The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 10

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Weekly Quiz Competition

Question 1
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The correct order of unpaired electrons present in d orbitals for elements $$Fe,\ Ti,\ Cr$$ and $$Co$$ is :

A
$$Cr=Fe>Co>Ti$$

B
$$Fe>Co>Cr>Ti$$

C
$$Cr>Fe>Co>Ti$$

D
$$Cr=Fe>Ti>Co$$

Solution

$$Cr= [Ar]\ 4s^1\ 3d^{5};$$ 6 unpaired electrons

$$Fe= [Ar]\ 4s^2\ 3d^{6};$$ 4 unpaired electrons

$$Co= [Ar]\ 4s^2\ 3d^{7};$$ 3 unpaired electrons

$$Ti= [Ar]\ 4s^2\ 3d^{2};$$ 2 unpaired electrons

$$Cr > Fe > Co > Ti$$

Hence, option $$C$$ is correct.
Question 2
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The actinoids exhibit more oxidation states in general than the lanthanoids. This is because:

A
the 5f orbitals are more buried than the 4f orbitals.

B
there is a similarity between 4f and 5f orbitals in their angular part of the wave function.

C
the actinoids are more reactive than the lanthanoids.

D
the 5f orbitals extend further from the nucleus than the 4f orbitals.

Solution

Hint: Variability in oxidation states corresponds to degeneracy in the outer orbitals.
Correct Answer: Option D
Explanation:
In actinoids, the outermost $$5f$$ orbitals are away from the nucleus. This makes them similar in energies to $$6d$$ and $$7s$$ orbitals. Thus, electrons can easily transition between these orbitals, and once they reach the valence orbital they can be ionized, This leads to the variability in oxidation states. In the case of lanthanoids, the outermost $$4f$$ orbitals are too close to the nucleus.

Question 3

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Match the catalysts to the correct processes:

Catalyst	Process
A. $$TiCl_3$$	i. Wacker process
B. $$PdCl_2$$	ii. Ziegler - Natta polymerization
C. $$CuCl_2$$	iii. Contact process
D. $$V_2O_5$$	iv. Deacon's process

(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)

(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)

(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)

(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)

Solution

A Ziegler–Natta catalyst, named after Karl Ziegler and Giulio Natta, is a catalyst used in the synthesis of polymers of 1-alkenes (alpha-olefins). The catalyst $$TiCl_3$$ is used for Ziegler - Natta polymerization.

The Wacker Oxidation is an industrial process, which allows the synthesis of ethanal from ethene by palladium-catalyzed oxidation with oxygen. The catalyst $$PdCl_2$$ is used for the Wacker process.

The Deacon process is a process used during the manufacture of alkalis (the initial end product was sodium carbonate) by the Leblanc process. The catalyst $$CuCl_2$$ is used for Deacon's process.

The contact process is the current method of producing sulfuric acid in the high concentrations needed for industrial processes. The catalyst $$V_2O_5$$ is used for the Contact process.

Hence, option $$B$$ is correct.

Question 4
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The transition metal ions responsible for color in Ruby and Emerald are, respectively :

A
$${Co}^{3+}$$ and $${Cr}^{3+}$$

B
$${Co}^{3+}$$ and $${Co}^{3+}$$

C
$${Cr}^{3+}$$ and $${Cr}^{3+}$$

D
$${Cr}^{3+}$$ and $${Co}^{3+}$$

Solution

Ruby is $${ Al }_{ 2 }{ O }_{ 3 }$$ in which Red colour is obtained when $${ C }r^{ +3 }$$ is replacing $${ Al }^{ +3 }$$ ions in octahedral sites.

Emerald is $${ Be }_{ 3 }{ Al }_{ 2 }{ \left( Si{ O }_{ 3 } \right) }_{ 6 }$$ in which green colour is obtained when $${ Cr }^{ +3 }$$ is replacing $${ Al }^{ +3 }$$ in octahedral site.

Hence, the correct option is $$C$$.
Question 5
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The correct order of spin-only magnetic moments among the following is:

(Atomic number: $$Mn=25, Co=27, Ni=28, Zn=30)$$.

A
$$[ZnCl_4]^{2-} > [NiCl_4]^{2-} > [CoCl_4]^{2-} > [MnCl_4]^{2-}$$

B
$$[CoCl_4]^{2-} > [MnCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$$

C
$$[NiCl_4]^{2-} > [CoCl_4]^{2-} > [MnCl_4]^{2-} > [ZnCl_4]^{2-}$$

D
$$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$$

Solution

The complex having higher number of unpaired electrons will have higher value of spin-only magnetic moment.

The correct order of spin-only magnetic moments is $$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$$.

In these complexes, the central metal ion is in $$\displaystyle +2$$ oxidation state.

$$\displaystyle Zn^{2+}$$ has $$\displaystyle 3d^{10}$$ outer electronic configuration with $$\displaystyle 0$$ unpaired electrons.

$$\displaystyle Ni^{2+}$$ has $$\displaystyle 3d^{8}$$ outer electronic configuration with $$\displaystyle 2$$ unpaired electrons.

$$\displaystyle Co^{2+}$$ has $$\displaystyle 3d^{7}$$ outer electronic configuration with $$\displaystyle 3$$ unpaired electrons.

$$\displaystyle Mn^{2+}$$ has $$\displaystyle 3d^{5}$$ outer electronic configuration with $$\displaystyle 5$$ unpaired electrons.

Hence, the correct option is D.

Question 6

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The correct order of the spin-only magnetic moment of metal ions in the following low spin complexes, $$[V(CN)_6]^{4-}, [Fe(CN)_6]^{4-}$$, $$[Ru(NH_3)_6]^{3+}, $$ and $$[Cr(NH)_3)_6]^{2+}$$, is:

$$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$

$$V^{2+} > Ru^{3+} > Cr^{2+} > Fe^{2+}$$

$$Cr^{2+} > V^{2+} > Ru^{3+} > Fe^{2+}$$

$$Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+}$$

Solution

Solution:- (A) $$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$

According to equations, all the complexes are low spin.

Complex	Configuration	No. of unpaired electrons
$$[V(CN)_6]^{4-}$$	$$t_{2g} ^3e_g^0$$	3
$$[Cr(NH)_3)_6]^{2+}$$	$$t_{2g} ^4e_g^0$$	2
$$[Ru(NH_3)_6]^{3+} $$	$$t_{2g} ^5e_g^0$$	1
$$[Fe(CN)_6]^{4-}$$	$$t_{2g} ^6e_g^0$$	0

Question 7
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The larger number of oxidation states are exhibited by the actinides than those by lanthanoids, the main reason being:

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

B
more reactive nature of the actinoids than the lanthanoids

C
4f orbitals more diffused than the 5f orbitals

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

Solution

The energy difference between 5f and 6d is lesser than that between 4f and 5d orbitals. Thus, in actinides, the electrons can be removed from 5f as well as 6d, so more number of oxidation states are exhibited by them.
Question 8
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Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code :

Column I
Column II
a. $$Co^{3+}$$ i. $$\sqrt {8}B.M.$$
b. $$Cr^{3+}$$ ii. $$\sqrt {35}B.M.$$
c. $$Fe^{3+}$$ iii. $$\sqrt {3}B.M.$$
d. $$Ni^{2+}$$ iv. $$\sqrt {24}B.M.$$
v. $$\sqrt {15}B.M.$$

A
$$a - iv, b - i, c - ii, d - iii$$

B
$$a - iv, b - v, c - ii, d - i$$

C
$$a - iii, b - v, c - i, d - ii$$

D
$$a - i, b - ii, c - iii, d - iv$$

Solution

Let us calculate the magnetic moments of the ions given :

The formula of magnetic moment $$(\mu)$$ is $$\sqrt {n(n+2)}$$, where n is the number of unpaired electrons.

(i) $${Co}^{3+}$$: It has $$d^6$$ electrons which means 4 unpaired electrons.
$$\mu=\sqrt {4(4+2)} = \sqrt {24}$$

(ii) $${Cr}^{3+}$$: It has $$d^3$$ electrons which means 3 unpaired electrons.
$$\mu=\sqrt {3(3+2)} = \sqrt {15}$$

(iii) $${Fe}^{3+}$$: It has $$d^5$$ electrons which means 5 unpaired electrons.
$$\mu=\sqrt {5(5+2)} = \sqrt {35}$$

(iv) $${Ni}^{2+}$$: It has $$d^8$$ electrons which means 2 unpaired electrons.
$$\mu=\sqrt {2(2+2)} = \sqrt {8}$$

Hence, the correct match is given by option B.
Question 9
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The calculated spin only magnetic moment of $$Cr^{2+}$$ ion is

A
$$4.90 BM$$

B
$$5.92 BM$$

C
$$2.84 BM$$

D
$$3.87 BM$$

Solution

Electronic configuration of the ion:

$$Cr^{2+} \Rightarrow [Ar] 4s^0 3d^4$$

$$n=4$$

Spin only magnetic moment $$\Rightarrow \sqrt {n(n+2)} = \sqrt {4 \times (4+2)}=4.9 BM$$

Option A is correct.
Question 10
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Which one of the following ions exhibits d-d transition and paramagnetism as well?

A
$$MnO_{4}^{-}$$

B
$$CrO_{4}^{2-}$$

C
$$MnO_{4}^{2-}$$

D
$$Cr_{2}O_{7}^{2-}$$

Solution

In case of:
(i) $${MnO_4}^-$$, Mn is in the $$+7$$ oxidation state which means it has zero d $$(d^o)$$ electrons. Hence, there is no d-d transition and no paramagnetism. Its colour is due to charge transfer spectra.

(ii) $${CrO_4}^{2-}$$, Cr is in the $$+6$$ oxidation state which means it has zero d $$(d^o)$$ electrons. Hence, there is no d-d transition and no paramagnetism.

(iii) $${MnO_4}^{2-}$$, Mn is in the $$+6$$ oxidation state which means there is one d $$(d^1)$$ electron which gives rise to d-d transition and paramagnetism.

(iv) $$Cr_2O_7^{2-}$$, Cr is in the $$+6$$ oxidation state which means it has zero d $$(d^o)$$ electrons. Hence, there is no d-d transition and no paramagnetism. Its colour is due to charge transfer spectra.

Hence, option C is the correct answer.

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Re-Attempt Weekly Quiz Competition

	Column I		Column II
a.	$$Co^{3+}$$	i.	$$\sqrt {8}B.M.$$
b.	$$Cr^{3+}$$	ii.	$$\sqrt {35}B.M.$$
c.	$$Fe^{3+}$$	iii.	$$\sqrt {3}B.M.$$
d.	$$Ni^{2+}$$	iv.	$$\sqrt {24}B.M.$$
		v.	$$\sqrt {15}B.M.$$

The d- and f- Block Elements Test - 10

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The d- and f- Block Elements Test - 10

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Question 1

$$Cr=Fe>Co>Ti$$

$$Fe>Co>Cr>Ti$$

$$Cr>Fe>Co>Ti$$

$$Cr=Fe>Ti>Co$$

Solution

Question 2

the 5f orbitals are more buried than the 4f orbitals.

there is a similarity between 4f and 5f orbitals in their angular part of the wave function.

the actinoids are more reactive than the lanthanoids.

the 5f orbitals extend further from the nucleus than the 4f orbitals.

Solution

Question 3

(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)

(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)

(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)

(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)

Solution

Question 4

$${Co}^{3+}$$ and $${Cr}^{3+}$$

$${Co}^{3+}$$ and $${Co}^{3+}$$

$${Cr}^{3+}$$ and $${Cr}^{3+}$$

$${Cr}^{3+}$$ and $${Co}^{3+}$$

Solution

Question 5

$$[ZnCl_4]^{2-} > [NiCl_4]^{2-} > [CoCl_4]^{2-} > [MnCl_4]^{2-}$$

$$[CoCl_4]^{2-} > [MnCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$$

$$[NiCl_4]^{2-} > [CoCl_4]^{2-} > [MnCl_4]^{2-} > [ZnCl_4]^{2-}$$

$$[MnCl_4]^{2-} > [CoCl_4]^{2-} > [NiCl_4]^{2-} > [ZnCl_4]^{2-}$$

Solution

Question 6

$$V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}$$

$$V^{2+} > Ru^{3+} > Cr^{2+} > Fe^{2+}$$

$$Cr^{2+} > V^{2+} > Ru^{3+} > Fe^{2+}$$

$$Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+}$$

Solution

Question 7

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

more reactive nature of the actinoids than the lanthanoids

4f orbitals more diffused than the 5f orbitals

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

Solution

Question 8

$$a - iv, b - i, c - ii, d - iii$$

$$a - iv, b - v, c - ii, d - i$$

$$a - iii, b - v, c - i, d - ii$$

$$a - i, b - ii, c - iii, d - iv$$

Solution

Question 9

$$4.90 BM$$

$$5.92 BM$$

$$2.84 BM$$

$$3.87 BM$$

Solution

Question 10

$$MnO_{4}^{-}$$

$$CrO_{4}^{2-}$$

$$MnO_{4}^{2-}$$

$$Cr_{2}O_{7}^{2-}$$

Solution

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