Coordination Compounds Test

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Coordination Compounds Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The primary and secondary valency of a central metal ion in $$[Co(NH_3)_4CO_3]$$$$Cl$$ complex is _________ and _________ respectively.

A
$$2, 6$$

B
$$4, 6$$

C
$$4, 5$$

D
$$3, 6$$

Solution

The primary and secondary valency of cobalt ion is 3 and 6 respectively.

The charge on cobalt ion is $$+3$$ which balances $$-2$$ charge on carbonate ion and $$-1$$ charge on chloride ion. Hence, the primary valency of cobalt ion is 3.

Cobalt ion is surrounded by four monodentate and one bidentate ligand (4 ammonia and on carbonate) in the coordination sphere. Hence, its coordination number (secondary valency) is 6.
Question 2
1 / -0

By which of the following, poisoning of lead in the body can be removed?

A
$$Pt$$

B
EDTA

C
$$Ox^{2-}$$

D
$$Pn$$

Solution

EDTA is a hexadentate ligand and forms the water-soluble stable chelate complex with metals such as lead. Hence, it can be used to remove lead from the body. In this way, EDTA helps in removing poisoning of lead from the body.
Question 3
1 / -0

Which of following is the incorrect match?

A
Insulin - Zinc

B
Haemoglobin - Iron

C
Vitamin $$B_{12}$$ - Cobalt

D
Chlorophyll - Chromium

Solution

Chlorophyll contains magnesium ion, encased in a large ring structure known as chlorine, derived from pyrrole.
Question 4
1 / -0

The correct energy level diagram for $$[Co(CN)_6]^{3-}$$

A

B

C

D

Solution

Since cyanide ion is a strong field ligand, electrons pair up. All six electrons are present in lower $$t_{2g}$$ energy level.
Question 5
1 / -0

The synthetic steroid ethynylestardiol (1) is a compuond used in the birth control pill.
How many $$sp$$ hybridised carbon atoms are present in compound (1) ?

A
2

B
4

C
6

D
8

E
10

Solution

$$sp$$ hybridized carbon are those who forms triple bond.
(Refer to Image)
These are two carbons which are $$sp$$ hybridized.
Question 6
1 / -0

$$CuSO_4\cdot 5H_2O$$ is blue is colour while $$CuSO_4$$ is colourless due to:

A
presence of strong field ligand in $$CuSO_4\cdot 5H_2O$$

B
due to absence of water (ligand), d - d transitions are not possible in $$CuSO_4$$

C
anhydrous undergoes d - d transitions due to crystal field splitting

D
colour is lost due to unpaired electrons.

Solution

In $$CuSO_4 \cdot 5H_2O$$, water acts as ligand and causes crystal filed splitting. This makes d - d transitions possible. On the other hand, in $$CuSO_4$$, due to absence of ligand crystal filed splitting is not possible. Hence no colour is observed.
Question 7
1 / -0

Energy profile diagram for an exothermic reaction, $$A\xrightarrow { 1 } B\xrightarrow { 2 } C\xrightarrow { 3 } D$$, is given below.

The rate determining step of the reaction is :

A
$$A\longrightarrow B$$

B
$$B\longrightarrow C$$

C
$$C\longrightarrow D$$

D
can not preditc

Solution

In a chemical reaction, the overall rate of a reaction is determined by the slowest step, known as the rate determining step. The rate equation is simplified by using the approximation of rate determining step. Since, the slowest step in this reaction is when it has highest potential energy. Thus the rate determining step of reaction is $$A \longrightarrow B$$.
Question 8
1 / -0

Which of the following does not show optical isomerism?

A
$$[Co(NH_3)_3Cl_3]$$

B
$$[Co(en)Cl_2(NH_3)_2]^+$$

C
$$[Co(en)_3]^{3+}$$

D
$$[Co(en)_2Cl_2]^+$$

Solution

The compound $$[Co(NH_3)_3Cl_3]$$ does not show optical isomerism.
This is a coordination compound where coordination number is 6 and geometry octahedral symmetrical. Compounds with an octahedral geometry having bidentate ligands show optical isomerism.
The given compound does not contain any bidentate ligands, also it has symmetrical geometry, thus, it does not show optical isomerism.
Question 9
1 / -0

Which of the following order is correct for the IR vibrational frequency of CO?

A
$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }<\left[ Co{ \left( CO \right) }_{ 4 } \right]^- <\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$

B
$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }>\left[ Co{ \left( CO \right) }_{ 4 } \right]^- >\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$

C
$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }>\left[ Co{ \left( CO \right) }_{ 4 } \right]^- <\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$

D
$${ \left[ Fe{ \left( CO \right) }_{ 4 } \right] }^{ 2- }<\left[ Co{ \left( CO \right) }_{ 4 } \right]^- >\left[ Ni{ \left( CO \right) }_{ 4 } \right] $$

Solution

The IR vibrational frequency depends on the strength of CO bond, More is the bond strength more will be the IR vibrational frequency.Since, CO is a pi acceptor ligand and accepts electron in antibonding pi orbital. More electrons in antibonding orbital implies less bond strength of CO. In the given complexes the oxidation state of metals is Fe(2-), Co(1-), Ni(0). Therefore more electrons on metal implies more will be the pi backbonding more stronger will be the Metal-C bond weaker will be the C-O bond hence, the IR vibrational frequency order will be $$[Fe(CO)_4]^{2-} < [Co(CO)_4]^{-} < [Ni CO)_4]$$
Question 10
1 / -0

Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy ?

A
$$SrSO_4$$

B
$$MgSO_4$$

C
$$CaSO_4$$

D
None of these