The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

Increasing order of magnetic moment among the following species is:
$$\displaystyle Na^{+},Fe^{+3},Co^{+2},Cr^{+2}$$

A
$$Na^{+} < Co^{2+} < Cr^{2+} < Fe^{3+}$$

B
$$Na^{+} < Fe^{3+} < Cr^{2+} < Co^{2+}$$

C
$$Cr^{2+} < Co^{2+} < Na^{+} < Fe^{3+}$$

D
$$Co^{2+}< Na^{+} < Cr^{2+} < Fe^{3+}$$

Solution

Magnetic moment is $$\mu = \sqrt { n(n+2) } $$, where n is no. of unpaired electron so more no. of unpaired electron means more magnetic moment.

$$Na^{+}$$ has zero unpaired electron
$$Co^{2+}$$ has 3 unpaired electrons
$$Cr^{2+}$$ has 4 unpaired electrons
$$Fe^{3+}$$ has 5 unpaired electron

The order is $$Na^{+} < Co^{2+} < Cr^{2+} < Fe^{3+}$$.
Question 2
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The yellow colour solution of $$Na_2CrO_4$$ changes to orange red on passing $$CO_2$$ gas due to the formation of:

A
$$CrO_5$$

B
$$CrO_3$$

C
$$Cr_2O_7^{2-}$$

D
$$Cr_2O_3$$

Solution

The yellow coloured solution of $$Na_2CrO_4$$ changes to orange red on passing $$CO_2$$ due to the formation of dichromte ion, $$Cr_2O_7^{2-}$$.
The reaction is as follows:
$$CO_2 + H_2O \rightarrow H_2CO_3 \rightleftharpoons 2H^+ + CO_3^{2-}$$
$$2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} (\text{orange red}) + H_2O$$
Question 3
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The transition metals are mostly :

A
diamagnetic

B
paramagnetic

C
neither diamagnetic nor paramagnetic

D
both diamagnetic and paramagnetic

Solution

Most of trasition metls have unpaired electron (d-configuration) in their electronic configuration so they are paramagnetic in nature.
Question 4
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Transition elements form complexes very readily because of:

A
small cation size

B
vacant d-orbitals

C
large nuclear charge

D
all of the above

Solution

As transition metals have vacant $$d$$-orbitals, they exhibt variable valencies and form complexes. Besides, they have smaller size and larger nuclear charge. Hence, they have large charge density which can stablize charged complexes. Therefore, transition metals form complexes very readily.
Question 5
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Select the correct order of ions with respect to their ionic radii.

A
$$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$$

B
$$Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$$

C
$$Lu^{3+} < Eu^{3+} < La^{3+} < Y^{3+}$$

D
$$La^{3+} < Eu^{3+} < Lu^{3+} < Y^{3+}$$

Solution

We know that as we move down the group, radii increases and as we move across a period, radii decreases. The correct order of ionic radii is as follows:

$$Y^{3+}< Lu^{3+}< Eu^{3+} < La^{3+}$$

This is because $$Eu$$ and $$Lu$$ are the members of the lanthanide series which have smaller radii than the transition metals.

$$Y^{3+}$$ ion has a lower radius as comparison to $$La^{3+}$$ because it lies immediately above it in the periodic table and hence, it has a lower effective nuclear charge and a larger size.
Question 6
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When $$CO_2$$ is passes into aqueous $$K_2CrO_4$$ solution, this changes to another colour. What is the another colour?

A
Black

B
Yellow

C
Pink

D
Orange

Solution

$$ CO_2+ H_2O \rightleftharpoons H_2 CO_3$$,

$$H_2CO_3 \rightleftharpoons H^+ + HCO^-_3$$,

$$H^+ $$change $$K_2CRO_4$$ into $$K_2Cr_2O_7$$,

$$ \underset {yellow}{2CrO^{2-}_4} + 2H^+ \longrightarrow \underset {orange}{Cr_2O^{2-}_7} + H_2O$$
Question 7
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Which of the following compounds is not colored?

A
$${Na}_{2}Cu{Cl}_{4}$$

B
$${Na}_{2}Cd{Cl}_{4}$$

C
$${Na}_{3}[Fe{(CN)}_{6}]$$

D
$${K}_{3}[Fe{(CN)}_{6}]$$

Solution

Electronic configuration: $$[Kr] 4d^{10} 5s^{2}$$

Cadmium is not considered as transition element which has completely filled d - configuration and is not having unpaired electrons to exhibit color property.
So $$Na_2CdCl_4$$ do not exhibit color
Hence option B is correct.
Question 8
1 / -0

What would happen when as solution of potassium chromate is treated with an excess of dilute nitric acid?

A
$$Cr^{3+}$$ and $$Cr_2O^{2-}_7$$ are formed

B
$$Cr_2O^{2-}_7$$ang $$H_2O$$ are formed

C
$$CrO^{2-}_4$$ is reduced to $$Cr^{3+}$$

D
None of the above
Question 9
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Spin only magnetic moment of the compound $$Hg[Co{(SCN)}_{4}]$$ is :

A
$$\sqrt 3$$

B
$$\sqrt 15$$

C
$$\sqrt 24$$

D
$$\sqrt 8$$

Solution

Cobalt is present as $$Co^{+2}$$ which has [Ar] $$ 4s^03d^7$$ configuration, which means it has 3 unpaired electrons. So the spin only magnetic moment of the compound is $$ \sqrt{3(3+2)} = \sqrt{15}$$
Question 10
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Least paramagnetic property is shown by:

A
$$Fe$$

B
$$Mn$$

C
$$Ni$$

D
$$Cu$$

Solution

Elements which have least number of unpaired electrons will be least paramagnetic.

In the given option, Cu has only one unpaired electron. Others have more than 1 unpaired electron in the valence shell.

So, option D is correct.

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 58

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The d- and f- Block Elements Test - 58

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Question 1

$$Na^{+} < Co^{2+} < Cr^{2+} < Fe^{3+}$$

$$Na^{+} < Fe^{3+} < Cr^{2+} < Co^{2+}$$

$$Cr^{2+} < Co^{2+} < Na^{+} < Fe^{3+}$$

$$Co^{2+}< Na^{+} < Cr^{2+} < Fe^{3+}$$

Solution

Question 2

$$CrO_5$$

$$CrO_3$$

$$Cr_2O_7^{2-}$$

$$Cr_2O_3$$

Solution

Question 3

diamagnetic

paramagnetic

neither diamagnetic nor paramagnetic

both diamagnetic and paramagnetic

Solution

Question 4

small cation size

vacant d-orbitals

large nuclear charge

all of the above

Solution

Question 5

$$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$$

$$Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$$

$$Lu^{3+} < Eu^{3+} < La^{3+} < Y^{3+}$$

$$La^{3+} < Eu^{3+} < Lu^{3+} < Y^{3+}$$

Solution

Question 6

Black

Yellow

Pink

Orange

Solution

Question 7

$${Na}_{2}Cu{Cl}_{4}$$

$${Na}_{2}Cd{Cl}_{4}$$

$${Na}_{3}[Fe{(CN)}_{6}]$$

$${K}_{3}[Fe{(CN)}_{6}]$$

Solution

Question 8

$$Cr^{3+}$$ and $$Cr_2O^{2-}_7$$ are formed

$$Cr_2O^{2-}_7$$ang $$H_2O$$ are formed

$$CrO^{2-}_4$$ is reduced to $$Cr^{3+}$$

None of the above

Question 9

$$\sqrt 3$$

$$\sqrt 15$$

$$\sqrt 24$$

$$\sqrt 8$$

Solution

Question 10

$$Fe$$

$$Mn$$

$$Ni$$

$$Cu$$

Solution

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