Coordination Compounds Test

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Coordination Compounds Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Secondary valencies correspond to ____________ of the metal atom and are satisfied by ligands.

A
coordination number

B
valency

C
element

D
molecules

Solution

Secondary valency corresponds to the coordination number of central metal atom or ion. This may be satisfied by either negative ions or neutral molecules called ligands.
Ligands satisfying the secondary valencies are always shown in square brackets, i.e. they form the coordination sphere of the metal atom.

Hence,option A is correct.
Question 2
1 / -0

Developed photography film is due to :

A
$$AgBr$$

B
$$[Ag(S_2O_3)_2]^{3+}$$

C
$$Ag$$

D
$$Ag_2O$$

Solution

The developed photography film is due to Ag
The treatment of the exposed photgraphic film with reducing agent is called developing of the film.
A developer is usually an alkaline solution of hydroquinone or alkaline solution of pyrogallol.
The pairs activated by light are reduced to deposit more of black silver.
Question 3
1 / -0

Total number of stereoisomers of $$[Co(acac)_{2}BrCl]^{\ominus}$$ are.

A
$$4$$

B
$$3$$

C
$$6$$

D
$$2$$

Solution

Option (B) is correct.
Total number of stereo isomers of $$[Co(acac)_{2}BrCl]^{-}$$ are 3 i.e. cis,trans,optical.
Question 4
1 / -0

Hypo is used in photography because it is :

A
A strong reducing agent

B
A strong oxidizing agent

C
A strong Complexing agent

D
Photo sensitive Compound

Solution

Correct Answer: Option C
Explanation:

Hypo forms the complex $$[Ag(S_2O_3)_{2}]^{3-}$$ from insoluble $$Ag$$.

$$[Ag(S_2O_3)_{2}]^{3-}$$ is formed by treatment of $$Na_2S_2O_3. 5H_2O$$ with $$AgBr$$.

Therefore, Hypo is used in photography because it is a strong Complexing agent.

Hence, the correct answer is option $$C$$.
Question 5
1 / -0

Chlorobenzene is o, p -directing in electrophilic substitution reaction. The directing influence is explained by

A
+M of Ph

B
+I of Cl

C
+M of Cl

D
-I of Ph

Solution
Question 6
1 / -0

A hybrid orbital formed from s-and p-orbital can contribute to:

A
a $$\sigma $$ bond only

B
$$\pi$$ bond only

C
either $$\sigma \;or\;\pi$$ bond

D
none of these

Solution

A hybrid orbital formed from s-and p-orbital can contribute to a σ bond only. This is because sideways or lateral overlap with s orbital is not possible.
Question 7
1 / -0

Which of the following statements is not correct ?

A
Hybridization is the mixing of atomic orbitals of large energy difference.

B
$$sp^2$$-hybrid orbitals are formed from two p - atomic orbitals and ones - atomic orbitals.

C
$$dsp^2$$-hybtrid orbitals are all at 90 to one another.

D
$$d^2sp^3$$-hybrid orbitals are directed towards the corners of a regular octahedron.

Solution

Hybridisation is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds
Question 8
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Amongest $$N{ O }_{3}^{-}\,, AsO_{3}^{3-}\, CO_{3}^{2-},\, Cl{ O }_{3}^{-},\, SO_{3}^{2-}$$ and $$BO_{3}^{3-}$$, the non - planar species are:

A
$$CO_{3}^{2-}, SO_{3}^{2-}$$ and $$BO_{3}^{3-}$$

B
$$AsO_{3}^{3-},\, ClO_{3}^{-}$$ and $$SO_{3}^{2-}$$

C
$$N{ O }_{3}^{-}\, CO_{3}^{2-},\, CO_{3}^{2-}$$ and $$BO_{3}^{3-}$$

D
$$AsO_{3}^{3-},\, Cl{ O }_{3}^{-}$$ and $$BO_{3}^{3-}$$

Solution

$$AsO_{3}^{3-},\, Cl{ O }_{3}^{-}$$ and $$SO_{3}^{2-}$$ are non - planar species.
In these species, the central atom has $$sp^3$$ hybridization due to presence of $$3$$ bonding domains and one lone pair of electrons. The molecular geometry is pyramidal and the electronic geometry is tetrahedral.
Question 9
1 / -0

The colour of blood is red due to the presence of:

A
magnesium

B
cytoglobin

C
haemoglobin

D
all of these

Solution

Since haemoglobin is red pigment present in blood which is red due to intra ligand charge transfer along porphyrin ring leads to red colouration.
Question 10
1 / -0

In the electronic configurations of elements $$A\;and\;B$$ are $$1s^2,\,2s^2\;2p^6,\,3s^1\;and\;1s^2,\,2s^2\;2p^6,\,3s^2\;3p^4$$ respectively, then the formula of the compound formes by the combination of these elements will be

A
$$AB$$

B
$$AB_3$$

C
$$AB_2$$

D
$$A_2B$$

Solution

Electronic configuration of $$A$$ is $$1s^2\,2s^2\,2p^6\,3s^1$$
Electronic configuration of $$B$$ is $$1s^2\,2s^2\,2p^6\,3s^2\,3p^4$$
Valency of $$A=1$$
Valency of $$B=8-6=2$$
$$\therefore$$ The compound formed by $$A\;\&\;B$$ is

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Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 21

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Coordination Compounds Test - 21

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Question 1

coordination number

valency

element

molecules

Solution

Question 2

$$AgBr$$

$$[Ag(S_2O_3)_2]^{3+}$$

$$Ag$$

$$Ag_2O$$

Solution

Question 3

$$4$$

$$3$$

$$6$$

$$2$$

Solution

Question 4

A strong reducing agent

A strong oxidizing agent

A strong Complexing agent

Photo sensitive Compound

Solution

Question 5

+M of Ph

+I of Cl

+M of Cl

-I of Ph

Solution

Question 6

a $$\sigma $$ bond only

$$\pi$$ bond only

either $$\sigma \;or\;\pi$$ bond

none of these

Solution

Question 7

Hybridization is the mixing of atomic orbitals of large energy difference.

$$sp^2$$-hybrid orbitals are formed from two p - atomic orbitals and ones - atomic orbitals.

$$dsp^2$$-hybtrid orbitals are all at 90 to one another.

$$d^2sp^3$$-hybrid orbitals are directed towards the corners of a regular octahedron.

Solution

Question 8

$$CO_{3}^{2-}, SO_{3}^{2-}$$ and $$BO_{3}^{3-}$$

$$AsO_{3}^{3-},\, ClO_{3}^{-}$$ and $$SO_{3}^{2-}$$

$$N{ O }_{3}^{-}\, CO_{3}^{2-},\, CO_{3}^{2-}$$ and $$BO_{3}^{3-}$$

$$AsO_{3}^{3-},\, Cl{ O }_{3}^{-}$$ and $$BO_{3}^{3-}$$

Solution

Question 9

magnesium

cytoglobin

haemoglobin

all of these

Solution

Question 10

$$AB$$

$$AB_3$$

$$AB_2$$

$$A_2B$$

Solution

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