The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 23

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

Ammonium dichromate on heating gives:

A
chromic oxide and nitrogen

B
chromic acid and ammonia

C
chromic acid and nitrogen

D
chromic oxide and ammonia

Solution

$$\underbrace {(NH_4)_2Cr_2O_7}_{Ammonium\quad Dichromate\quad(orange\quad colour)}\longrightarrow \underbrace {Cr_2O_3}_{chromic\quad oxide\quad(green\quad colour)}+4H_2O+\underbrace {N_2}_{Nitrogen\quad(colourless\quad gas)}$$
Question 2
1 / -0

Of the following metals, the most reactive metal is:

A
$$Fe$$

B
$$Ni$$

C
$$Pt$$

D
$$Au$$

Solution

According to reactive series,
Most reactive metal is Iron $$(Fe)$$
Reactivity order is: $$Fe>Ni>Au>Pt$$
Question 3
1 / -0

A compound of vanadium possesses a magnetic moment of 1.73 BM. The oxidation state of vanadium in this compounds is :

A
1

B
2

C
4

D
can not be predicted.

Solution

$$1.73=\sqrt {n(n+2})$$
$$n=1$$, where n is the number of unpaired electrons.
Thus, vanadium in oxidation state +4 has one unpaired electron.
Question 4
1 / -0

Knowing that the chemistry of lanthanids (Ln) is dominated by its +3 oxidation state, which of the following statements are correct?
(I) The ionic sized of Ln (III) decreases in general with increasing atomic number.
(II) Ln (III) compounds are generally colourless.
(III) Ln (III) hydroxides are mainly basic in character.
(IV) Because of the large size of the Ln (III) ions, the bonding in its compounds is predominently ionic in character.

A
I, III and III

B
II, III and IV

C
I, II and IV

D
I, III and IV

Solution

The ionic radii of trivalent lanthanides steadily decreases with increase in atomic number due to lanthanide contraction.
In lanthanides (At no. of elements 58 to 71) the electronic configuration of three outermost shelts are $$(n-2)^{f-14}, (n-1)s^2 p^6d^{10 - 1}ns^2$$.
So, most of the trivalent lanthanoid compounds except that of $$La^{+3}$$ and $$Lu^{+3}$$ are coloured, both in the solid state and in the aqueous solution. These ions are coloured due to the presence of unpaired f-electrons.
Due to small electronegativity and large size of the Ln (III) ions the bonding in its compounds is predominently ionic in character.
Question 5
1 / -0

The ionic radius of $$_{57}La^{3+}$$ is 1.06 A. Which one of the following given values will be closest to the ionic radius of $$_{71}Lu^{3+}?$$

A
1.06 A

B
1.56 A

C
0.85 A

D
1.35 A

Solution

As we move along a row radii decreases so ionic radius of Lu$$^{3+}$$ should be smaller than that of La$$^{3+}$$ because of lanthanide contraction across the period.
Question 6
1 / -0

Lanthanide contraction implies:

A
decrease in density

B
decrease in mass

C
decrease in ionic radii

D
decrease in radioactivity

Solution

Lanthanoid contraction effect results from poor shielding of nuclear charge (nuclear attractive force on electrons) by 4f electrons; the 6s electrons are drawn towards the nucleus, thus resulting in a smaller atomic radius.
Question 7
1 / -0

Directions For Questions

Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the Contact Process, vanadium compounds in the $$+5$$ state ($${ V }_{ 2 }{ O }_{ 5 }$$ or $${ VO }_{ 3 }^{ - }$$) are used to oxidise $${ SO }_{ 2 }$$ to $${ SO }_{ 3 }$$:

$${ { SO }_{ 2 } }+\cfrac { 1 }{ 2 } { O }_{ 2 }\xrightarrow [ ]{ { V }_{ 2 }{ O }_{ 5 } } { SO }_{ 3 }$$

It is thought that the actual oxidation process takes place in two stages. In the first step, $${ V }^{ 5+ }$$ is reduced to $${ V }^{ 4+ }$$.

$$2{ V }^{ 5+ }+{ O }^{ 2- }+{ { SO }_{ 2 } }\longrightarrow 2{ V }^{ 4+ }+{ SO }_{ 3 }\quad $$

In the second step, $${ V }^{ 5+ }$$ is regenerated from $${ V }^{ 4+ }$$ by oxygen:

$$2{ V }^{ 4+ }+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow 2{ V }^{ 5+ }+{ O }^{ 2- }$$

The overall process is, of course, the sum of these two steps:

$${ { SO }_{ 2 } }+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { SO }_{ 3 }$$

...view full instructions

Which of the following ion involved in the above process will show paramagnetism?

A
$${ V }^{ 5+ }$$

B
$${ V }^{ 4+ }$$

C
$${ O }^{ 2- }$$

D
$${ VO }_{ 3 }^{ - }$$

Solution

$${ V }^{ 4+ }$$ has $$3d^1$$ electronic configuration. It contains on unpaired electron. Hence, it is paramagnetic in nature.
Question 8
1 / -0

All $$Zn(+II)$$ compounds are white because:

A
$${ Zn }^{ 2+ }$$ has a $${d}^{10}$$ configuration and the $$d$$-subshell is full

B
$${ Zn }^{ 2+ }$$ shows $$d-d$$ transition

C
$${ Zn }^{ 2+ }$$ has no electron in the $$4s-$$subshell

D
$$Zn$$ is not a transition element

Solution

The electronic configuration of $$Zn^{2+}$$ is $$3d^{10}4s^0$$. All the electrons are paired.
Hence, all $$Zn^{2+}$$ compounds are colorless or white.
Question 9
1 / -0

Which of the following factors favour the formation of complexes by d-block elements?

A
Small size.

B
High nuclear charge.

C
Presence of low energy vacant orbitals to accept lone pair of electrons donated by ligands.

D
All of the above

Solution

The d-block elements are small in size and have high electropositive density. They consist of $$(n-1)d$$ free orbitals to accept the free electrons from the ligand and hence, form complexes easily.
Question 10
1 / -0

In $$Cr_2O^{2-}_7$$ every Cr is linked to :

A
two O atoms

B
three O atoms

C
four O atoms

D
five O atoms

Solution

In $$Cr_2O^{2-}_7$$ ion, every chromium atom is linked to four oxygen atoms. Two chromium atoms are present at the centre with $$Cr-O-Cr$$ linkage and each chromium atom is bonded to three more oxygen atom.So, each chromium atom is bonded to four $$O$$ atom.

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 - 23

Result

The d- and f- Block Elements Test - 23

Score

-

Rank

-

SHARING IS CARING

Question 1

chromic oxide and nitrogen

chromic acid and ammonia

chromic acid and nitrogen

chromic oxide and ammonia

Solution

Question 2

$$Fe$$

$$Ni$$

$$Pt$$

$$Au$$

Solution

Question 3

1

2

4

can not be predicted.

Solution

Question 4

I, III and III

II, III and IV

I, II and IV

I, III and IV

Solution

Question 5

1.06 A

1.56 A

0.85 A

1.35 A

Solution

Question 6

decrease in density

decrease in mass

decrease in ionic radii

decrease in radioactivity

Solution

Question 7

$${ V }^{ 5+ }$$

$${ V }^{ 4+ }$$

$${ O }^{ 2- }$$

$${ VO }_{ 3 }^{ - }$$

Solution

Question 8

$${ Zn }^{ 2+ }$$ has a $${d}^{10}$$ configuration and the $$d$$-subshell is full

$${ Zn }^{ 2+ }$$ shows $$d-d$$ transition

$${ Zn }^{ 2+ }$$ has no electron in the $$4s-$$subshell

$$Zn$$ is not a transition element

Solution

Question 9

Small size.

High nuclear charge.

Presence of low energy vacant orbitals to accept lone pair of electrons donated by ligands.

All of the above

Solution

Question 10

two O atoms

three O atoms

four O atoms

five O atoms

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.