The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 11

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

Which one of the following does not correctly represent the correct order of the property indicated against it?

A
$${Ti}^{3+}< {V}^{3+}< {Cr}^{3+}< {Mn}^{3+}$$; increasing magnetic moment

B
$$Ti< V< Cr< Mn$$; increasing melting point

C
$$Ti< V< Mn< Cr$$; increasing 2nd ionization enthalpy

D
$$Ti< V< Cr< Mn$$; increasing number of oxidation states

Solution

The following does not correctly represent the correct order of the property indicated against it.
$$Ti< V< Cr< Mn$$; increasing melting point

The correct order of the melting point is $$Ti< V< Cr> Mn$$.
Along the 3rd series melting point increases up to the $$Cr$$ and then decreases.

This is because the number of unpaired electrons increases up to $$Cr$$ and then decreases.

$$Cr$$ has a maximum number of unpaired electrons (6).
Question 2
1 / -0

Which of the following arrangements does not represent the correct order of the property stated against it?

A
$${ V }^{ 2+ }<{ Cr }^{ 2+ }<{ Mn }^{ 2+ }<{ Fe }^{ 2+ }$$: Paramagnetic behavior

B
$${ Ni }^{ 2+ }<{ Co }^{ 2+ }<{ Fe }^{ 2+ }<{ Mn }^{ 2+ }$$: Ionic size

C
$${ Co }^{ 3+ }<{ Fe }^{ 3+ }<{ Cr }^{ 3+ }<{ Sc }^{ 3+ }$$: Stability in aqueous solution

D
$$Sc< Ti< Cr< Mn$$: Number of oxidation states
Solution
Explanation:
- Option A, more the number of unpaired electrons more is paramagnetic behaviour.
  Where, $$n$$ is number of unpaired electrons
$$V^{2+}(3d^3)$$ i.e. $$n=3$$

$$Cr^{2+}(3d^4)$$ i.e.$$n=4$$

$$Mn^{2+}(3d^5)$$ i.e. $$n=5$$

$$Fe^{2+}(3d^4)$$ i.e. $$n=4$$

Hence, the order of paramagnetic behavior is $$Mn^{2+}>Fe^{2+}>Cr^{2+}>V^{2+}$$
- Option B, the ionic size decrease as the charge on the atom increases. All species have the same charge but a different atomic number.
$$Ni^{2+}<Co^{2+}<Fe^{2+}<Mn^{2+}$$ is correct sequence of ionic size.
- Option C, stability in an aqueous solution is dependent on electrode potential. More negative the electrode potential more stable solution. Electrode potential increases across the period in the transition series.
$$Co^{3+}<Fe^{3+}<Cr^{3+}<Sc^{3+}$$ is correct.
- Option D, the number of oxidation states increases up to the middle of the transition series.
$$Sc<Ti<Cr<Mn$$ is correct.

Hence, the correct answer is option $$A$$.
Question 3
1 / -0

Which of the following has the highest paramagnetism?

A
$$[Cr(H_2O)_6]^{3+}$$

B
$$[Fe(H_2O)_6]^{2+}$$

C
$$[Cu(H_2O)_6]^{2+}$$

D
$$[Zn(H_2O)_6]^{2+}$$

Solution

Paramagnetism $$\propto$$ Number of unpaired electrons

In $$\displaystyle [Cr(H_2O)_6]^{3+} $$, Cr(III) has $$\displaystyle 3d^3$$ outer electronic configuration with 3 unpaired electrons.

In $$\displaystyle [Fe(H_2O)_6]^{2+} $$, Fe(II) has $$\displaystyle 3d^6 $$ outer electronic configuration with 4 unpaired electrons.

In $$\displaystyle [Cu(H_2O)_6]^{2+}$$, Cu(II) has $$\displaystyle 3d^9$$ electronic configuration with 1 unpaired electron.

In $$\displaystyle [Zn(H_2O)_6]^{2+}$$, Zn(II) has $$\displaystyle 3d^{10}$$ electronic configuration with 0 unpaired electrons.

$$\displaystyle [Fe(H_2O)_6]^{2+} $$ has the highest paramagnetism as it contains a maximum number of unpaired electrons.

Hence, option $$B$$ is correct.
Question 4
1 / -0

Which of the following transition metal ions will have definite value of magnetic moment?

A
$$Sc^{3+}$$

B
$$Ti^{3+}$$

C
$$Cu^+$$

D
$$Zn^{2+}$$

Solution

Value of magnetic moment depends upon number of unpaired electrons. All given metal ions (except $$Ti^{3+} \: (3d^1)$$) have either fully filled d-subshell ( $$Zn^{2+}, Cu^+$$) or empty d-subshell ( $$Sc^{3+}$$). As such only $$Ti^{3+}$$ has a net value of magnetic moment.

Magnetic moment of $$Ti^{3+}=\sqrt {n(n+2)}\ BM=\sqrt {1(1+2)}\ BM=\sqrt {3} \ BM=1.73\ BM$$.
Question 5
1 / -0

Which of the following group belongs to actinide series ?

A
Th, Pa, U

B
Ce Pr, Nd

C
Ba, La, Hf

D
Pt, Au, Ag

Solution

$$\displaystyle Th (thorium)$$, $$\displaystyle Pa (proactinium)$$ and $$\displaystyle U (uranium)$$ belongs to actinoid series.

$$\displaystyle Ce Pr, Nd$$ are lanthanoids.

$$\displaystyle Ba, La, Hf$$ belongs to the sixth period.

$$\displaystyle Pt, Au, Ag$$ are transition elements.

Hence, option $$A$$ is correct.
Question 6
1 / -0

The atomic numbers of elements of second inner transition elements lie in the range of :

A
88 to 101

B
89 to 102

C
90 to 103

D
91 to 104

Solution

Second inner transition elements are from thorium to lawrencium, i.e., from atomic numbers 90 to 103.
Question 7
1 / -0

The Lanthanide contraction relates to:

A
oxidation states

B
magnetic state

C
atomic radius

D
valence electrons

Solution

Due to the poor shielding effect of 4f electrons the outer electrons experience more attractive force from nucleus.

Due to this attraction, the size of elements contracts. This contraction is known as lanthanide contraction. In this way, lanthanide contraction is related to atomic radii.

Option C is correct.
Question 8
1 / -0

Transition metals are good electrical conductor because ___________.

A
they are metals

B
they are solids

C
they have free electrons in outer energy levels

D
they are hard

Solution

Transition metals have free electrons in outer energy levels because d-orbitals shields poorly and due to this they acts as good conductor of electricity.
Question 9
1 / -0

Which of the following is used as Catalyst in the hydrogenation of oil?

A
$$V_{2}O_{5}$$

B
Pd

C
Fe

D
Ni

Solution

Non-precious metal catalysts, especially those based on nickel (such as Raney Nickel and Urushibara Nickel) are generally used as catalyst in the hydrogenation of oils. Palladium can also be used but it is very costly.
Question 10
1 / -0

Catalytic activity of transition elements and their compounds is due to their _______.

A
small size

B
vacant d-orbitals

C
higher densities

D
colour

Solution

The catalytic activity of transition elements is ascribed to their ability to adopt multiple oxidation states and to form complexes.

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 - 11

Result

The d- and f- Block Elements Test - 11

Score

-

Rank

-

SHARING IS CARING

Question 1

$${Ti}^{3+}< {V}^{3+}< {Cr}^{3+}< {Mn}^{3+}$$; increasing magnetic moment

$$Ti< V< Cr< Mn$$; increasing melting point

$$Ti< V< Mn< Cr$$; increasing 2nd ionization enthalpy

$$Ti< V< Cr< Mn$$; increasing number of oxidation states

Solution

Question 2

$${ V }^{ 2+ }<{ Cr }^{ 2+ }<{ Mn }^{ 2+ }<{ Fe }^{ 2+ }$$: Paramagnetic behavior

$${ Ni }^{ 2+ }<{ Co }^{ 2+ }<{ Fe }^{ 2+ }<{ Mn }^{ 2+ }$$: Ionic size

$${ Co }^{ 3+ }<{ Fe }^{ 3+ }<{ Cr }^{ 3+ }<{ Sc }^{ 3+ }$$: Stability in aqueous solution

$$Sc< Ti< Cr< Mn$$: Number of oxidation states

Solution

Question 3

$$[Cr(H_2O)_6]^{3+}$$

$$[Fe(H_2O)_6]^{2+}$$

$$[Cu(H_2O)_6]^{2+}$$

$$[Zn(H_2O)_6]^{2+}$$

Solution

Question 4

$$Sc^{3+}$$

$$Ti^{3+}$$

$$Cu^+$$

$$Zn^{2+}$$

Solution

Question 5

Th, Pa, U

Ce Pr, Nd

Ba, La, Hf

Pt, Au, Ag

Solution

Question 6

88 to 101

89 to 102

90 to 103

91 to 104

Solution

Question 7

oxidation states

magnetic state

atomic radius

valence electrons

Solution

Question 8

they are metals

they are solids

they have free electrons in outer energy levels

they are hard

Solution

Question 9

$$V_{2}O_{5}$$

Pd

Fe

Ni

Solution

Question 10

small size

vacant d-orbitals

higher densities

colour

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.