The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids, is:

A
the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals

B
more energy difference between 5f and 6d orbitals than between 4f and 5d orbitals

C
greater reactive nature of the actinoids than the lanthanoids

D
larger atomic size of actinoids than the lanthanoids

Solution

Actinoids shows larger oxidation states due to poor shielding of both $$4f$$ and $$5f$$ orbital electrons as a result these orbital have almost similar energy and hence take part in bond formation. Also, there is very small energy between $$5f, 6d$$ and $$7s$$ subshells.
Question 2
1 / -0

Out of $$TiF^{2-}_6$$ and $$CoF^{3-}_6$$ , $$Cu_2Cl_2$$ and $$NiCl^{2-}_4$$ (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are:

A
$$CoF^{3-}_6\; and \;NiCl^{2-}_4$$

B
$$TiF^{2-}_6 \; and \;CoF^{3-}_6$$

C
$$Cu_2Cl_2 \; and \;NiCl^{2-}_4$$

D
$$TiF^{2-}_6 \; and \;Cu_2Cl_2$$

Solution

$$TiF_6^{2-}\quad Ti^{+4}\rightarrow [Ar]\quad 0$$ unpair $$e^-\Rightarrow $$ Dimagnetic/ colourless
$$CoF_6^{3-}\quad Co^{+3}\rightarrow [Ar]3d^6\quad 4$$ unpair $$e^-\Rightarrow $$ Paramagnetic/ coloured
$$Cu_2Cl_2\quad Cu^{+}\rightarrow [Ar]3d^{10}\quad 0$$ unpair $$e^-\Rightarrow $$ Dimagnetic/ colourless
$$NiCl_4^{2-}\quad Ni^{+2}\rightarrow [Ar]3d^8\quad 2$$ unpair $$e^-\Rightarrow $$Paramagnetic/ coloured
Question 3
1 / -0

Which one of the following ions is the most stable in aqueous solution?
(At. No. are: $$ Ti = 22, V = 23, Cr = 24, Mn = 25$$)

A
$$Mn^{3+}$$

B
$$Cr^{3+}$$

C
$$V^{3+}$$

D
$$Ti^{3+}$$

Solution

In aqueous solution the configuration of the ions are following:

$$Cr^{3+}(21)= 3d^34s^0 \Rightarrow {t_{2g}}^3 e_g^0$$

$$V^{3+}(20)= 3d^24s^0 \Rightarrow {t_{2g}}^2 e_g^0$$

$$Ti^{3+}(19)= 3d^14s^0\Rightarrow {t_{2g}}^1 e_g^0$$

$$Mn^{3+}(22)= 3d^44s^0\Rightarrow {t_{2g}}^4 e_g^0$$

We know that half-filled and fully filled orbitals are more stable than another half nor fully filled orbitals. Hence among the given ions the only ion which occupies half-filled $$t_{2g}$$ orbitals is Chromium (III) ion.

$$Note:$$ Although $$Mn^{3+}$$ ion has half-filledr $$t_{2g}$$ orbital but there is one more electron filled in $$e_{g}$$ orbital which reduces the stability.

Hence option A is the correct answer.
Question 4
1 / -0

Which one of the following ions will be colourless in an aqueous solution?

A
$$Fe^{2+}$$

B
$$Mn^{2+}$$

C
$$Ti^{3+}$$

D
$$Sc^{3+}$$

Solution

$$(A)$$ $$Fe= [Ar]3d^64s^2$$

$$Fe^{2+}= [Ar]3d^6$$

$$4$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

$$(B)$$ $$Mn= [Ar]3d^54s^2$$

$$Mn^{2+}= [Ar]3d^5$$

$$5$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

$$(C)$$ $$Ti= [Ar]3d^24s^2$$

$$Ti^{3+}= [Ar]3d^1$$

$$1$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

$$(D)$$ $$Sc= [Ar]3d^14s^2$$

$$Sc^{+3}= [Ar]$$

$$0$$ unpaired $$e^-s\longrightarrow$$ diamagnetic.

Note: Ions which are paramagnetic show colour in an aqueous solution.
Question 5
1 / -0

Which ion is colourless?

A
$$C{ r }^{ 4+ }$$

B
$$Sc^{ 3+ }$$

C
$$Ti^{ 3+ }$$

D
$${ V }^{ 3+ }$$

Solution

$$(A)$$ $$Cr\quad [Ar]3d^54s^1$$
$$Cr^{4+}\quad [Ar]3d^2$$
$$2$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

$$(B)$$ $$Sc\quad [Ar]3d^14s^2$$
$$Sc^{3+}\quad [Ar]3d^0$$
$$0$$ unpaired $$e^-s\longrightarrow$$ diamagnetic.

$$(C)$$ $$Ti\quad [Ar]3d^24s^2$$
$$Ti^{3+}\quad [Ar]3d^1$$
$$1$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

$$(D)$$ $$V\quad [Ar]3d^34s^2$$
$$V^{3+}\quad [Ar]3d^2$$
$$2$$ unpaired $$e^-s\longrightarrow$$ paramagnetic.

Ions which do not have unpaired electrons do not show colour in aqueous solution. These are known as diamagnetic ions.
Question 6
1 / -0

Which of the following statement is not correct?

A
$$La(OH)_3$$ is less basic than $$Lu(OH)_3$$

B
In Lanthanide series ionic radius of $$Ln^{+3}$$ ions decreases

C
La is actually an element of transition series rather Lanthanide

D
Aomic radius of Zr and Hf are same because of Lanthanide contraction

Solution

$$La(OH)_3$$ is less basic than $$Lu(OH)_3$$ is an incorrect statement.
Basic strength $$La(OH)_3>Lu(OH)_3$$
Due to Lanthanide contraction as the size of the lanthanide ions decreases from $$La^{+3}$$ to $$Lu^{+3}$$, the covalent character of hydroxides increases and hence basic strength decreases.
Question 7
1 / -0

Which of the following does not corrode when exposed to the atmosphere?

A
iron

B
copper

C
gold

D
silver

Solution

Gold is very least reactive hence does not corrode at all.
And Au is a highly stable element because of its electronic distribution , hence it doesn't react easily.
Question 8
1 / -0

The number of oxidation states is exhibited by the actinoids more than by the lanthanide. The main reason for this is:

A
more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals.

B
the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.

C
the greater metallic character of the lanthanoids than that of the corresponding actinoids.

D
more active nature of the actinoids.

Solution

Actinides show larger oxidation states due to poor shielding of both $$4f$$ and $$5f$$ orbital electrons, as a result, their orbitals have almost similar energy and hence take part in bond formation. Thus the energy gap between $$5f, 6d$$ and $$7s$$ become very small.
Question 9
1 / -0

Which of the following shows highest magnetic moment ?

A
$$Co^{+3}$$

B
$$Cr^{+3}$$

C
$$Fe^{+3}$$

D
$$Ni^{+2}$$

Solution

$$Cr^{3+}$$ $$\rightarrow$$ $$3d^0$$
$$Co^{3+}$$ $$\rightarrow$$ $$3d^6$$
$$Fe^{3+}$$ $$\rightarrow$$ $$3d^5$$
$$Ni^{2+}$$ $$\rightarrow$$ $$3d^8$$
therefore $$Fe^{3+}$$ has the highest number of unpaired electrons and is always paramagnetic and shows highest magnetic moment.
Question 10
1 / -0

Variable valency is characteristic of:

A
halogen

B
transition elements

C
alkali metals

D
noble gas

Solution

In transition metals, $$(n-1)$$ d and ns electrons have nearly same energy level. Hence, most of these electrons take part in chemical bonding. Hence, transition metals show variable valency.

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The d- and f- Block Elements Test - 22

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The d- and f- Block Elements Test - 22

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Question 1

the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals

more energy difference between 5f and 6d orbitals than between 4f and 5d orbitals

greater reactive nature of the actinoids than the lanthanoids

larger atomic size of actinoids than the lanthanoids

Solution

Question 2

$$CoF^{3-}_6\; and \;NiCl^{2-}_4$$

$$TiF^{2-}_6 \; and \;CoF^{3-}_6$$

$$Cu_2Cl_2 \; and \;NiCl^{2-}_4$$

$$TiF^{2-}_6 \; and \;Cu_2Cl_2$$

Solution

Question 3

$$Mn^{3+}$$

$$Cr^{3+}$$

$$V^{3+}$$

$$Ti^{3+}$$

Solution

Question 4

$$Fe^{2+}$$

$$Mn^{2+}$$

$$Ti^{3+}$$

$$Sc^{3+}$$

Solution

Question 5

$$C{ r }^{ 4+ }$$

$$Sc^{ 3+ }$$

$$Ti^{ 3+ }$$

$${ V }^{ 3+ }$$

Solution

Question 6

$$La(OH)_3$$ is less basic than $$Lu(OH)_3$$

In Lanthanide series ionic radius of $$Ln^{+3}$$ ions decreases

La is actually an element of transition series rather Lanthanide

Aomic radius of Zr and Hf are same because of Lanthanide contraction

Solution

Question 7

iron

copper

gold

silver

Solution

Question 8

more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals.

the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.

the greater metallic character of the lanthanoids than that of the corresponding actinoids.

more active nature of the actinoids.

Solution

Question 9

$$Co^{+3}$$

$$Cr^{+3}$$

$$Fe^{+3}$$

$$Ni^{+2}$$

Solution

Question 10

halogen

transition elements

alkali metals

noble gas

Solution

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