The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 69

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Question 1

1 / -0

Match the complex ions gives in Column-I with the colors given in Column-II and assign the correct code.

Column-I(Complex ion)	Column-II(Color)
(a) $$[Co(NH_3)_6]^{3+}$$	(p) Violet
(b) $$[Ti(H_2O)_6]^{3+}$$	(q) Green
(c) $$[Ni(H_2O)_6]^{2+}$$	(r) Pale blue
(d) $$[Ni(H_2O)_4(en)]^{2+}(aq)$$	(s) Yellowish orange
	(t)Blue

a(p) b(q) c(s) d(t)

a(s) b(r) c(q) d(p)

a(r) b(q) c(s) d(p)

a(s) b(p) c(q) d(r)

Solution

Question 2
1 / -0

The 'spin-only' magnetic moment [in units of Bohr magneton, $$(\mu_B)$$] of $$Ni^{2+}$$ in aqueous solution would be (atomic number of $$Ni=28$$).

A
$$6$$

B
$$1.73$$

C
$$2.84$$

D
$$4.90$$

Solution

$$Ni=[Ar]3d^84s^2;\quad Ni^{2+}=[Ar]3d^8$$

$$\mu_B=\sqrt{n(n+2)}$$

In $$Ni^{2+}$$, unpaired electron is $$2$$.

$$\therefore \mu_B=\sqrt{2(2+2)}=\sqrt{8}$$

$$\mu_B=2.828$$

$$\mu_B\approx 2.84\ BM$$.
Question 3
1 / -0

Among $$Ni(CO)_4, [Ni(CN)_4]^{2-}$$, and $$[NiCl_4]^{2-}$$.

A
$$Ni(CO)_4$$ and $$[NiCl_4]^{2-}$$ are diamagnetic and $$[Ni(CN)_4]^{2-}$$ is paramagnetic

B
$$[NiCl_4]^{2-}$$ and $$[Ni(CN)_4]^{2-}$$ are diamagnetic and $$Ni(CO)_4$$ is paramagnetic

C
$$Ni(CO)_4$$ and $$[Ni(CN)_4]^{2-}$$ are diamagnetic and $$[NiCl_4]^{2-}$$ is paramagnetic

D
$$Ni(CO)_4$$ is diamagnetic, and $$[NiCl_4]^{2-}$$ and $$[Ni(CN)_4]^{2-}$$ are paramagnetic

Solution

a). $$[NiCl_4]^{2-}$$

The oxidation number of Ni is $$+2$$

Atomic number $$=28$$

$$Ni=[Ar]3d^84s^2$$

$$Ni^{2+}=[Ar]3d^8$$

Chlorido is a weak field ligand, no pairing.

$$sp^3$$ - Hybridization (tetrahedral)

There are two unpaired electrons, so the complex is paramagnetic.

Spin magnetic moment

$$\mu =\sqrt{n(n+2)}BM$$

$$=\sqrt{2(2+2)}BM=\sqrt{8}BM$$

b). $$[Ni(CN)_4]^{2-}$$

The oxidation number of Ni is $$+2$$.

Atomic number $$=28$$

$$Ni=[Ar]3d^84s^2$$

$$Ni^{2+}=[Ar]3d^8$$

Cyano is a strong field ligand, it causes pairing.

$$dsp^2$$ - Hybridization (square planar)

There are no unpaired electrons so, the complex is diamagnetic.

Spin magnetic moment $$=$$ zero

c). $$[Ni(CO)_4]$$

The oxidation number of Ni is $$0$$.

Atomic number $$=28$$

$$Ni=[Ar]3d^84s^2$$

$$Ni=[Ar]3d^{10} \to$$ in the complex.

CO is a strong field ligand, so pairing occurs

$$sp^3$$ - Hybridization (tetrahedral)

There are no unpaired electrons so, the complex is diamagnetic.

Spin magnetic moment $$=$$ zero
Question 4
1 / -0

Wilkinson's catalyst contains :

A
rhodium

B
iron

C
aluminium

D
cobalt

Solution
Question 5
1 / -0

Among the following complexes $$(K-P)$$
$$K_3[Fe(CN)_6](K), [Co(NH_3)_6]Cl_3(L), $$

$$Na_3[Co(oxalate)_3](M), [Ni(H_2O)_6]Cl_2(N), $$

$$K_2[Pt(CN)_4](O)$$ and $$[Zn(H_2O)_6](NO_3)_2(P)$$

The diamagnetic are:

A
K, L, M, N

B
K, M, O, P

C
L, M, O, P

D
L, M, N, O

Solution
Question 6
1 / -0

Colour in transition metal compounds is attributed to:

A
small size of metal ions

B
absorption of light in $$UV$$ region

C
moderate ionisation energy

D
incomplete $$(n-1)d$$ subshell

Solution
Question 7
1 / -0

Which of the following ions is coloured in solution?

A
$$Zn^{2+}$$

B
$$Ti^{4+}$$

C
$$Cu^{+}$$

D
$$V^{3+}$$

Solution

Transition elements form different coloured compounds.The compound in which metal ion has an unpaired electron that compund will be coloured.

A. The electronic configuration of $${Zn}^{+2}$$ is [Ar]$${3d}^{10}$$.
No. of unpaired electron = 0.

B. The electronic configuration of $${Ti}^{+4}$$ is [Ar]$${3d}^{0}$$.
No. of unpaired electron = 0.

C. The electronic configuration of $${Cu}^{+}$$ is [Ar]$${3d}^{10}$$.
No. of unpaired electron = 0.

D. The electronic configuration of $${V}^{3+}$$ is [Ar]$${3d}^{2}$$.
No. of unpaired electron = 2.
As $${V}^{3+}$$ contains 2 unpaired electron it will show colour.
$${V}^{3+}$$ shows green coloured solution.

Hence, Correct option is (D).
Question 8
1 / -0

Directions For Questions

When crystals of $$CuS{ O }_{ 4 }\cdot 4N{ H }_{ 3 }$$ are dissolved in water, there is hardly any evidence for the presence of $$Cu^{2+}$$ ions or ammonia molecules. A new ion, $${ \left[ Cu{ \left( N{ H }_{ 3 } \right) }_{ 4 } \right] }^{ 2+ }$$, is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, the aqueous solution of $$Fe\left( CN \right) _{ 2 }\cdot 4KCN$$ does not gives tests of $$Fe^{2+}$$ and $$CN^{-}$$ ions but gives test of a new ion. $${ \left[ Fe{ \left( { CN }\right) }_{ 6 } \right] }^{ 4- }$$. The ions $${ \left[ Cu{ \left( N{ H }_{ 3 } \right) }_{ 4 } \right] }^{ 2+ }$$ and $${ \left[ Fe{ \left( { CN }\right) }_{ 6 } \right] }^{ 4- }$$ are called complex ions. These ions are represented as :

...view full instructions

Which one of the following statements is correct?

A
$$[Cu(NH_3)_4]^{2+}$$ is diamagnetic while $$[Fe(CN)_6]^{4-}$$ is paramagnetic

B
$$[Cu(NH_3)_4]^{2+}$$ is paramagnetic while $$[Fe(CN)_6]^{4-}$$ is diamagnetic

C
Both are paramagnetic

D
Both are diamagnetic

Solution
Question 9
1 / -0

Transition metals show paramagnetic :

A
due to characteristic configuration

B
high lattice energy

C
due to variable oxidation states

D
due to unpaired electrons

Solution
Question 10
1 / -0

Match List-I with List -II and select the correct answer using codes given ahead in the lists:
List-I
Metal ions List-II
Magnetic moments (B.M)
$$Cr^{3+}$$ $$\sqrt{35}$$
$$Fe^{2+}$$ $$\sqrt{30}$$
$$Ni^{2+}$$ $$\sqrt {24}$$
$$Mn^{2+}$$ $$\sqrt{15}$$
$$\sqrt 8$$

A
$$A-1, B-3, C-5, D-4$$

B
$$A-2, B-3, C-5, D-1$$

C
$$A-4, B-3, C-5, D-1$$

D
$$A-4, B-5, C-3, D-1$$

Solution

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Re-Attempt Weekly Quiz Competition

List-I Metal ions	List-II Magnetic moments (B.M)
$$Cr^{3+}$$	$$\sqrt{35}$$
$$Fe^{2+}$$	$$\sqrt{30}$$
$$Ni^{2+}$$	$$\sqrt {24}$$
$$Mn^{2+}$$	$$\sqrt{15}$$
	$$\sqrt 8$$

The d- and f- Block Elements Test - 69

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The d- and f- Block Elements Test - 69

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Question 1

a(p) b(q) c(s) d(t)

a(s) b(r) c(q) d(p)

a(r) b(q) c(s) d(p)

a(s) b(p) c(q) d(r)

Solution

Question 2

$$6$$

$$1.73$$

$$2.84$$

$$4.90$$

Solution

Question 3

$$Ni(CO)_4$$ and $$[NiCl_4]^{2-}$$ are diamagnetic and $$[Ni(CN)_4]^{2-}$$ is paramagnetic

$$[NiCl_4]^{2-}$$ and $$[Ni(CN)_4]^{2-}$$ are diamagnetic and $$Ni(CO)_4$$ is paramagnetic

$$Ni(CO)_4$$ and $$[Ni(CN)_4]^{2-}$$ are diamagnetic and $$[NiCl_4]^{2-}$$ is paramagnetic

$$Ni(CO)_4$$ is diamagnetic, and $$[NiCl_4]^{2-}$$ and $$[Ni(CN)_4]^{2-}$$ are paramagnetic

Solution

Question 4

rhodium

iron

aluminium

cobalt

Solution

Question 5

K, L, M, N

K, M, O, P

L, M, O, P

L, M, N, O

Solution

Question 6

small size of metal ions

absorption of light in $$UV$$ region

moderate ionisation energy

incomplete $$(n-1)d$$ subshell

Solution

Question 7

$$Zn^{2+}$$

$$Ti^{4+}$$

$$Cu^{+}$$

$$V^{3+}$$

Solution

Question 8

$$[Cu(NH_3)_4]^{2+}$$ is diamagnetic while $$[Fe(CN)_6]^{4-}$$ is paramagnetic

$$[Cu(NH_3)_4]^{2+}$$ is paramagnetic while $$[Fe(CN)_6]^{4-}$$ is diamagnetic

Both are paramagnetic

Both are diamagnetic

Solution

Question 9

due to characteristic configuration

high lattice energy

due to variable oxidation states

due to unpaired electrons

Solution

Question 10

$$A-1, B-3, C-5, D-4$$

$$A-2, B-3, C-5, D-1$$

$$A-4, B-3, C-5, D-1$$

$$A-4, B-5, C-3, D-1$$

Solution

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