+}, Ti^{4+}, V^{5+}$
0%
$Cu^{2+}, Cr^+, Mn^{2+}$
0%
$Ni^{2+}, Cu^+, Hg^{2+}$

Explanation

Ion is coloured if there are electrons in d-suborbit. Paramagnetic nature is also due to unpaired electrons. Thus, every coloured ion is also paramagnetic.

$Cu^{2+}=[Ar]3d^9$ , one unpaired electron in

$3$ d.

$Cr^+=[Ar]3d^5$ , five unpaired electrons in

$3$ d.

$Mn^{2+}=[Ar]3d^5$ five unpaired electrons in

$3d$ .

Experimental value of magnetic momentum of

${ Mn }^{ 2+ }$ complex is

$5.96$ B.M. This indicates:

0%
Axial and orbital motion of electron in same direction
0%
Axial and orbital motion of electron in opposite direction
0%
Electron does not exhibit orbital motion, it only exhibits axial motion
0%
Electron does not exhibit axial motion, it only exhibits orbital motion

Explanation

$\sqrt {n(n+2)}= 5.96$

$n=5$

The axial motion of electron is clockwise because

$S=+1/2$ and the orbital motion will also be in clockwise direction.

The highest magnetic moment is shown by the transition metal ion with outer electronic configuration:

0%
${ d }^{ 9 }$
0%
${ d }^{ 7 }$
0%
${ d }^{ 5 }$
0%
${ d }^{ 3 }$

Explanation

The highest magnetic moment is shown by the transition metal ion with outer electronic configuration

${ d }^{ 5 }$ as it has maximum (five) number of unpaired electrons. Higher is the number of unpaired electrons, higher is the magnetic moment.

$\displaystyle \mu_{eff}=\sqrt {n(n+2)}$ B.M.

${ d }^{ 3 }$ ,

${ d }^{ 7 }$ and

${ d }^{ 9 }$ have 3, 3 and 1 unpaired electrons respectively.

Ammonia will not form complex with:

0%
${ Ag }^{ 2+ }$
0%
${ Pb }^{ 2+ }$
0%
${ Cu }^{ 2+ }$
0%
${ Cd }^{ 2+ }$

There are

$14$ elements in actinoid series. Which of the following elements does not belong to this series?

0%
$U$
0%
$Np$
0%
$Tm$
0%
$Fm$

Explanation

Explanation:

Actinoid series is a series of $14$ elements.
The name of the series is given by the name of the first element actinium.
Actinoid series has elements from atomic number $90$ to $103$ .
Uranium $U$ has atomic number $92$ .
Neptunium $Np$ has atomic number $93$ .
Fermium $Fm$ has atomic number $100$ .
Thulium $Tm$ has atomic number $69$ .
Hence, $Tm$ does not belong to the actinoid series.

Hence, the correct answer is option $C$ .

Due to lanthanide contraction :

0%
$Fe, Co, Ni$ have equal size
0%
$Zr$ and $Hf$ have equal size
0%
all $f-$ block ions have equal size
0%
all isoelectronic ions have equal size

Explanation

The lanthanide contraction cancels almost exactly the normal size increase on descending a group of transition elements. Thus,

$Zr$ and

$Hf$ have equal size.

Generally transition elements for coloured salts due to the presence of unpaired electrons. Which of the following compound will be coloured in solid state?

0%
$Ag_2SO_4$
0%
$CuF_2$
0%
$Cu_2Cl_2$
0%
$ZnF_2$

Explanation

A salt is coloured if transition ion contains unpaired d-electrons.

$Ag^+ - [Kr] 4d^{10}$

$Cu^{2+} - [Ar] 3d^9$

$Zn^{2+} - [Ar] 3d^{10}$

$Cu^+ - [Ar] 3d^{10}$

$CuF_2, Cu$ is

$+ 2$ contains an unpaired electron.

Hence, it is a coloured compound.

Which of the following ions has a magnetic moment of 5.93 BM?

[At . No. : V = 23 , Cr = 24, Mn = 25, Fe = 26]

0%
$Mn^{2+}$
0%
$Fe^{2+}$
0%
$Cr^{2+}$
0%
$V^{3+}$

Explanation

Magnetic moment

$= \sqrt{n(n+2)} BM$

where n = number of unpaired electrons

$5.93 = \sqrt{n(n+2)}$

$n = 5$

$Mn^{2+}$ ion

$(3d^5)$ has

$5$ unpaired electrons and thus, magnetic moment is

$5.93 BM$

Hence, option

$A$ is correct.

The type of magnetism exhibited by

$[Mn(H_2O)_6]^{2+}$ ion is:

0%
paramagnetism
0%
diamagnetism
0%
both (a) and (b)
0%
none of these

Explanation

The given complex is [Mn (H

$_2O)_6]^{2+}$

The electronic configuration of Mn is [Ar]

$3d^5 4s^2$

for Mn

$^{2+}$

$\rightarrow$ {Ar]

$3d^5 4s^0$

[Ar]

$\upharpoonleft$

3d 4s

3d 4s 4p 4d

So, [Mn (H

$_2O)_6]^{2+}$

$\rightarrow$ [Ar]

$\upharpoonleft$

$\upharpoonleft$ xx xx xx xx xx xx

$H_2O$

$^3 d^2$ hybridised

So, due to presence of 5 unpaired d

$e^-$ s, it is paramagnetic.

Which of the following compound is expected to be coloured?

0%
$Ag_2SO_4$
0%
$CuF_2$
0%
$MgF_2$
0%
CuCl

Explanation

$Ag_2SO_4$ contain

$Ag^+$ ion having outer electronic configuration of

$4d^{10}$ with 0 unpaired electrons and is colourless.

$CuF_2$ contains

$Cu^{2+}$ ion having outer electronic configuration of

$3d^9$ with 1 unpaired electron and is coloured.

$MgF_2$ contains

$Mg^{2+}$ with 0 unpaired electrons having outer electronic configuration of

$3s^2$ and is colourless.

$CuCl$ contains

$Cu^+$ ion having outer electronic configuration of

$3d^{10}$ with 0 unpaired electrons and is colourless.

Hence, option

$B$ is correct.

The spin only magnetic moment of

$Fe^{3+}$ ion (in BM) is approximately:

0%
4
0%
7
0%
5
0%
6

Explanation

$_{26}Fe=[Ar]3d^64s^2$

$Fe^{3+}=[Ar]3d^5$

Number of unpaired electrons

$= 5$

$\mu=\sqrt{n(n+2)}$

$\mu =\sqrt{5\times (5+2)}$

$\mu=\sqrt{35}= 5.91\, BM$

$\mu= 6\, BM$

Magnetic moment of

$[Cu(NH_3)_4]^{2+}$ ion is?

0%
1.414
0%
1.73
0%
2.23
0%
2.38

Explanation

The given complex is

$[Cu(NH_3)_4]^{2+}$

The electronic configuration of Cu is [Ar]

$3d^{10} 4s^1$

for Cu

$^{2+}$

$\rightarrow$ [Ar]

$3d^{9} 4s^0$

[Ar]

$\upharpoonleft \downharpoonright$

$\upharpoonleft$

3d 4s

3d 4s 4p

So, [Cu(NH

$_3)_4]^{2+}$

$\rightarrow$ [Ar]

$\upharpoonleft \downharpoonright$

$\upharpoonleft \downharpoonright$ xx xx xx xx

$\upharpoonleft$

$NH_3$

$NH_2$

$NH_3$

dsp

$^2$ hybridisation

No. of unpaired

$e^-$ s =1

Magnetic moment=

$\sqrt{n(n+2)}$ BM

Where, n= no.of unpaired

$e^-$ s

So, magnetic moment =

$\sqrt{1(1+2)} =\sqrt{3}$ = 1.73 BM

Which of the following is paramagnetic in nature?

0%
$V(CO)_6$
0%
$FeCO_5$
0%
$Fe_2(CO)_9$
0%
$Cr(CO)_6$

Explanation

i) V(CO)

$_6$

Oxidation state of V is 0.

So, electronic configuration [Ar]

$3d^34s^2$ .

[ 6 coligands ]

[Ar]

$\upharpoonleft$

$\downharpoonright$

$\rightarrow$ [Ar]

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$ xx xx xx xx xx xx

3d 4s 3d 4s 4p

Because of the presence of 1 unpaired e

$^-$ they are paramagnetic.

ii)

$Fe(CO)_5$

The oxidation state of Fe is 0.

So, electronic configuration [Ar]

$3d^64s^2$ .

[Ar]

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$

3d 4s

When co (ligands) approaches: [ 5 (CO) co-ligands ]

[Ar]

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$ xx xx xx xx xx

3d 4s 4p

So , no unpaired e

$^-$ present and it is diamagnetic.

iii)

$Cr(CO)_6$

Electronic configuration of Cr is [Ar]

$3d^54s^2$

[Ar]

$\upharpoonleft$

3d 4s

When co (ligands) approach:

[Ar]

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$

$\upharpoonleft$

$\downharpoonright$ xx xx xx xx xx xx

3d 4s 4p

So, no unpaired e

$^-$ present and it is diamagnetic.

Metallic radii of some transition elements are given below. Which of these elements will have the highest density?
Element

$Fe$

$CO$

$Ni$

$Cu$
Metallic radii/pm 126 125 125 128

0%
same
0%
$Ni$
0%
$Co$
0%
$Cu$

Explanation

As we all know that on moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.
Hence, among the given four choices

$Cu$ belongs to right side of Periodic Table in transition metal, and it must have highest density. So the correct option is D

Which of the following arrangement does not represent the correct order of property stated against it ?

0%
$V^{2+} < Cr^{3+} < Mn^{2+} < Fe^{2+}$ : Paramagnetic behaviour
0%
$Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$ Ionic size
0%
$Co^{3+} < Fe^{3+} , Cr^{3+} < Sc^{3+}$ : Stability in aqueous solution
0%
$Sc < Ti < Cr < Mn$ : Number of oxidation states

Explanation

Paramagnetic property depends upon the number of unpaired electrons.

As the number of unpaired electrons unpaired electron increases, the paramagnetic property increases.

${ Fe }^{ 2+ }\Rightarrow n=4\quad \quad { Mn }^{ 2+ }\Rightarrow n=5$

$\Rightarrow \quad { Mn }^{ 2+ }>{ Fe }^{ 2+ }$ paramagnetic behaviour.

Which of the following example is not correctly matched?

0%
Two most abundant element -Fe, Al.
0%
Two metals which occur in native state -Au, Pt.
0%
Two metals which can occur in combined and native state both - Zn, Fe.
0%
None of these.

Explanation

A few elements like carbon, sulphur, gold, and noble gases, occur in the free state while others are found in combined forms in the earth’s crust.
Elements vary in abundance. Among metals, aluminium is the most abundant. Iron is the second most abundant metal in the earth’s crust. It forms a variety of compounds and their various uses make it a very important element. It is one of the essential elements in biological systems as well.
$Fe$ and $Zn$ is not available in both combined and native state and hence option C is the correct answer.

Some compounds turn dark when exposed to light. They are, therefore, used in photography. A metal used for making such compounds is:

0%
aluminium
0%
nickel
0%
chromium
0%
silver

Magnetic moment of

${ x }^{ n+ }$ is

$\sqrt { 24 }$ B.M. Hence No. of unpaired electron and value of n is _____ respectively.
(Atomic number =26)

0%
4, 3
0%
3, 5
0%
4, 2
0%
4, 1

Explanation

We know that spin magnetic moment is given by

$\sqrt{n(n+1)}$

where ,n is the number of unpaired electron in the element

So, according to the question

$\sqrt{n(n+1)}=\sqrt{24}$

Squaring both side

${n(n+1)}={24}$

$n^2+2n-24=0$

$(n-4)(n+6)=0$

$n=4$

$\therefore$ Number of unpaired electron

$^{26}X:1s^22s^22p^63s^23p^63d^64s^2$

Since number of unpaired electrons are 4 so, from

$^{26}X$ , 2 electrons need to be removed.

then

$X^+:1s^22s^22p^63s^23p^63d^6$

Generally metal iron and their salts are coloured due to presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

0%
$Ce(SO_{4})_{2}$
0%
$TiCl_{4}$
0%
$Cu_{2}Cl_{2}$
0%
None of these

Explanation

${ Ce\left( { SO }_{ 4 } \right) }_{ 2 }$ is coloured. If it has unpaired electrons

${ Ce }^{ 4+ }$ is central metal atom

$\downarrow$

${ Ce }^{ 4+ }\left( Z=58 \right) \rightarrow \left[ Ar \right] { 3d }^{ 10 }{ 4P }^{ 6 }{ 5S }^{ 2 }{ 4d }^{ 10 }{ 5P }^{ 6 }{ 6S }^{ 2 }{ 4f }^{ 2 }$

$\left[ Ar \right] { 3d }^{ 10 }{ 4P }^{ 6 }{ 5S }^{ 2 }{ 4d }^{ 10 }{ 5P }^{ 4 }{ 6S }^{ 0 }{ 4f }^{ 2 }$

unpaired electrons

$=n=4$

$\rightarrow$ Coloured due to presence of unpaired

${ e }^{ - }s$ .

1044949_877271_ans_dbff5e0b7abb49fa9734e3cfbf1383fc.jpg

When

$MnO_{2}$ is fused with KOH in the presence of air, a coloured compound is formed, the product and its colour is :

0%
$K_{2}MnO_{4}$ , dark green
0%
$KMnO_{4}$ , purple
0%
$Mn_{2}O_{3}$ , brown
0%
$Mn_{3}O_{4}$ , black.

Explanation

The reaction between MnO

$_2$ and KOH in presence of air is given as:

$2MNO_2 +4KOH +O_2$

$\rightarrow$

$2K_2 MnO_4 +2H_2O$

(potassium manganate)

The

$K_2Mn O_4$ formed is dark green in colour.

Which of the following reactions are disproportionation reactions?
(I)

$Cu^+ \, \rightarrow \, Cu^{2+} \, + \, Cu$
(II)

$\displaystyle 3Mn{ O }{_{4}{^-}} \, + \, 4H^+ \, \rightarrow \, 2Mn{ O }{_{4}^{-}} \, + \, MnO_2 \, + \, 2H_2O$
(III)

$2KMnO_4 \, \rightarrow \, K_2MnO_4 \, + \, MnO_2 \, + \, O_2$
(IV)

$2Mn{ O }{_{4}{^-}} \, + \, 3Mn^{2+} \, + \, 2H_2O \, \rightarrow \, 5MnO_2 \, + 4H{^+}$

0%
(I), (II)
0%
(I), (II), (III)
0%
(II), (III), (IV)
0%
(I), (IV)

Explanation

Disproportionation reaction are the reactions in which the same element/compound get oxidized and reduced simultaneously.

Since I and II reactions are as follows

$2Cu^{+}\rightarrow Cu^{2+}+ Cu^{0}$

And

$3MnO_{4}^{2-}4H^{+} \rightarrow 2 MnO_{4}^{-} + MnO_{2}+ 2 H_{2}O$

Other two reactions are correct and belong to comproportionation reaction.

Thus option A is correct.

There are 14 elements in actinoid series. Which of the following elements does not belong to this series?

0%
$U$
0%
$Np$
0%
$Tm$
0%
$Pm$

Explanation

Thulium

$(Tm)$ has atomic number 69, and actinoid series has elements from atomic number 90-103.

Thulium belongs to lanthanoid series .

Hence, option C is correct.

When

$KMnO_4$ solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because:

0%
$CO_2$ is formed as the product
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reaction is exothermic
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$Mn{ O }_{4}^{-}$ catalyses the reaction
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$Mn^{2+}$ acts as autocatalyst

Explanation

Here, KMnO4 is a reactant in the reaction which forms

$Mn^{2+}$ which acts as an auto-catalyst. The reaction is slow initially because

${MnO_4}^-$ is getting converted to

$Mn^{2+}$ but once

$Mn^{2+}$ forms, the reaction becomes faster due to the catalytic action.

Hence, option D is correct.

Zr and Hf have almost equal atomic and ionic radii because of:

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diagonal relationship.
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lanthanoid contraction.
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actinoid contraction.
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belonging to the same group.

Explanation

This is because of poor screening effect by 4f electrons,which lead to the lanthanoid contraction.Hence,is the reason of

$Zn$ and

$Hf$ have almost equal atomic size.

Due to lanthanoid contraction which of the following properties is not expected to be similar in the same vertical columns of second and third row transition elements?

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Atomic radii
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Ionisation energies
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Magnetic moments
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Lattice energies

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 7 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 7 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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