The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 25

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

The highest magnetic moment is shown by

A
$$Ni$$

B
$$Co$$

C
$$Fe$$

D
$$Sc$$

Solution

The expression for magnetic moment is $$\mu = \sqrt {n(n+2)}$$, where $$n$$ is the number of unpaired electrons. Greater is the number of unpaired electrons, greater is the magnetic moment. $$Sc,\ Ni,\ Fe$$ and $$Co$$ have $$1,\ 2,\ 4$$ and $$3$$ electrons respectively. Therefore, $$Fe$$ has the highest magnetic moment.
Question 2
1 / -0

Which plays a major role in the formation of complex compound ?

A
Transition metal

B
Lanthanides and actinides

C
Representative elements

D
p-block element

Solution

Because of the presence of unpaired electrons in the inner d-orbital, transition element forms coordination complexes with appropriate ligand.
$$K_3Fe(CN)_6$$
$$Ni(CO)_4$$

Question 3

1 / -0

The colour imparted by $$Co(II)$$ compounds to glass is

green

deep blue

yellow

red

Solution

Vanadium	yellow-green
Chrome	emerald green
Iron	Coke bottle green
Manganese	amethyst
Cobalt	violet blue
Copper	greenish blue to blue
Nickel	grayish brown
Selenium	salmon pink
Cerium- Titanium	yellow
Neodymium	dichroic violet-pink
Uranium	yellow (fluorescent

Question 4
1 / -0

The main reason for larger number of oxidation state exhibited by the actinides than that corresponding lanthanide, is:

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

B
larger atomic size of actinides than the lanthanides

C
more energy differenc between 5f and 6d orbitals than between 4f and 5d-orbitals

D
greater reactive nature of the actinides than the lanthanides

Solution

Unlike lanthanides, actinides show a variety of oxidation states from +3 to +6 due to the very small energy gap between 5f, 6d, and 7s subshells.

The principal oxidation states are +3 and +4. The +3 oxidation state is the most stable. The +4 oxidation state is the most stable in Th and Pu. +5 in Pa and Np and +6 is seen in U. In actinides, the distributions of oxidation states are uneven.
Question 5
1 / -0

Misch metal is:

A
an alloy of lathanide and copper

B
an alloy of lanthanide and nickel

C
an alloy of lathanide,iron and carbon

D
an alloy of calcium and copper

Solution

Misch metal, alloy consisting of about 50 percent cerium, 25 percent lanthanum, 15 percent neodymium, and 10 percent other rare-earth metals.
Question 6
1 / -0

Transition elements does not show :

A
paramagnetism

B
colour

C
fixed valency

D
all of these

Solution

Transition element shows variable valencey and due to unpaired electron they shows colour and magnetic properties.
Question 7
1 / -0

The magnetic moment of a transition metal of $$3d$$-series is $$6.92BM$$. Its electronic configuration will be:

A
$${ \left( 3d \right) }^{ 4 }{ \left( 4s \right) }^{ 2 }$$

B
$${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 1 }$$

C
$${ \left( 3d \right) }^{ 10 }$$

D
$${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 0 }$$

Solution

The magnetic moment, $$ \mu_s = \sqrt {n(n+2)} B M$$ where n is the number of unpaired electrons.

Option B has 6 unpaired electrons and the corresponding magnetic moment is $$ 6.92\mu_B$$.
Question 8
1 / -0

$$\displaystyle [{ Fe\left( { H }_{ 2 }{ O } \right) }_{ 5 }{ No{ ] }^{ 2+ } }$$ is formed as brown ring in $$\displaystyle { NO }_{ 3 }^{ + }$$ test. $$\displaystyle Fe$$ in this complex has _________ unpaired electrons.

A
one

B
two

C
three

D
four

Solution

The magnetic moment of nitroso ferrous sulphate is 3.87 BM. This indicates presence of 3 unpaired electrons. This is due to $$3d^7$$ electronic configuration of $$Fe^+$$ ion.
Question 9
1 / -0

The magnetic moment order is correctly given in:

A
$$Fe^{2+} < Fe^{3+} < Mn^{4+} < Cr^{3+}$$

B
$$Cr^{3+} = Mn^{4+} < Fe^{2+} < Fe^{3+}$$

C
$$Cr^{3+} = Mn^{4+} < Fe^{3+} < Fe^{2+}$$

D
$$Fe^{2+} < Cr^{3+} < Fe^{3+} < Mn^{4+}$$

Solution

As we know,
magnetic moment $$=\sqrt {n(n+2)}$$ where n is no of unpaired electron
here,
$$Cr^{3+}, Mn^{4+}, Fe^{2+}$$ and $$Fe^{3+}$$ have $$3,3,4,5$$ unpaired electrons.
so order is $$Cr^{3+} = Mn^{4+} < Fe^{2+} < Fe^{3+}$$
Question 10
1 / -0

The position in the periodic table to which $$Fe$$ and $$Zn$$ elements belong to __________ .

A
Alkali metal

B
Alkaline earth metal

C
d-block

D
None of these

Solution

$$Fe$$ and $$Zn$$ elements belong to d-block because the valence electron enters into the d-orbital.
Hence option C is correct.

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 - 25

Result

The d- and f- Block Elements Test - 25

Score

-

Rank

-

SHARING IS CARING

Question 1

$$Ni$$

$$Co$$

$$Fe$$

$$Sc$$

Solution

Question 2

Transition metal

Lanthanides and actinides

Representative elements

p-block element

Solution

Question 3

green

deep blue

yellow

red

Solution

Question 4

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

larger atomic size of actinides than the lanthanides

more energy differenc between 5f and 6d orbitals than between 4f and 5d-orbitals

greater reactive nature of the actinides than the lanthanides

Solution

Question 5

an alloy of lathanide and copper

an alloy of lanthanide and nickel

an alloy of lathanide,iron and carbon

an alloy of calcium and copper

Solution

Question 6

paramagnetism

colour

fixed valency

all of these

Solution

Question 7

$${ \left( 3d \right) }^{ 4 }{ \left( 4s \right) }^{ 2 }$$

$${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 1 }$$

$${ \left( 3d \right) }^{ 10 }$$

$${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 0 }$$

Solution

Question 8

one

two

three

four

Solution

Question 9

$$Fe^{2+} < Fe^{3+} < Mn^{4+} < Cr^{3+}$$

$$Cr^{3+} = Mn^{4+} < Fe^{2+} < Fe^{3+}$$

$$Cr^{3+} = Mn^{4+} < Fe^{3+} < Fe^{2+}$$

$$Fe^{2+} < Cr^{3+} < Fe^{3+} < Mn^{4+}$$

Solution

Question 10

Alkali metal

Alkaline earth metal

d-block

None of these

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.