The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 21

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

Variable valency of transition elements is on account of :

A
incomplete s - orbitals

B
incomplete d - orbitals

C
completely filled d - orbitals

D
incomplete p-orbitals

Solution

D-block elements show variable valency due to their incomplete d-orbitals.
Question 2
1 / -0

Which of the following pair has the same size?

A
Fe$$^{2+}$$, Ni$$^{2+}$$

B
Zr$$^{4+}$$, Ti$$^{4+}$$

C
Zr$$^{4+}$$, Hf$$^{4+}$$

D
Zn$$^{2+}$$, Hf$$^{4+}$$

Solution

$$Zr^{+4}$$ and $$Hf^{4+}$$ have same size due to lanthanide contraction.
Lanthanide contraction is caused due to poor shielding effect of the $$4f$$ electrons.
Question 3
1 / -0

Colour in transition metal compounds is attributed to :

A
small size of metal ions

B
absorption of light in UV region

C
complete $$(ns)$$ subshell

D
incomplete $$(n-1)d$$ subshell

Solution

Any compound or ion showing colour is due to presence of unpaired electron.
Transition metal compounds have incomplete (n-1)d sub shell and because of that they have unpaired electron and thus they show colour.
Question 4
1 / -0

Transition metals are less reactive because of their :

A
High ionization potential and low melting point

B
High ionization potential and high melting point

C
Low ionization potential and low melting point

D
Low ionization potential and high melting point

Solution

Transition metals are less reactive relative to I and II group due to higher ionization potential and high melting point (due to greater no of bonding electrons).
Question 5
1 / -0

The magnetic moment of an ion is close to $$36\times 10^{-24}$$Joule /Tesla. The number of unpaired electrons of the ion are ____________.

A
4

B
2

C
1

D
3

Solution

As magnetic moment is $$\sqrt{n(n+2)}$$ Bohr magneton, where n is no. of unpair electron (Bohr Magneton is $$9.273\times 10^{-24}J.tesla^{-1}$$) , magnetic moment is $$36\times 10^{-24}J.tesla^{-1} = 3.89 B.M.$$ so unpair electron is 3.
Question 6
1 / -0

Statement 1: Lanthanides have much less tendency to form complexes than actinides.
Statement 2: Compared to actinides, the lanthanides have relatively larger size of atoms and less nuclear charge.

A
Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1

B
Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1

C
Statement 1 is true, Statement 2 is False

D
Statement 1 is False, Statement 2 is True

Solution

Because of higher effective nuclear charge and smaller size of atoms the actinides have higher charge density and because of that they have greater tendency to form complexes than the lanthanides.
Question 7
1 / -0

The purple colour of $$[Ti(H_{2}O)_{6}]^{3+}$$ ions is due to ______________?

A
unpaired d-electron

B
charge transfer

C
intermolecular vibrations

D
presence of paired electrons

Solution

Any compound or ion showing colour is due to presence of unpair electron. $$[Ti(H_{2}O)_{6}]^{3+}$$ has $$d^1$$ configuration and have one unpair electron and because of this it has purple colour.
Question 8
1 / -0

The radius of $$La^{3+}$$ (At. No. of La = 57) is 1.06 A$$^{0}$$. Which one of the following given values will be closest to the radius of $$Lu^{3+}$$?( Atomic number of Lu=71)

A
$$1.40 A^0$$

B
$$1.06 A^0$$

C
$$0.85 A^0$$

D
$$1.60 A^0$$

Solution

Because of lanthanoid contraction as we move from left to right in a row, radii decreases.
So radii of $$Lu^{+3}$$ should be lesser than $$1.06A^0$$.
Question 9
1 / -0

The catalytic activity of transition metals and their compounds is ascribed mainly to:

A
their unfilled d-orbitals

B
their ability of adopt variable oxidation states

C
their chemical reactivity

D
their magnetic behaviour

Solution

Transition metals show variable oxidation states, due to which they can perform as a catalyst.
For example: Contact process, i.e., manufacturing of sulphuric acid.
$$SO_3+\overset {+5}{V_2}O_5\longrightarrow SO_3+\overset {+4}{V_2}O_4$$
$$V_2O_4+O_2\longrightarrow SO_3+\overset {+4}{V_2}O_4$$
$$V_2O_4+O_2\longrightarrow \overset {+5}{V_2}O_5$$
Question 10
1 / -0

In which of the following pairs are both the ions coloured in aqueous solution?
[At. No. : $$Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27$$]

A
$$Ni^{2+}, Ti^{3+}$$

B
$$Sc^{3+}, Ti^{3+}$$

C
$$Sc^{3+}, Co^{2+}$$

D
$$Ni^{2+}, Cu^{+}$$

Solution

$$(I)$$ $$Ni\quad [Ar]3d^84s^2$$
$$Ni^{+2}\quad [Ar]3d^8$$
$$2$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.
$$(II)$$ $$Cu\quad [Ar]3d^{10}4s^1$$
$$Cu^{+}\quad [Ar]3d^10$$
$$0$$ unpaired $$e^-s\longrightarrow$$ dimagnetic.
$$(III)$$ $$Ti\quad [Ar]3d^24s^2$$
$$Ti^{+3}\quad [Ar]3d^1$$
$$1$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.
$$(IV)$$ $$Sc\quad [Ar]3d^14s^2$$
$$Sc^{+3}\quad [Ar]$$
$$0$$ unpaired $$e^-s\longrightarrow$$ dimagnetic.
$$(V)$$ $$Co\quad [Ar]3d^74s^2$$
$$Co^{+2}\quad [Ar]3d^7$$
$$3$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.
Ions which are paramagnetic show colour in aqueous solution.

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 - 21

Result

The d- and f- Block Elements Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

incomplete s - orbitals

incomplete d - orbitals

completely filled d - orbitals

incomplete p-orbitals

Solution

Question 2

Fe$$^{2+}$$, Ni$$^{2+}$$

Zr$$^{4+}$$, Ti$$^{4+}$$

Zr$$^{4+}$$, Hf$$^{4+}$$

Zn$$^{2+}$$, Hf$$^{4+}$$

Solution

Question 3

small size of metal ions

absorption of light in UV region

complete $$(ns)$$ subshell

incomplete $$(n-1)d$$ subshell

Solution

Question 4

High ionization potential and low melting point

High ionization potential and high melting point

Low ionization potential and low melting point

Low ionization potential and high melting point

Solution

Question 5

4

2

1

3

Solution

Question 6

Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1

Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1

Statement 1 is true, Statement 2 is False

Statement 1 is False, Statement 2 is True

Solution

Question 7

unpaired d-electron

charge transfer

intermolecular vibrations

presence of paired electrons

Solution

Question 8

$$1.40 A^0$$

$$1.06 A^0$$

$$0.85 A^0$$

$$1.60 A^0$$

Solution

Question 9

their unfilled d-orbitals

their ability of adopt variable oxidation states

their chemical reactivity

their magnetic behaviour

Solution

Question 10

$$Ni^{2+}, Ti^{3+}$$

$$Sc^{3+}, Ti^{3+}$$

$$Sc^{3+}, Co^{2+}$$

$$Ni^{2+}, Cu^{+}$$

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.