The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 55

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

Chloride of which of the following elements will be coloured:

A
Silver

B
Mercury

C
Zinc

D
Cobalt

Solution

Chloride of Co elements will be coloured due to presence of unpaired electron.
Question 2
1 / -0

In which of the following. tendency towards formation of coloured ions is maximum

A
s-block elements

B
d-block elements

C
p-block elements

D
f-block elements

Solution

Tendency towards formation of coloured ions is maximum is maximum in d-block elements.
Hence, Option "B" is the correct answer.
Question 3
1 / -0

Arrange $Ce^{3+} , La^{3+} ,Pm^{3+}$ and $Yb^{3+}$ in increasing order of their ionic radii

A
$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$

B
$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$

C
$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$

D
$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$

Solution

Due to Lanthanoid contraction order will be: $Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+} .$
Question 4
1 / -0

Transition metals show paramagnetism

A
due to characteristic configuration

B
high lattice energy

C
due to variable oxidation states

D
due to unpaired electrons

Solution

Paramagnetic character is actually due to the presence of unpaired electrons. Transition metals show paramagnetism due to one or more number of unpaired electrons.
Question 5
1 / -0

Among the following which exhibits hydrogen, covalent, and coordinate bonding?

A
$MnSO_4 . 4H_2 O$

B
$CuSO_4 . 5H_2 O$

C
$FeSO_4 . 6H_2 O$

D
$FeSO_4. 5H_2 O$

Solution

$CuSO_4$ exhibit coordinate, covalent and hydrogen bonding.
Hence, Option "B" is the correct answer.
Question 6
1 / -0

Which of the following pair of elements cannot form an alloy ?

A
$Zn - Hg$

B
$Fe - Hg$

C
$Hg - Cr$

D
$Na - Hg$

Solution

Hg forms alloys with all metals except Iron, Platinum and Tungsten.
Hence, Option "B" is the correct answer.
Question 7
1 / -0

Iron loses magnetic property at

A
Melting point

B
1000K

C
Curie point

D
Boiling point

Solution

Iron loses magnetic property at Curie point. The temperature at which a certain material loses its permanent magnetic properties is known as the Curie temperature.
Question 8
1 / -0

The term platforming is

A
Platinum painting

B
Flat sheet of platinum

C
Platinum manufacturing

D
Platinum used as a catalyst

Solution

The term platforming is referred to use of Platinum as a catalyst in petrochemical plant.
Question 9
1 / -0

If $CdI_2$ is pink in colour, the $CdCl_2$ will be '________' coloured.

A
yellow

B
red

C
blue

D
cannot be predicted

Solution

In $CdI_2$ Pink colour is due to polarization of $I^-$ by $Cd^{2+}$ has pseudonoble gas configuration. In $CdCl_2$ , extent of polarization of $Cl^-$ by $Cd^{2+}$ is relatively less, hence specific colour cannot be predicted.
Question 10
1 / -0

Which metal absorbs hydrogen?

A
$Al$

B
$Fe$

C
$Pd$

D
$K$

Solution

Palladium is a unique material with a strong affinity to hydrogen owing to both its catalytic and hydrogen absorbing properties.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 55

Result

The d- and f- Block Elements Test - 55

Score

-

Rank

-

SHARING IS CARING

Question 1

Silver

Mercury

Zinc

Cobalt

Solution

Question 2

s-block elements

d-block elements

p-block elements

f-block elements

Solution

Question 3

Yb3+<Pm3+<Ce3+<La3+ Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+} Yb3+<Pm3+<Ce3+<La3+

Ce3+<Yb3+<Pm3+<La3+ Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+} Ce3+<Yb3+<Pm3+<La3+

Yb3+<Pm3+<La3+<Ce3+ Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+} Yb3+<Pm3+<La3+<Ce3+

Pm3+<La3+<Ce3+<Yb3+ Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+} Pm3+<La3+<Ce3+<Yb3+

Solution

Question 4

due to characteristic configuration

high lattice energy

due to variable oxidation states

due to unpaired electrons

Solution

Question 5

MnSO4.4H2O MnSO_4 . 4H_2 O MnSO4​.4H2​O

CuSO4.5H2O CuSO_4 . 5H_2 O CuSO4​.5H2​O

FeSO4.6H2O FeSO_4 . 6H_2 O FeSO4​.6H2​O

FeSO4.5H2O FeSO_4. 5H_2 O FeSO4​.5H2​O

Solution

Question 6

Zn−Hg Zn - HgZn−Hg

Fe−Hg Fe - Hg Fe−Hg

Hg−Cr Hg - Cr Hg−Cr

Na−Hg Na - Hg Na−Hg

Solution

Question 7

Melting point

1000K

Curie point

Boiling point

Solution

Question 8

Platinum painting

Flat sheet of platinum

Platinum manufacturing

Platinum used as a catalyst

Solution

Question 9

yellow

red

blue

cannot be predicted

Solution

Question 10

AlAlAl

FeFeFe

PdPdPd

KKK

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.

$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$

$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$

$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$

$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$

$MnSO_4 . 4H_2 O$

$CuSO_4 . 5H_2 O$

$FeSO_4 . 6H_2 O$

$FeSO_4. 5H_2 O$

$Zn - Hg$

$Fe - Hg$

$Hg - Cr$

$Na - Hg$

$Al$

$Fe$

$Pd$

$K$