Coordination Compounds Test

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Coordination Compounds Test - 15

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Question 1
1 / -0

Transition elements form complexes readily because :

A
Small size of cation

B
Large ionic Charge

C
Vacant d orbitals

D
All the above
Solution
Transition elements form complexes readily because
Due to lower size and higher charge, they have high charge density.
They have vacant d orbital in which a ligand can donate its electron and form complex.
Question 2
1 / -0

The following ion shows colour not due to d-d transistion.

A
$$Cr_{2}O{_{7}}^{2-}$$

B
$$MnO{_{4}}^{-}$$

C
$$CrO{_{4}}^{2-}$$

D
All

Solution

All are coloured not due to d-d transition but due to charge transfer theory. They are coloured, though there are no unpaired electrons, because of "CHARGE TRANSFER SPECTRUM" (ie). Colour is associated with the electrons being promoted from one energy level to another, and absorbing or emitting the energy difference between the two levels. Thus transfer of an electron from M to L or viceverse results in charge transfer which gives rise to a spectra called charge transfer spectra.
Question 3
1 / -0

The splitting of degenerated d-orbitals takes place into which of the following two sets :

A
$$d_{xy},d_{z^{2}},d_{xz}$$and $$d_{yz},d_{x^{2}-y^{2}}$$

B
$$d_{xy},d_{yz},d_{zx}$$ and $$d_{x^{2}-y^{2}},d_{z^{2}}$$

C
$$d_{xy},d_{x^{2}-y^{2}},d_{z^{2}}$$ and $$d_{yz},d_{xz}$$

D
$$d_{xy},d_{x^{2}-y^{2}},d_{xz}$$ and $$d_{yz},d_{z^{2}}$$

Solution

The d-orbitals split into two sets with an energy difference, the crystal-field splitting parameter where the $$d_{xy}, d_{xz} \ and \ d_{yz}$$ orbitals will be lower in energy than the $$d_{z^2} \ and \ d_{x^2-y^2}$$, which will have higher energy, because the former group is farther from the ligands than the latter and therefore experience less repulsion. The three lower-energy orbitals are collectively referred to as $$t_{2g}$$, and the two higher-energy orbitals as $$e_g$$.
Question 4
1 / -0

During the splitting of degenerate d-orbitals under the influence of ligand the average d-orbital energy

A
remains same

B
increases

C
decreases

D
may increase or decrease

Solution

When examining a single transition metal ion, the five d-orbitals have the same energy. When ligands approach the metal ion, some experience more opposition from the d-orbital electrons than others based on the geometric structure of the molecule. Since ligands approach from different directions, not all d-orbitals interact directly. These interactions, however, create a splitting due to the electrostatic environment.The average energy of the five d orbitals remains the same.
Question 5
1 / -0

In the presence of strong eletrical field, the following set of orbitals are not degenerate :

A
$$3d_{xy} and 3d_{z}2$$

B
$$3d_{xy} and 3d_{yz}$$

C
$$3d_{xy}, 3d_{yz} and 3d_{zx}$$

D
all the above

Solution

Due to their different axis orientation; in presence of strong electrical field, $$3d_{xy}$$ and $$3d_{z^2}$$ orbitals are not degenerated.
Question 6
1 / -0

Number of $$Cl^{-}$$ ions satisifying both primary and secondary valency are in $$CoCl_{3}.5NH_{3}$$.

A
1

B
2

C
3

D
4

Solution

There is one $$Cl^{-}$$ ion which satisfies both primary and secondary valency in $$CoCl_{3}.5NH_{3}$$.

Option A is correct.
Question 7
1 / -0

1 mole of amino cobalt chloride on treating with an excess of $$AgNO$$$$_{3}$$ solution gives $$2$$ moles of $$AgCl$$ precipitate. The number of $$Cl^-$$ ions which satisfies both primary and secondary valency of cobalt in the complex is:

A
$$2$$

B
$$3$$

C
$$1$$

D
$$0$$

Solution

1 mole of amino cobalt chloride gives 2 moles of $$AgCl$$ so number of $$Cl^-$$ ions which satisfies both primary and secondary valency of cobalt in the complex is 1.
Question 8
1 / -0

When mole of $$CoCl_{3}.6NH_{3}$$ is treated with excess of AgNO$$_{3}$$ solution. The number of moles of AgCl precipitated is:

A
3

B
0

C
1

D
2

Solution

One mole of $$CoCl_{3}.6NH_{3}$$ gives 3 moles of $$Cl$$ ion so when mole of $$CoCl_{3}.6NH_{3}$$ is treated with excess of $$AgNO_{3}$$ solution, it gives 3 moles of $$AgCl$$.

$$CoCl_3.6NH_3(aq) + excess AgNO_3(aq) \rightarrow 3AgCl + Co(NO_3)_3.6NH_3(aq)$$
Question 9
1 / -0

IUPAC name for the complex $$[Cu(NH_{3})_{4}]SO_{4}$$ is:

A
cupramonium sulphate

B
copper sulphate tetraamonia

C
tetrammine copper (II) sulphate

D
copper amonium (IV) sulphate

Solution

Since 4 amino groups are present, it will be called as tetrammine.
Since copper(II) is the central metal, it will be tetrammine copper(II). Note that we can determine the oxidation number of copper by charge balance.
Now finally since sulphate is the counter ion, it's full name will be tetrammine copper(II) sulphate.
C is the correct answer.
Question 10
1 / -0

An excess of AgNO$$_{3}$$ is added to 100 ml of a 0.1 M solution of dichlorotetraaquachromium (III) chloride. The number of mol of AgCl precipitate would be:

A
0.001

B
0.002

C
0.003

D
0.01

Solution

dichlorotetra aquachromium (III) cholride is $$[Cr(H_2O)_4Cl_2]Cl$$, so one mole of it will give one mole of $$Cl^-$$ ion so 100 ml of 0.1M will give 0.01 mole of $$Cl^-$$ ion and 0.01 mole of $$AgCl$$ formed.

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Coordination Compounds Test - 15

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Coordination Compounds Test - 15

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Question 1

Small size of cation

Large ionic Charge

Vacant d orbitals

All the above

Solution

Question 2

$$Cr_{2}O{_{7}}^{2-}$$

$$MnO{_{4}}^{-}$$

$$CrO{_{4}}^{2-}$$

All

Solution

Question 3

$$d_{xy},d_{z^{2}},d_{xz}$$and $$d_{yz},d_{x^{2}-y^{2}}$$

$$d_{xy},d_{yz},d_{zx}$$ and $$d_{x^{2}-y^{2}},d_{z^{2}}$$

$$d_{xy},d_{x^{2}-y^{2}},d_{z^{2}}$$ and $$d_{yz},d_{xz}$$

$$d_{xy},d_{x^{2}-y^{2}},d_{xz}$$ and $$d_{yz},d_{z^{2}}$$

Solution

Question 4

remains same

increases

decreases

may increase or decrease

Solution

Question 5

$$3d_{xy} and 3d_{z}2$$

$$3d_{xy} and 3d_{yz}$$

$$3d_{xy}, 3d_{yz} and 3d_{zx}$$

all the above

Solution

Question 6

1

2

3

4

Solution

Question 7

$$2$$

$$3$$

$$1$$

$$0$$

Solution

Question 8

3

0

1

2

Solution

Question 9

cupramonium sulphate

copper sulphate tetraamonia

tetrammine copper (II) sulphate

copper amonium (IV) sulphate

Solution

Question 10

0.001

0.002

0.003

0.01

Solution

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