The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 44

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Weekly Quiz Competition

Question 1
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Which element is a transition metal?

A
melting point in $$^{o}C$$ - 98; density in $$g/{cm}^{3}$$- 1.0; colour of oxide - white

B
melting point in $$^{o}C$$ - 328; density in $$g/{cm}^{3}$$- 11.3; colour of oxide - yellow

C
melting point in $$^{o}C$$ - 651; density in $$g/{cm}^{3}$$- 1.7; colour of oxide - white

D
melting point in $$^{o}C$$ - 1240; density in $$g/{cm}^{3}$$- 7.4; colour of oxide - black

Solution

$$\text{A transition metal must have high melting point with low density.}$$
$$\text{So option C is correct.}$$
Question 2
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Compare magnetic moment of $$Mn,{ Mn }^{ + },{ Mn }^{ 2+ },{ Mn }^{ 3+ }:$$

A
$${ Mn }^{ 3+ }>Mn>{ Mn }^{ 2+ }>{ Mn }^{ + }$$

B
$${ Mn }^{ + }>Mn^{ 2+ }>{ Mn }^{ 3+ }>{ Mn }$$

C
$$Mn={ Mn }^{ 2+ }>Mn^{ 3+ }>{ Mn }^{ + }$$

D
$${ Mn }^{ + }>Mn=Mn^{ 2+ }>{ Mn }^{ 3+ }$$

Solution
Question 3
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Which of the following pair has similar magnetic moment?

A
$$Cr^{3+}, Mn^{3+}$$

B
$$Fe^{3+}, Mn^{2+}$$

C
$$Fe^{2+}, Mn^{2+}$$

D
$$Ni^{2+}, Co^{2+}$$

Solution

The magnetic moment is given as $$\mu=\sqrt{n(n+2)}$$
where, $$n=$$ number of unpaired electrons

$$Fe^{3+}$$ and $$Mn^{2+}$$ have an equal number of 5 unpaired electrons.

Thus, they have a similar magnetic moment.
Question 4
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How many electron present in penultimate shell of d-block element?

A
$$ 9-18 $$

B
$$ 8-18 $$

C
$$ 1-10 $$

D
$$ 1-32 $$
Question 5
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Transition metal compounds are usually colored. This is due to the electronic transition

A
from p-orbital to s-orbital

B
from d-orbital to s-orbital

C
from d-orbital to p-orbital

D
within the d-orbitals

Solution

Colour in transition series metal compounds arises due to two types of transitions. they are:
(a) Charge transfer transitions [Ligand-to-metal charge transfer (LMCT), Metal-to-ligand charge transfer]
(b) $$d-d$$ transition

In transition metal complexes, all the $$d-$$ orbitals do not possess same energy. So, electrons can jump from a lower energy $$d-$$ orbital to a higher energy $$d-$$ orbital, by absorbing energy. When it returns to the ground state, excess energy is released, and a corresponding wavelength is found in the visible region.

Therefore, transition metals are colored due to the electronic transition within the $$d-$$orbitals.

So, option D is the correct answer.
Question 6
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Which of the following statements is correct for the complex, $$K_4[Fe(CN)_5O_2]$$ having $$t^6_{2g}, e^0_g$$ electronic configuration?

A
$$d^2sp^3$$ hybridized and diamagnetic

B
$$sp^3d^2$$ hybridized and paramagnetic

C
$$sp^3d^2$$ hybridized and diamagnetic

D
$$d^2sp^3$$ hybridized and paramagnetic

Solution
Question 7
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Directions For Questions

A metal complex having composition $$Cr(NH_3)_4Cl_2Br$$ has been isolated in two forms, (X) and (Y). The form (X) reacts with $$AgNO_3$$ to give a white precipitate readily soluble in dilute aqueous ammonia, whereas (Y) gives a pale yellow precipitate soluble in concentrate ammonia.

...view full instructions

Choose the true statement regarding compounds X and Y.

A
Both compounds X and Y have different magnetic moment(spin only)

B
Both complexes X and Y have same magnetic moment(spin only)

C
Magnetic moment of both X and Y is calculated by assuming they are low-spin complex

D
Both compounds X and Y have different primary valency

Solution
Question 8
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Among the following the lowest degree of paramagnetism per mole of the compound at $$298$$K will be shown by?

A
$$MnSO_4\cdot 4H_2O$$

B
$$FeSO_4\cdot H_2O$$

C
$$CuSO_4\cdot 5H_2O$$

D
$$NiSO_4\cdot 6H_2O$$

Solution

The degree of paramagnetism will depend on the number of unpaired electrons because spin magnetic moment =$$\sqrt{n(n+2)}$$B.M
Water is a weak field ligand, so there will be no pairing of electrons as the interaction of the central metal with water molecules is very weak. All the metal ions are in $$+2$$ oxidation state. The valence shell electronic configuration of $$Cu^{+2}$$ is $$3d^{9}4s^{0}$$. So, $$Cu^{2+}$$ has only one unpaired electron.
$$Mn^{2+}$$ has four unpaired electrons.
$$Fe^{2+}$$ has four unpaired electrons.
$$Ni^{+2}$$ has two unpaired electrons.

Therefore, the lowest degree of paramagnetism per mole of the compound is shown by $$CuSO_4.5H_2O$$.

Hence, the correct option is (C).
Question 9
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Which of the following statements is correct?

A
With $$d^2sp^3$$ hybridization, $$[FeCl(CN)_4(O_2)]^{4-}$$ complex is diamagnetic.

B
$$[NiCl_4]^{2-}$$ complex is more stable than $$[Ni(dmg)_2]$$ due to higher C.F.S.E. value.

C
$$[V(CO)_6]$$ is not very stable and easily reduces to $$[V(CO)_6]^-$$.

D
Ligands such as $$CO, CN^-, NO^+$$ are $$\pi e^-$$ donor due to the presence of filled $$\pi$$-molecular orbital.

Solution

The complex $$[Fe(CN)_{4}(O_{2})]^{-4}$$ is an octahedral complex. So, it may have the hybridization of $$sp^{3}d^{2}$$ or $$d^{2}sp^{3}$$. As $$CN^{-}$$ is a strong field ligand, the electronic configuration of $$Fe^{+3}$$ in the complex will be $$t_{2g}^{6}e_{g}^{0}$$ due to the pairing of electrons.

So, it will be an inner-orbital complex involving the inner $$3d$$ orbitals ($$d^{2}sp^{3}$$ hybridisation). It is also a diamagnetic complex as all the $$t_{2g}$$ electrons are paired up.

So, option (A) is correct.
Question 10
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An aqueous solution of titanium chloride, when subjected to magnetic measurement, measured zero magnetic moment. Assuming the complex as octahedral in aqueous solution, the formula of the complex is?

A
$$[Ti(H_2O)_6]Cl_2$$

B
$$[Ti(H_2O)_6]Cl_4$$

C
$$[TiCl_6]^{3-}$$

D
$$[Ti(H_2O)_4Cl_2]$$

Solution

Magnetic moment of any complex arises due to the presence of unpaired electron(s) in the central metal ion or atom. An octahedral complex is a complex surrounded by six monodentate ligands inside the coordination sphere. In the compound $$[Ti(H_{2}O)_{6}]Cl_{4}$$, $$Ti$$ is in $$+4$$ oxidation state. It has an electronic configuration of $$[Ar]3d^{0}$$. Hence, it has zero magnetic moment. So, the formula $$[Ti(H_2O)_6]Cl_4$$ fits for the conditions mentioned in the question.

Therefore, the correct option is (B).

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The d- and f- Block Elements Test - 44

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The d- and f- Block Elements Test - 44

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Question 1

melting point in $$^{o}C$$ - 98; density in $$g/{cm}^{3}$$- 1.0; colour of oxide - white

melting point in $$^{o}C$$ - 328; density in $$g/{cm}^{3}$$- 11.3; colour of oxide - yellow

melting point in $$^{o}C$$ - 651; density in $$g/{cm}^{3}$$- 1.7; colour of oxide - white

melting point in $$^{o}C$$ - 1240; density in $$g/{cm}^{3}$$- 7.4; colour of oxide - black

Solution

Question 2

$${ Mn }^{ 3+ }>Mn>{ Mn }^{ 2+ }>{ Mn }^{ + }$$

$${ Mn }^{ + }>Mn^{ 2+ }>{ Mn }^{ 3+ }>{ Mn }$$

$$Mn={ Mn }^{ 2+ }>Mn^{ 3+ }>{ Mn }^{ + }$$

$${ Mn }^{ + }>Mn=Mn^{ 2+ }>{ Mn }^{ 3+ }$$

Solution

Question 3

$$Cr^{3+}, Mn^{3+}$$

$$Fe^{3+}, Mn^{2+}$$

$$Fe^{2+}, Mn^{2+}$$

$$Ni^{2+}, Co^{2+}$$

Solution

Question 4

$$ 9-18 $$

$$ 8-18 $$

$$ 1-10 $$

$$ 1-32 $$

Question 5

from p-orbital to s-orbital

from d-orbital to s-orbital

from d-orbital to p-orbital

within the d-orbitals

Solution

Question 6

$$d^2sp^3$$ hybridized and diamagnetic

$$sp^3d^2$$ hybridized and paramagnetic

$$sp^3d^2$$ hybridized and diamagnetic

$$d^2sp^3$$ hybridized and paramagnetic

Solution

Question 7

Both compounds X and Y have different magnetic moment(spin only)

Both complexes X and Y have same magnetic moment(spin only)

Magnetic moment of both X and Y is calculated by assuming they are low-spin complex

Both compounds X and Y have different primary valency

Solution

Question 8

$$MnSO_4\cdot 4H_2O$$

$$FeSO_4\cdot H_2O$$

$$CuSO_4\cdot 5H_2O$$

$$NiSO_4\cdot 6H_2O$$

Solution

Question 9

With $$d^2sp^3$$ hybridization, $$[FeCl(CN)_4(O_2)]^{4-}$$ complex is diamagnetic.

$$[NiCl_4]^{2-}$$ complex is more stable than $$[Ni(dmg)_2]$$ due to higher C.F.S.E. value.

$$[V(CO)_6]$$ is not very stable and easily reduces to $$[V(CO)_6]^-$$.

Ligands such as $$CO, CN^-, NO^+$$ are $$\pi e^-$$ donor due to the presence of filled $$\pi$$-molecular orbital.

Solution

Question 10

$$[Ti(H_2O)_6]Cl_2$$

$$[Ti(H_2O)_6]Cl_4$$

$$[TiCl_6]^{3-}$$

$$[Ti(H_2O)_4Cl_2]$$

Solution

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