The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 35

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following is paramagnetic in nature?

A
$V(CO)_6$

B
$FeCO_5$

C
$Fe_2(CO)_9$

D
$Cr(CO)_6$

Solution

i) V(CO) $_6$
Oxidation state of V is 0.
So, electronic configuration [Ar] $3d^34s^2$ .
[ 6 coligands ]
[Ar] $\upharpoonleft$    $\upharpoonleft$    $\upharpoonleft$ $\upharpoonleft$ $\downharpoonright$ $\rightarrow$   [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] $\upharpoonleft$ $\downharpoonright$    $\upharpoonleft$    $\upharpoonleft$    $\upharpoonleft$    $\upharpoonleft$     $\upharpoonleft$ $\downharpoonright$
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] $\upharpoonleft$ $\upharpoonleft$ $\upharpoonleft$ $\upharpoonleft$ $\upharpoonleft$ $\upharpoonleft$
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.
Question 2
1 / -0

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

A
(I), (II)

B
(I), (II), (III)

C
(II), (III), (IV)

D
(I), (IV)

Solution

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.
Question 3
1 / -0

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

A
$CO_2$ is formed as the product

B
reaction is exothermic

C
$Mn{ O }_{4}^{-}$ catalyses the reaction

D
$Mn^{2+}$ acts as autocatalyst

Solution

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.
Question 4
1 / -0

Which of the following example is not correctly matched?

A
Two most abundant element -Fe, Al.

B
Two metals which occur in native state -Au, Pt.

C
Two metals which can occur in combined and native state both - Zn, Fe.

D
None of these.
Solution
- 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.
Question 5
1 / -0

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

A
aluminium

B
nickel

C
chromium

D
silver
Solution
- A silver halide (or silver salt) is one of the chemical compounds that can form between the element silver and one of the halogens.
- Silver halides are used in photographic film and photographic paper, including graphic art film and paper, where silver halide crystals in gelatin are coated on to a film base, glass or paper substrate. The gelatin is a vital part of the emulsion as the protective colloid of appropriate physical and chemical properties.
- Hence option D is correct answer.
Question 6
1 / -0

Which of the following is not an actinoid?

A
Curium (Z = 96)

B
Californium (Z = 98)

C
Uranium (Z = 92)

D
Terbium (Z = 65)
Question 7
1 / -0

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

A
4, 3

B
3, 5

C
4, 2

D
4, 1

Solution

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
$X^+:1s^22s^22p^63s^23p^63d^6$
Question 8
1 / -0

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

A
$Ag_2SO_4$

B
$CuF_2$

C
$ZnF_2$

D
$Cu_2Cl_2$

Solution

The respective transition elements exists in
$Ag^{+}-{4d}^{10}5s^0$ , (have completely filled d orbital)
$Cu^{+2}-3d^{9}4s^0$ , (have incompletely filled d orbital)
$Zn^{+2}-3d^{10}4s^0$ , (have completely filled d orbital)
$Cu^{+1}-3d^{10}4s^0$ (have completely filled d orbital),
Since, $Cu^{+2}$ has unpaired electron it will show electron transitions.
Hence will be coloured.
Question 9
1 / -0

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

A
$Fe$

B
$Ni$

C
$Co$

D
$Cu$

Solution

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 $Cu$ will have highest density.
Question 10
1 / -0

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

A
$U$

B
$Np$

C
$Tm$

D
$Pm$

Solution

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.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Element	$Fe$	$Co$	$Ni$	$Cu$
Metallic radii/pm	126	125	125	128

The d- and f- Block Elements Test - 35

Result

The d- and f- Block Elements Test - 35

Score

-

Rank

-

SHARING IS CARING

Question 1

V(CO)6V(CO)_6V(CO)6​

FeCO5FeCO_5FeCO5​

Fe2(CO)9Fe_2(CO)_9Fe2​(CO)9​

Cr(CO)6Cr(CO)_6Cr(CO)6​

Solution

Question 2

(I), (II)

(I), (II), (III)

(II), (III), (IV)

(I), (IV)

Solution

Question 3

CO2CO_2CO2​ is formed as the product

reaction is exothermic

MnO4−Mn{ O }_{4}^{-}MnO4−​ catalyses the reaction

Mn2+Mn^{2+}Mn2+ acts as autocatalyst

Solution

Question 4

Two most abundant element -Fe, Al.

Two metals which occur in native state -Au, Pt.

Two metals which can occur in combined and native state both - Zn, Fe.

None of these.

Solution

Question 5

aluminium

nickel

chromium

silver

Solution

Question 6

Curium (Z = 96)

Californium (Z = 98)

Uranium (Z = 92)

Terbium (Z = 65)

Question 7

4, 3

3, 5

4, 2

4, 1

Solution

Question 8

Ag2SO4Ag_2SO_4Ag2​SO4​

CuF2CuF_2CuF2​

ZnF2ZnF_2ZnF2​

Cu2Cl2Cu_2Cl_2Cu2​Cl2​

Solution

Question 9

FeFeFe

NiNiNi

CoCoCo

CuCuCu

Solution

Question 10

UUU

NpNpNp

TmTmTm

PmPmPm

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$V(CO)_6$

$FeCO_5$

$Fe_2(CO)_9$

$Cr(CO)_6$

$CO_2$ is formed as the product

$Mn{ O }_{4}^{-}$ catalyses the reaction

$Mn^{2+}$ acts as autocatalyst

$Ag_2SO_4$

$CuF_2$

$ZnF_2$

$Cu_2Cl_2$

$Fe$

$Ni$

$Co$

$Cu$

$U$

$Np$

$Tm$

$Pm$