Coordination Compounds Test

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Coordination Compounds Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Select the correct IUPAC name for $$[Co(NH_3)_6]^{2+}$$.

A
Hexammoniacobaltate(II) ion

B
Hexaamminecobaltate(II) ion

C
Hexammoniacobalt(II) ion

D
Hexaamminecobalt(II) ion

Solution

In $$[CO(NH_3)_6]^{2+}$$ the oxidation state of CO is 2. So the correct IUPAC name is hexaamminecobalt(II) ion.
Question 2
1 / -0

The shapes of $$IF_3$$ and $$IF_7$$ are respectively:

A
square pyramidal and pentagonal bipyramidal

B
octahedral and pyramidal

C
trigonal bipyramidal and square antiprismatic

D
distorted square planar and distorted octahedral

Solution

As refer to IMAGE $$01$$
$$I\rightarrow 5s^25p^5$$
at ground state.
For $$IF_3$$;
$$I$$ at excited state.
Geometry is Trigonal bipyramidal as refer to IMAGE $$02$$
As refer to IMAGE $$03$$
For $$IF_7$$,
$$I$$ at excited state is $$sp^3d^3$$
Geometry $$=$$ Pentagonal bipyramidal
Shape $$=$$ Pentagonal bipyramidal as refer to IMAGE $$04$$
Question 3
1 / -0

The hybridisation and geometry of $$BrF_3$$ molecules are:

A
$$sp^3d$$ and $$T$$ shaped

B
$$sp^2d^2$$ and tetragonal

C
$$sp^3d$$ and bent

D
none of these

Solution

Steric no. of $$BrF_3=\cfrac{1}{2}[7+3]=5 \rightarrow sp^3d$$
Molecular geometry is trigonal bipyramidal.
Shape is $$T-$$shaped (Refer IMAGE $$01$$)
$$Br$$ at ground state (Refer IMAGE $$02$$)
$$Br- 4s^{2}4p^{5}$$
$$Br$$ at excited state (Refer IMAGE $$03$$)
$$Br- sp^{3}d$$ with $$2lp$$
Question 4
1 / -0

The geometry of $$[Ni(CN)_4]^{2-}$$ is:

A
tetrahedral

B
square planar

C
triangular

D
octahedral

Solution

The geometry of $${ \left[ Ni{ (CN) }_{ 4 } \right] }^{ 2- }$$ is square planar as shown in the image.
Question 5
1 / -0

The IUPAC name of $$[Ni(CO)_4]$$ is :

A
Tetracarbonyl nickelete (0)

B
Tetracarbonyl nickelete (II)

C
Tetracarbonyl nickel (0)

D
Tetracarbonyl nickel (II)

Solution

The IUPAC name of $$Ni{ (CO) }_{ 4 }$$ is Tetracarbonyl nickel (0) since the oxidation state of carbonyl ligand is 0, so $$Ni$$ will also be 0 and four ligand are bonded to central metal atom.
Question 6
1 / -0

What is the angle between any two bonds in methane molecule?

A
$$105^o54'$$

B
$$109^o28'$$

C
$$119^o28'$$

D
$$190^o28'$$

Solution

The angle between any two bonds to the central carbon atom in the methane molecule is $$109^o28'$$. The molecule of methane has a tetrahedral geometry.
Question 7
1 / -0

Which of the following biomolecules contain non-transition metal ion?

A
Vitamin $$B_{12}$$

B
Chlorophyll

C
Haemoglobin

D
Insulin

Solution

Chlorophyll contains a non-transition metal ion. Chlorophyll is magnesium porphyrin complex. It is a green pigment present in plant leaves and is used in photosynthesis.

Hence, option $$B$$ is correct.
Question 8
1 / -0

Which compound is formed when excess of $$KCN$$ is added to an aqueous solution of copper sulphate?

A
$$Cu(CN)_2$$

B
$$K_2[Cu(CN)_2]$$

C
$$K[Cu(CN)_2]$$

D
$$K_3[Cu(CN)_4]$$

Solution

When excess of $$KCN$$ is added to an aqueous solution of $$CuSO_4$$ then-

1. Initially, cupric cyanide is formed, $$Cu(CN)_2$$.
$$CuSO_4+2KCN\rightarrow K_2CO_4+Cu(CN)_2$$

2.Cupric Cyanide will decompose to produce cuprous cyanide,$$Cu_2(CN)_2$$.
$$2Cu(CN)_2\rightarrow 2Cu_2(CN)_2 + 2(CN)$$

3. Cuprous cyanide reacts with an excess of $$KCN$$ to form a complex, $$K_3[Cu(CN)_4]$$, which is a stable complex.
$$CuCN+3Ka\rightarrow K_3[Cu(CN)_4]$$.

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

The isomerism exhibited by following compounds $$[Co(NH_3)_6][Cr(CN)_6]$$ and $$[Cr(NH_3)_6][Co(CN)_6]$$ is:

A
ionization isomerism

B
polymerization isomerism

C
linkage isomerism

D
coordination isomerism

Solution

Coordination isomerism occurs in those complexes in which the total ratio of ligand to metal remains the same but the ligands attached to the specific metal ion varies.

Hence, the correct option is $$\text{D}$$
Question 10
1 / -0

An octahedral complex is formed when hybrid orbitals of the following type are involved:

A
$${sp}^{3}$$

B
$$d{sp}^{2}$$

C
$${d}^{2}{sp}^{3}$$

D
$${sp}^{2}{d}^{2}$$

Solution

Hybridisation Geometry of
complex
$$sp_3$$ Tetrahedral
$$dsp^2$$ Square planar
$$d^2sp^3$$ Octahedral
$$sp^3d^2$$ Not possible
Hence option C is correct.

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Hybridisation	Geometry of complex
$$sp_3$$	Tetrahedral
$$dsp^2$$	Square planar
$$d^2sp^3$$	Octahedral
$$sp^3d^2$$	Not possible

Coordination Compounds Test - 23

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Coordination Compounds Test - 23

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Question 1

Hexammoniacobaltate(II) ion

Hexaamminecobaltate(II) ion

Hexammoniacobalt(II) ion

Hexaamminecobalt(II) ion

Solution

Question 2

square pyramidal and pentagonal bipyramidal

octahedral and pyramidal

trigonal bipyramidal and square antiprismatic

distorted square planar and distorted octahedral

Solution

Question 3

$$sp^3d$$ and $$T$$ shaped

$$sp^2d^2$$ and tetragonal

$$sp^3d$$ and bent

none of these

Solution

Question 4

tetrahedral

square planar

triangular

octahedral

Solution

Question 5

Tetracarbonyl nickelete (0)

Tetracarbonyl nickelete (II)

Tetracarbonyl nickel (0)

Tetracarbonyl nickel (II)

Solution

Question 6

$$105^o54'$$

$$109^o28'$$

$$119^o28'$$

$$190^o28'$$

Solution

Question 7

Vitamin $$B_{12}$$

Chlorophyll

Haemoglobin

Insulin

Solution

Question 8

$$Cu(CN)_2$$

$$K_2[Cu(CN)_2]$$

$$K[Cu(CN)_2]$$

$$K_3[Cu(CN)_4]$$

Solution

Question 9

ionization isomerism

polymerization isomerism

linkage isomerism

coordination isomerism

Solution

Question 10

$${sp}^{3}$$

$$d{sp}^{2}$$

$${d}^{2}{sp}^{3}$$

$${sp}^{2}{d}^{2}$$

Solution

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