Coordination Compounds Test

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Coordination Compounds Test - 34

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Question 1
1 / -0

The sum of coordination number and oxidetion number of the metal M in the complex $$\left [ M\left ( en \right )_{2}\left ( C_{2}O_{4} \right ) \right ]$$ Cl (where en is ethylenediamine) is:

A
6

B
7

C
8

D
9

Solution

$$[M(en)_2(C_2O_4)]Cl$$
this can be written as $$[M(en)_2(C_2O_4)]^{+1}$$
Consider $$X$$ as oxidation number,
$$X-4-2=+1$$
$$X=+7$$
so oxidation number is $$+7$$
Question 2
1 / -0

Which of the following are not true about hexagonal close packing?

A
It has a coordination number of $$6$$

B
It has $$26$$% empty space

C
It is $$ABCABC......$$ type of arrangement

D
It is as closely packed as body centred cubic packing

Solution

Hexagonal closed pack has the third layer with the same arrangement of spheres as the first layer covers all the tetrahedral holes. They efficiently occupy $$74\%$$ of space while $$26\%$$ is empty. It has a co-ordination number of $$12$$ and contains $$6$$ $$atoms/unit$$ cell.
Question 3
1 / -0

Which of the following molecules are formed by $$p-p$$ over lapping?

A
$$Cl_2$$

B
$$HCl$$

C
$$H_2O$$

D
$$NH_3$$

Solution

Solution:- (A) $${Cl}_{2}$$
$$p-p$$ overlapping is produced by the overlapping of the $$p$$-orbitals of the atoms. In the case of chlorine molecule, it is produced by the overlapping of the $$3{p}_{z}$$ orbitals of two chlorine atoms.
Question 4
1 / -0

The number of isomers represented by the M.F $${C}_{2}{D}_{2}(Cl)(Br)$$ is:

A
2

B
3

C
5

D
none of these

Solution

There are $$2$$ isomers possible.
Question 5
1 / -0

In aqueous solution 1 mole of $$CoCl_3.4NH_3$$ gives 1 mole of AgCl precipitate with excess $$AgNO_3$$, then what is the secondary valency of metal?

A
3

B
4

C
5

D
6

Solution

$$1$$ Mole of $$CoC{l_3}.4HN{O_3}$$ gives $$1$$ mole of $$AgCl$$
According to Alfred Werner theory of co-ordination compound the secondarty valence of metal is $$6$$
Hence, option $$(D)$$ is the correct answer.
Question 6
1 / -0

A six-coordinate complex of formula $$CrCl_3. 6H_2O $$ has a green colour. A 0.1 M solution of the complex, when treated with excess $$AgNO_3$$, gave 28.7 g of white precipitate. The formula of the complex would be:

A
$$[Cr(H_2O)_6]Cl_3 $$

B
$$[Cr(H_2O)_5Cl]Cl_2.H_2O$$

C
Both A & B

D
None of these

Solution
Question 7
1 / -0

With a change in hybridization of the carbon bearing the charge, the stability of a carbanion decreases in the order

A
$$sp < { sp }^{ 2 } < { sp }^{ 3 }$$

B
$$sp < { sp }^{ 3 } < { sp }^{ 2 }$$

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

D
$${ sp }^{ 2 } < sp < { sp }^{ 3 }$$

Solution

Stability of carbanions increases with increase in $$s-$$ character of hybrid orbital of carbon bearing charge hence correct order is:

$$ sp^{3} < sp^{2} < sp$$
Question 8
1 / -0

The complex ion$$\left[ Fe({ CN) }_{ 6 } \right] ^{ 4- }$$ contains :

A
$${ sp }^{ 3 }{ d }^{ 2 }$$ hybrid orbitals with octahedral structure

B
total 104 electrons on $${ Fe }^{ +2 }$$

C
six sigma Bonds

D
total of 36 electrons on $${ Fe }^{ 2+ }$$ ions.

Solution

The hybridisation of $$[Fe(Cn)_6]_4^{3+}$$ is $$d^2sp^3$$ and the shape is octahedral due to the strong field ligand $$CN^-$$
* $$Fe^{2+}$$ has already 24 electrons in electronic configuration and $$6C\bar N$$ molecules donate 12 electrons combine gives 36 electrons on $$fe^{2+}$$ ions.
Question 9
1 / -0

According to Werner's theory the primary valency and secondary valency in complex $$[Co(NH_3)_4Cl_2]Cl$$ is respectively:

A
3,3

B
3, 6

C
6, 3

D
4, 2

Solution

The primary valency is the oxidation state of the metal ion and the coordination number of the metal is the secondary valency of the metal ion.

The oxidation number of $$Co = x + 0 -3 = 0$$

$$x = 3$$

The primary valency is 3 and the secondary valency is 6. Therefore, the correct option is B.
Question 10
1 / -0

Which of the following is an EAN of iron in potassium ferricyanide?

A
$$28$$

B
$$36$$

C
$$35$$

D
$$39$$

Solution

$$EAN$$= Atomic number of central atom $$-$$ Oxidation state+ (Number of ligands $$\times 2$$)
$$EAN= 26-3+(6 \times 2)= 26+12-3=38-3$$
Hence, $$EAN=35$$ .

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Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 34

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Coordination Compounds Test - 34

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SHARING IS CARING

Question 1

6

7

8

9

Solution

Question 2

It has a coordination number of $$6$$

It has $$26$$% empty space

It is $$ABCABC......$$ type of arrangement

It is as closely packed as body centred cubic packing

Solution

Question 3

$$Cl_2$$

$$HCl$$

$$H_2O$$

$$NH_3$$

Solution

Question 4

2

3

5

none of these

Solution

Question 5

3

4

5

6

Solution

Question 6

$$[Cr(H_2O)_6]Cl_3 $$

$$[Cr(H_2O)_5Cl]Cl_2.H_2O$$

Both A & B

None of these

Solution

Question 7

$$sp < { sp }^{ 2 } < { sp }^{ 3 }$$

$$sp < { sp }^{ 3 } < { sp }^{ 2 }$$

$${ sp }^{ 3 } < { sp }^{ 2 } < sp$$

$${ sp }^{ 2 } < sp < { sp }^{ 3 }$$

Solution

Question 8

$${ sp }^{ 3 }{ d }^{ 2 }$$ hybrid orbitals with octahedral structure

total 104 electrons on $${ Fe }^{ +2 }$$

six sigma Bonds

total of 36 electrons on $${ Fe }^{ 2+ }$$ ions.

Solution

Question 9

3,3

3, 6

6, 3

4, 2

Solution

Question 10

$$28$$

$$36$$

$$35$$

$$39$$

Solution

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