Coordination Compounds Test

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Coordination Compounds Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The cobalt hexammine cation has an atom of cobalt in the centre, surrounded by six molecules of ammonia that are attached to it. The shape of this ion is?

A
Tetrahedral

B
Square

C
Linear

D
Octahedral

Solution

REF.Image.
$$ CO^{3+} \Rightarrow [Ar]d^{6} \Rightarrow $$
$$ \Rightarrow SP^{3}d^{2}$$
$$ \Rightarrow [Co(NH_{3})_{6}]^{3P}$$
octagonal
$$ \Rightarrow PP^{3}d^{2}\rightarrow $$ hydrogenation.
Question 2
1 / -0

Pair of compounds which is planar:

A
$${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 }$$

B
$${ \left[ Ni{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }\quad $$

C
$${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Rh{ \left( CO \right) }_{ 2 }{ \left( P{ Ph }_{ 3 } \right) }_{ 2 } \right] }^{ + }$$

D
$${ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }$$
Question 3
1 / -0

The total number of possible isomer for the complex comp. :- $$\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{I_4}} \right]$$

A
3

B
6

C
5

D
4

Solution

Total 4 isomers are possible [Cu(NH3)3Cl][Pt(Cl3)NH3] , [Cu(NH3)2Cl2][PtCl2(NH3)2], [CuNH3Cl3][PtCl(NH3)3], [CuCl4][Pt(NH3)4]
Question 4
1 / -0

Oxidation number of $$Cr$$ in $$[Cr[NH_3]_4 Cl_2]^+$$ is ____.

A
$$+4$$

B
$$+2$$

C
$$+3$$

D
$$+6$$

Solution

Oxidation number of $$Cr$$ in $$[Cr[NH_3]_4 Cl_2]^+$$ is $$+3$$
$$x + (4 \times 0) + (-1 \times 2) = +1$$
$$x - 2 = +1$$
$$x = +3$$
Question 5
1 / -0

In which of the following octahedral complexes of Co (at no. 27), will the magnitude of $$\triangle _0$$ be the highest?

A
$$[Co(C_2O_4)_3]^{3-}$$

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

C
$$[Co(NH_3)_6]^{3+}$$

D
$$[Co(CN)_6]^{3-}$$

Solution
Question 6
1 / -0

Which of the following will give maximum number of isomers?

A
$$\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]$$

B
$$\left [ Ni\left ( en \right )\left ( NH_{3} \right )_{4} \right ]^{2+}$$

C
$$\left [ Ni\left ( C_{2}O_{4} \right )\left ( en \right )_{2} \right ]$$

D
$$\left [ Cr\left ( SCN \right )_{2}\left ( NH_{3} \right )_{4} \right ]^{2+}$$

Solution

$$[Cr(SCN)_2(NH_3)_4]^{2+}$$ has the maximum number of isomers (geometrical + linkage). $$\left [ Ni\left ( en \right )\left ( NH_{3} \right )_{4} \right ]^{2+}$$, $$\left [ Ni\left ( C_{2}O_{4} \right )\left ( en \right )_{2} \right ]$$ and $$\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]$$ have 2 geometrical isomers each.
Question 7
1 / -0

The correct IUPAC name of $$Mn_{3}(CO)_{12}$$ is:

A
mangesedodecylcarbonyl(0)

B
dodecacarbonylmanganate(0)

C
dodecacarbonylmanganic(II)

D
dodecacarbonyltrimanganese(0)

Solution

The correct IUPAC name of $$Mn_{3}(CO)_{12}$$ is $$dodecacarbonyltrimanganese(0)$$. The central metal atom is a cobalt atom in zero oxidation state. The oxidation state is written in parenthesis. Three such $$Mn$$ atoms are present. Hence, the prefix tri is used. $$12$$ carbonyl groups are present. Hence, the prefix $$dodeca$$ is used.
Question 8
1 / -0

Which of the following metal ion is present in Ziegler-Natta catalyst?

A
$$Fe^{2+}$$

B
$$Zr^{4+}$$

C
$$Rh^{+}$$

D
$$Ti^{4+}$$

Solution

Ziegler-Natta catalyst is triethylaluminium $$Al(C_2H_5)_3$$ along with $$TiCl_4$$.

Hence, option D is correct.
Question 9
1 / -0

Which of the following ions is expected to be coloured?

A
$$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{3+}$$

B
$$\left [ Sc\left ( H_{2}O \right )_{6} \right ]^{3+}$$

C
$$\left [ Zn\left ( H_{2}O \right )_{6} \right ]^{2+}$$

D
$$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{4+}$$

Solution

A) Ti has the electronic configuration $$[Ar] 3d^2$$. In this complex, it is present as $$Ti^{3+}$$ which has one unpaired electron in 3d orbital. It undergoes $$d^2sp^3$$ (octahedral) hybridization to accommodate 6 pairs of water.
So, the complex is octahedral in shape, will be coloured and paramagnetic due to the presence of one unpaired electron.
B) Sc has the electronic configuration $$[Ar] 3d^1$$. Since, it is present in the complex as $$Sc^{3+}$$, it does not have any unpaired electron. It undergoes $$d^2sp^3$$ hybridization to accommodate 6 pairs of electrons from water. So, it is octahedral in shape, colourless and diamagnetic due to the absence of unpaired electrons.
C) In this complex, $$Zn$$ is present as $$Zn^{2+}$$ and there is no unpaired electron in the complex. So, it is colourless.
D) In this complex, $$Ti$$ is present as $$Ti^{0}$$ and there is no unpaired electron in the complex. So, it is colourless.
Question 10
1 / -0

The number of isomers exhibited by $$\left [ Cr\left ( NH_{3} \right )_{3}Cl_{3} \right ]$$ is:

A
2

B
3

C
4

D
5

Solution

The complex $$\left [ Cr\left ( NH_{3} \right )_{3}Cl_{3} \right ]$$ is of the type $$Ma_3b_3$$, has two isomers cis and trans or facial and meridional.

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Coordination Compounds Test - 13

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Coordination Compounds Test - 13

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Question 1

Tetrahedral

Square

Linear

Octahedral

Solution

Question 2

$${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 }$$

$${ \left[ Ni{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }\quad $$

$${ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 },{ \left[ Rh{ \left( CO \right) }_{ 2 }{ \left( P{ Ph }_{ 3 } \right) }_{ 2 } \right] }^{ + }$$

$${ \left[ Pt{ Cl }_{ 4 } \right] }^{ -2 },{ \left[ Ni{ \left( CN \right) }_{ 4 } \right] }^{ -4 }$$

Question 3

3

6

5

4

Solution

Question 4

$$+4$$

$$+2$$

$$+3$$

$$+6$$

Solution

Question 5

$$[Co(C_2O_4)_3]^{3-}$$

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

$$[Co(NH_3)_6]^{3+}$$

$$[Co(CN)_6]^{3-}$$

Solution

Question 6

$$\left [ Co\left ( NH_{3} \right )_{4} Cl_{2}\right ]$$

$$\left [ Ni\left ( en \right )\left ( NH_{3} \right )_{4} \right ]^{2+}$$

$$\left [ Ni\left ( C_{2}O_{4} \right )\left ( en \right )_{2} \right ]$$

$$\left [ Cr\left ( SCN \right )_{2}\left ( NH_{3} \right )_{4} \right ]^{2+}$$

Solution

Question 7

mangesedodecylcarbonyl(0)

dodecacarbonylmanganate(0)

dodecacarbonylmanganic(II)

dodecacarbonyltrimanganese(0)

Solution

Question 8

$$Fe^{2+}$$

$$Zr^{4+}$$

$$Rh^{+}$$

$$Ti^{4+}$$

Solution

Question 9

$$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{3+}$$

$$\left [ Sc\left ( H_{2}O \right )_{6} \right ]^{3+}$$

$$\left [ Zn\left ( H_{2}O \right )_{6} \right ]^{2+}$$

$$\left [ Ti\left ( H_{2}O \right )_{6} \right ]^{4+}$$

Solution

Question 10

2

3

4

5

Solution

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