The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 36

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Weekly Quiz Competition

Question 1
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Zr and Hf have almost equal atomic and ionic radii because of:

A
diagonal relationship.

B
lanthanoid contraction.

C
actinoid contraction.

D
belonging to the same group.

Solution

This is because of poor screening effect by 4f electrons,which lead to the lanthanoid contraction.Hence,is the reason of $$Zn$$ and $$Hf$$ have almost equal atomic size.
Question 2
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Due to lanthanoid contraction which of the following properties is not expected to be similar in the same vertical columns of second and third row transition elements?

A
Atomic radii

B
Ionisation energies

C
Magnetic moments

D
Lattice energies

Solution

Magnetic moment depends on the number of unpaired electrons, as the configuration changes the number of unpaired electrons changes. Hence magnetic moment changes.
Question 3
1 / -0

The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of $$Cr^{3+}$$ ion is :

A
2.87 B.M.

B
3.87 B.M.

C
3.47 B.M.

D
3.57 B.M.

Solution

$${ Cr }^{ +3 }$$ has an electronic configuration of $$\left[ Ar \right] 3{ d }^{ 3 }4{ s }^{ 0 }$$ ,having unpaired electrons as $$3$$.Hence magnetic moment of $${ Cr }^{ +3 }$$ is $$\sqrt { 3(3+2) }=3.87 BM $$
Question 4
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The second and third row elements of transition metals resemble each other much more than they resemble the first row because of:

A
lanthanoid contraction which results in almost same radii of second and third row metals

B
diagonal relationship between second and third row elements

C
similar ionisation enthalpy of second and third row elements

D
similar oxidation states of second and third row metals

Solution

The second and third row of transition elements resemble each other more than they do with the first row of transition elements due to lanthanide contraction calmost same atomic radius because of increase in nuclear charge.
Question 5
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Which of the following statements is not correct about magnetic behaviour of substances?

A
Diamagnetic substances are repelled by an applied magnetic field.

B
Paramagnetic substances are attracted by an applied magnetic field.

C
Magnetic moment of n unpaired electrons is given by $$\mu=\sqrt n(n-2)$$B.M.

D
Magnetic moment increases as the number of unpaired electrons increases.

Solution

Diamasnetic substances are repelled by an applied magnetic field.[since,it has no unpaired electrons]
paramagnetic substances,due to presence of unpaired free electrons in them,are attracted by an applied magnetic field.
Magnetic moment of nunpaired electrons is given by the formula,magnetic moment$$\mu=\sqrt{n(n+2)}$$.
Question 6
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In which of the following ions, the colour is not due to $$d-d$$ transition?

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

B
$$[Cu(NH_3)_4]^{2+}$$

C
$$[CoF_6]^{3-}$$

D
$$CrO_4^{2-}$$

Solution

Colour is due to d-d transtition ,then such complexes have colour due to d-d transition are surely octahedral complexes.
Among the given options,
$${ \left[ Ti{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$
$${ \left[ Cu{ \left( { NH }_{ 3 } \right) }_{ 4 } \right] }^{ 2+ }$$
$${ \left[ { COF }_{ 6 } \right] }^{ 3- }$$
This are octahedral complexes.
$$CrO_4^{2-}$$ is a salt,it colour is not due to d-d transition.
Question 7
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Directions For Questions

Two types of magnetic behaviour are found in substances :
(a) Diamagnetism and (b) Paramagnetism
Diamagnetic substances are those which are repelled by an applied magnetic field. Such substances have no unpaired electron. Paramagnetic substances are those which are attracted by an applied magnetic field. Transition metals and many of their compounds show paramagnetic behaviour where there are unpaired electron or electrons. The magnetic moment arise from the spin and orbital motions in ions or molecules. Magnetic moment of $$n$$ unpaired electrons is given as,
$$\mu =\sqrt { n\left( n+2 \right) } $$ Bor Magneton

Magnetic moment increases as the number of unpaired electrons increases.

...view full instructions

Arrange the following in increasing value of magnetic moments.
(i) $$[Fe(CN)_6]^{4-}$$
(ii) $$Fe(CN)_6]^{3-}$$
(iii) $$Cr(NH_3)_6^{ 3+}$$
(iv) $$Ni(H_2O)^4]^{ 2+}$$

A
$$(I) < (II) < (III) < (IV)$$

B
$$(I) < (II) < (IV) < (III)$$

C
$$(II) < (III) < (I) < (IV)$$

D
$$(IV) < (III) < (II) < (I)$$

Solution

1) [Fe(CN)$$_6]^{4-}$$
Here, Fe is present as Fe$$^{2+}$$
Electronic configuration of  Fe$$^{2+}$$ is  [Ar]3d$$^6 4s^0$$
Since CN$$^-$$ is a strong field ligand so when it approaches Fe$$^{2+}$$ it pair up the e$$^-$$ s and result in the formation of low spin complex.
No. of unpaired e$$^-$$ s =0
So, magnetic moment for n unpaired e$$^-$$s
= $$\sqrt{n(n+2)} BM$$
=0
2) [Fe(CN)$$_6]^{3-}$$
Here, Fe is present as Fe$$^{3+}$$
Electronic configuration of  Fe$$^{3+}$$ is  [Ar]3d$$^5 4s^0$$
Since , CN$$^-$$ is a strong field ligand so when it approaches Fe$$^{2+}$$ it pair up the e$$^-$$s so,
No. of unpaired e$$^-$$ s =1
So, magnetic moment for n unpaired e$$^-$$s
= $$\sqrt{1\times (1+2)} $$
=$$\sqrt{3} $$ BM.
3) [Cr(NH$$_3)_6]^{3+}$$
Electronic configuration of Cr$$^{3+}$$ is  [Ar]3d$$^3 4s^0$$
Since , NH$$_3$$ is a strong field ligand so it pairs up the electrons when it approaches when it approaches  Cr$$^{3+}$$
No. of unpaired e$$^-$$ s =3
So, magnetic moment = $$\sqrt{3(3+2)} $$
=$$\sqrt{1\times 5} $$
=$$\sqrt{15}$$ BM
4)[Ni(H$$_2O)_4]^{2+}$$
Electronic configuration of Ni$$^{2+}$$ is  [Ar]3d$$^8 4s^0$$
Since , H$$_2$$O is a weak field ligand. it does not pair up the e$$^-$$s and forms outer orbital complex.
No. of unpaired e$$^-$$ s =2
So, magnetic moment =$$\sqrt{2(2+2)} $$
=$$\sqrt{8}$$ BM
But because of orbital effect it will be little lesser. So, correct order is iii)>iv)> ii)> i)
Question 8
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Which of the following statements for the reaction, is correct?
$$Na_2CrO_4 + H_2SO_4 \rightarrow$$

A
It is a redox reaction in which green solution of $$[Cr(H_2O)_6]^{+3}$$ is produced.

B
One of the product is reaction has trigonal planar structure.

C
Dimeric bridged tetrahedral metal ion is produced.

D
Dark blue color in obtained in reaction.

Solution

$${ 2Na }_{ 2 }{ CrO }_{ 4 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { Na }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }+{ Na }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }O$$
$$\Downarrow $$
dimeric bridged tetrahedral
Question 9
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Match the column I with column II and mark the appropriate choice.
Column I Column II
(A) $$FeSO_4.7H_2O$$ (i) Green
(B) $$NiCl_2.4H_2O$$ (ii) Light Pink
(C) $$MNCl_2.4H_2O$$ (iii) Pale green
(D) $$CoCl_2.6H_20$$ (iv) Pink
(E) $$Cu_2Cl_2$$ (v) Colourless

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

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

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

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

Solution

The colour of the following compounds are:
(i) $$FeSO_4$$. & $$H_2O$$ - Pale green.
(II) $$NiCl_2$$.$$4H_2O$$ -Green
(iii) $$MnCl_2$$.$$4H_2O$$ - Light pink
(iv) $$CoCl_2$$.$$6H_2O$$ - Pink
(v) $$CuCl_2$$ - Colourless.
The colour of the transition metal compounds is due to partially filled d orbitals. Due to that, d-d transition its complementary colour is seen.
Hydrated complexes are coloured because in that come H$$_2$$O molecules act as a ligand and gets arranged around the central metal atom so as to bring d-d transition.
Question 10
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Which of the following arrangements does not represent the correct order of the property stated against it?

A
$$Cu^{3+}$$< $$Fe^{3+}$$ <$$Cr^{3+}$$<$$Sc^{3+}$$: stability in aqueous solution

B
$$V^{2+}$$< $$Cr^{2+} $$<$$Fe^{2+}$$<$$Mn^{2+} $$: paramagnetic behaviour

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

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

Solution

The electronic configuration of various ionic species are:
No. of unpaired e$$^-$$
V$$^{2+}$$ - [Ar] 3d$$^3 4s^0$$ 3
Cr$$^{2+}$$ - [Ar] 3d$$^4 4s^0$$ 4
Mn$$^{2+}$$ - [Ar] 3d$$^5 4s^0$$ 5
Fe$$^{2+}$$ - [Ar] 3d$$^6 4s^0$$ 4
So, the paramagnetic behavior depends on the no. of unpaired e$$^-$$ s. Hence more the no. of unpaired e$$^-$$s, more paramagnetic species are , so the correct order is :
Mn$$^{2+}$$ >Fe$$^{2+}$$ > Cr$$^{2+}$$ >V$$^{2+}$$

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	Column I		Column II
(A)	$$FeSO_4.7H_2O$$	(i)	Green
(B)	$$NiCl_2.4H_2O$$	(ii)	Light Pink
(C)	$$MNCl_2.4H_2O$$	(iii)	Pale green
(D)	$$CoCl_2.6H_20$$	(iv)	Pink
(E)	$$Cu_2Cl_2$$	(v)	Colourless

The d- and f- Block Elements Test - 36

Result

The d- and f- Block Elements Test - 36

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SHARING IS CARING

Question 1

diagonal relationship.

lanthanoid contraction.

actinoid contraction.

belonging to the same group.

Solution

Question 2

Atomic radii

Ionisation energies

Magnetic moments

Lattice energies

Solution

Question 3

2.87 B.M.

3.87 B.M.

3.47 B.M.

3.57 B.M.

Solution

Question 4

lanthanoid contraction which results in almost same radii of second and third row metals

diagonal relationship between second and third row elements

similar ionisation enthalpy of second and third row elements

similar oxidation states of second and third row metals

Solution

Question 5

Diamagnetic substances are repelled by an applied magnetic field.

Paramagnetic substances are attracted by an applied magnetic field.

Magnetic moment of n unpaired electrons is given by $$\mu=\sqrt n(n-2)$$B.M.

Magnetic moment increases as the number of unpaired electrons increases.

Solution

Question 6

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

$$[Cu(NH_3)_4]^{2+}$$

$$[CoF_6]^{3-}$$

$$CrO_4^{2-}$$

Solution

Question 7

$$(I) < (II) < (III) < (IV)$$

$$(I) < (II) < (IV) < (III)$$

$$(II) < (III) < (I) < (IV)$$

$$(IV) < (III) < (II) < (I)$$

Solution

Question 8

It is a redox reaction in which green solution of $$[Cr(H_2O)_6]^{+3}$$ is produced.

One of the product is reaction has trigonal planar structure.

Dimeric bridged tetrahedral metal ion is produced.

Dark blue color in obtained in reaction.

Solution

Question 9

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

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

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

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

Solution

Question 10

$$Cu^{3+}$$< $$Fe^{3+}$$ <$$Cr^{3+}$$<$$Sc^{3+}$$: stability in aqueous solution

$$V^{2+}$$< $$Cr^{2+} $$<$$Fe^{2+}$$<$$Mn^{2+} $$: paramagnetic behaviour

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

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

Solution

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