Coordination Compounds Test

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Coordination Compounds Test - 45

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Weekly Quiz Competition

Question 1
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Consider the following isomers
(i) $$[Pt(NH_{3})_{4}Cl_{2}]Br$$
(ii) $$[Pt(NH_{3})_{4}Br_{2}]Cl_{2}$$
(iii) $$[Co(NH_{3})_{4}Cl_{2}]NO_{2}$$

Which of the following observation is correct?

A
(i) will give a pale yellow and (ii) will give a white precipitate with $$AgNO_{3}$$ solution.

B
(iii) will give a white precipitate with $$AgNO_{3}$$ solution

C
(i), (ii) and (iii) will give white precipitate with $$AgNO_{3}$$ solution.

D
None of the above isomers will give white precipitate with $$AgNO_{3}$$ solution.

Solution

$$\bullet$$ $$\mathrm{[Pt(NH_{3})_{4}Cl_{2}]Br_{2}\rightarrow [Pt(NH_{3})_{4}Cl_{2}]^{2+}+2Br^{-}}$$

$$\mathrm{Br^{-}+AgNO_{3}\rightarrow \underset{pale\ yellow}{AgBr}+NO_{3}^{-}}$$

$$\bullet$$ $$\mathrm{[Pt(NH_{3})_{4}Br_{2}]Cl_{2}\rightarrow [Pt(NH_{3})_{4}Br_{2}]^{2+}+2Cl^{-}}$$

$$\mathrm{Cl^{-}+AgNO_{3}\rightarrow \underset{white}{AgCl}+NO_{3}^{-}}$$

$$\bullet$$ $$\mathrm{[Co(NH_3)_4Cl_2]NO_2 \rightarrow [Co(NH_3)_4Cl_2]^+ + NO_2}^-$$

$$\mathrm{NO_2^- + AgNO_3 \rightarrow \underset{Water \ Soluble}{AgNO_2}}$$

Hence, Option "A" is the correct answer.
Question 2
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Both geometrical and optical isomerism are shown by

A
$$[Co(en)_{2}Cl_{2}]^{+}$$

B
$$[Co(NH_{3})_{5}Cl]^{2+}$$

C
$$[Co(NH_{3})_{4}Cl_{2}]^{+}$$

D
$$[Co(ox)_{3}]^{3-}$$

Solution

Octahedral complexes of type $$\mathrm{[M(\widehat{aa}) B_2]}$$, show both geometrical as well as optical isomerism.
Of all given options, $$\mathrm{[Co(en)Cl_2]^+}$$ is of form $$\mathrm{[M(\widehat{aa}) B_2]}$$.

Hence, Option "A" is the correct answer.
Question 3
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One mole of a complex compound $$Co(NH_{3})_{5}Cl_{3}$$ gives three moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $$AgNO_{3}$$ and yields two moles of $$AgCl{(s)}$$. The complex is:

A
$$[Co(NH_{3})_{4}Cl_{2}]$$$$Cl.NH_{3}$$

B
$$[Co(NH_{3})_{4}Cl]$$$$Cl_{2}.NH_{3}$$

C
$$[Co(NH_{3})_{5}Cl]$$$$Cl_{2}$$

D
$$[Co(NH_{3})_{3}Cl]$$$$Cl_{3}.2NH_{3}$$

Solution

It is given that one mole of a complex compound $$Co(NH_{3})_{5}Cl_{3}$$ gives three moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $$AgNO_{3}$$ and yields two moles of $$AgCl{(s)}$$.

The complex ion is $$[Co(NH_{3})_{5}Cl]$$ $$Cl_{2}$$ as two $$Cl$$ ions are out of the coordination complex.

Therefore, on dissolution in water, three ions are produced and they are $$2Cl^-$$ and $$[Co(NH_3)_5Cl]^{2+}$$ ions.

The two $$Cl^-$$ ions which are free react with $$2$$ moles of $$AgNO_3$$ to give $$2$$ moles of $$AgCl$$.
Question 4
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Mixture $$[X] = 0.02$$ mole of $$ [Co(NH_3)_5SO_4]Br$$ and 0.02 mol of $$[Co(NH_3)_5Br]SO_4$$ was prepared in 2 litre of solution.
1 litre of mixture [X] + excess $$AgNO_3 \longrightarrow$$ [Y]
1 litre of mixture [X] + excess $$BaCl_2\longrightarrow$$ [Z]
No. of moles of [Y] and [Z] are :

A
$$0.01, 0.01$$

B
$$0.02, 0.01$$

C
$$0.01, 0.02$$

D
$$0.02,0.02$$

Solution

Total volume of the solution is 2 L,

Mixture X contains 0.02 moles of $$[Co(NH_3)_5 SO_4] Br$$

Therefore Molarity of $$[Co(NH_3)_5SO_4 ]Br,$$

$$\cfrac{\text {No. of moles of solute}} {\text {Volume of solution in Litres}} = \cfrac{0.02} {2} = 0.01 M$$

Mixture of X also contains 0.02 moles of $$[Co(NH_3)_5 Br] SO_4,$$

Therefore, Molarity of $$[Co(NH_3)_5 Br] SO_4=$$ $$\cfrac{\text {No. of moles of solute}} {\text {Volume of solution in Litres}} = \cfrac{0.02} {2} = 0.01 M$$

Now, according to the reaction,

$$\underset{(0.01M)}{[Co(NH_3)_5 SO_4] Br} + AgNO_3 \rightarrow [Co(NH_3)_5 SO_4] NO_3 (Soluble) +\underset{[Y]ppt}{ AgBr} \downarrow$$

According to Avogadro law, the molarity of $$AgBr$$ will also be 0.01, since molarity is moles per litre. Therefore, the number of moles of $$AgBr$$ will also be 0.01.

Similarly in the second reaction, $$BaSO_4$$ (Z) will be precipitated and $$[Co(NH_3)_5 Br] Cl_2$$ will be formed which is soluble. Therefore, no. of moles of $$BaSO_4$$ = 0.01
Question 5
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The solubility of silver bromide in hypo solution is due to the formation of:

A
$$Ag_{2}SO_{3}$$

B
$$Ag_{2}S_{2}O_{3}$$

C
$$[Ag(S_{2}O_{3})]^{-}$$

D
$$[Ag(S_{2}O_{3})_{2}]^{3-}$$

Solution

Unreacted silver bromide $$(AgBr)$$ can be dissolved from photographic emulsions using 'hypo' solution $$(Na_2S_2O_3)$$ (aq.). The reaction can be represented as:
$$AgBr(s)+3Na_2S_2O_3(aq)\rightarrow Na_5Ag(S_2O_3)_3(aq)+NaBr(aq)$$
Hence, option $$D$$ is correct.
Question 6
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Which one of the following has largest number of stereoisomers?

A
$$[Ru(NH_{3})_{4}Cl_{2}]^{+}$$

B
$$[Co(NH_{3})_{5}Cl]^{2+}$$

C
$$[Ir(PR_{3})_{2}H(CO)]^{2+}$$

D
$$[Co(en)_{2}Cl_{2}]^{+}$$

Solution

A) It has two geometrical isomers i.e., cis and trans form and no optical isomer is possible in this case.
B) No geometrical or optical isomer is shown by this compound.
C) It has two geometrical isomers i.e., cis and trans form and no optical isomer is possible in this case.
D) It has two geometric (cis and trans form) and two optical isomers. The optical isomerism is possible for cis form only.
Question 7
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Match the correct pairs of isomers with their examples.
A)Ionisation (a) $$Pt(NH_{3})_{2}Cl_{2}][Pt(NH_{3})_{4}]$$
$$[PtCl_{4}]$$
B) Linkage (b) $$[Cr(NH_{3})_{6}][Co(CN)_{6}]$$ & $$[Co(NH_{3})_{6}[Cr(CN)_{6}]$$
C)Coordination (c) $$[Co(NH_{3})_{5}(NO_{2})]Cl_{2}$$ & $$[Co(NH_{3})_{5}(ONO)]Cl_{2}$$
D)Polymerisation (d) $$[Co(SO_{4})(NH_{3})_{5}]Br$$ & $$[Co(Br (NH_{3})_{5}]SO_{4}$$

A
A -> a, B -> b, C -> c, D -> d

B
A -> b, B -> a, C -> d, D -> c

C
A -> d, B -> c, C -> b, D -> a

D
A -> d, B -> b, C -> c, D -> a

Solution

$$Polymerization\, Isomerism$$ : This is not true isomerism because it occurs between compounds having the same empirical formula, but different molecular weights.
For example, $$[Pt(NH_{3})_{2}CI_{2}],[Pt(NH_{3})_{4}][PtCI_{4}],[Pt(NH_{3})_{4}][Pt(NH_{3})CI_{3}]_{2}$$.
$$Inonization\, Isomerism$$ : This type of isomerism is due to the exchange of groups between the complex ion and the ions outside it. $$[Co(NH_{3})_{5}Br]SO_{4}$$ is red-violet. An aqueous solution gives a white precipitate of $$BaSO_{4}$$ with $$BaCI_{2}$$ solution, thus confirming the presence of free $$SO_{4}^{2-}$$ ions. In contrast $$[Co(NH_{3})_{5}SO_{4}]Br$$ is red. A solution of this complex does not give a positive sulphate test with $$BaCI_{2}$$.It does give a cream-coloured precipitate of $$AgBr$$ with $$AgNO_{3}$$, thus confirming the presence of free $$Br^{-}$$ ions.
$$Linkage\, Isomerism$$ : Certain ligands contain more than one atom which could donate an electron pair. In the $$NO_{2}^{-}$$ group in the complex ion. In the $$NO^-_2$$ ion, either $$N$$ or $$O$$ atoms could act as the electrons pair donor. Thus there is the possibility of isomerism. Two different complexes $$[Co(NH_3)_5NO)_2]Cl_2$$ and $$[Co(NH_3)_5(ONO)]Cl_2$$ have been prepared, each containing the $$NO^-_2$$ group in the complex ion.
$$Coordination\,Isomerism$$ : If the complex is a salt having both cation and anion as complex ions then the ligands can exchange position between the cation and the anion. This will result in the formation of coordinates isomers. For example
$$[Co(en)_{3}][Cr(C_{2}O_{4})_{3}]$$ and $$[Co(en)_{2}(C_{2}O_{4})][Cr(en)(C_{2}O_{4})_{2}]$$
$$[Cr(en)_{2}(C_{2}O_{4})][Co(en)(C_{2}O_{4})_{2}]$$ and $$[Cr(en)_{3}][Co(C_{2}O_{4})_{3}]$$
Question 8
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The ratio of pure and hybrid orbitals of valence shell in $$H_{2}C=CH-CH=CH_{2}$$ is:

A
$$7 : 12$$

B
$$14 : 13$$

C
$$6 : 5$$

D
$$5 : 6$$

Solution

Every double bonded carbon has $$3$$ $$sp^2$$ orbitals and one pure p orbital. And every hydrogen has one pure s orbital, so the number of hybrid orbitals are $$4 \times 3=12$$, and pure orbitals are $$1 \times 4+6 \times 1=10$$.
So, their ratio is $$10:12$$ or $$5:6$$.
Question 9
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The observed magnetic moment value $$(\mu _{abs})$$ is higher than calculated magnetic moment value $$(\mu _{cal})$$ for which of the following species?

A
Ti$$^{+3}$$

B
V$$^{+2}$$

C
Co$$^{+2}$$

D
Cr$$^{+2}$$

Solution

The observed magnetic moment value $$(\mu _{abs})$$ of $$Co^{2+}$$ is in the range of $$4.3-5.2$$ which is higher than calculated magnetic moment value $$(\mu _{cal})$$ which is $$3.87$$. Values are different enough that they usually confirm the number of unpaired spins (S) and indicate whether the complex is high spin or low spin.

Question 10

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Match list-I (Cooordination Compounds) with list-II (Type of Isomerism) and select the correct answer using the codes given below :

	List-I		List-II
A.	$$Co(NH_3)_4Cl_2$$	1.	Optical isomerism
B.	$$Cis-Co(en)_3Cl_2$$	2.	Ionization Isomerism
C.	$$Co(en)_2(NO)_2ClSCN$$	3.	Co-ordination Isomerism
D.	$$Co(NH_3)_6Cr(CN)_6$$	4.	Geometrical Isomerism

A-4, B-3, C-2, D-1

A-1, B-4, C-3, D-2

A-4, B-1, C-2, D-3

A-4, B-2, C-3, D-1

Solution

$$[Co(NH_3)_4Cl] $$ shows geometrical isomerism. It exists in cis and trans form.
$$Cis [Co(en)_3Cl_2] $$ shows optical isomerism It exists in $$d-$$ and $$l-$$ form. The trans isomer is optically active.
$$[Co(en)_2ClSCN$$ exhibits ionization isomerism.
$$[Co(NH_3)_6][Cr(CN)_6]$$ exhibits coordination isomerism.

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Coordination Compounds Test - 45

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Coordination Compounds Test - 45

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Question 1

(i) will give a pale yellow and (ii) will give a white precipitate with $$AgNO_{3}$$ solution.

(iii) will give a white precipitate with $$AgNO_{3}$$ solution

(i), (ii) and (iii) will give white precipitate with $$AgNO_{3}$$ solution.

None of the above isomers will give white precipitate with $$AgNO_{3}$$ solution.

Solution

Question 2

$$[Co(en)_{2}Cl_{2}]^{+}$$

$$[Co(NH_{3})_{5}Cl]^{2+}$$

$$[Co(NH_{3})_{4}Cl_{2}]^{+}$$

$$[Co(ox)_{3}]^{3-}$$

Solution

Question 3

$$[Co(NH_{3})_{4}Cl_{2}]$$$$Cl.NH_{3}$$

$$[Co(NH_{3})_{4}Cl]$$$$Cl_{2}.NH_{3}$$

$$[Co(NH_{3})_{5}Cl]$$$$Cl_{2}$$

$$[Co(NH_{3})_{3}Cl]$$$$Cl_{3}.2NH_{3}$$

Solution

Question 4

$$0.01, 0.01$$

$$0.02, 0.01$$

$$0.01, 0.02$$

$$0.02,0.02$$

Solution

Question 5

$$Ag_{2}SO_{3}$$

$$Ag_{2}S_{2}O_{3}$$

$$[Ag(S_{2}O_{3})]^{-}$$

$$[Ag(S_{2}O_{3})_{2}]^{3-}$$

Solution

Question 6

$$[Ru(NH_{3})_{4}Cl_{2}]^{+}$$

$$[Co(NH_{3})_{5}Cl]^{2+}$$

$$[Ir(PR_{3})_{2}H(CO)]^{2+}$$

$$[Co(en)_{2}Cl_{2}]^{+}$$

Solution

Question 7

A -> a, B -> b, C -> c, D -> d

A -> b, B -> a, C -> d, D -> c

A -> d, B -> c, C -> b, D -> a

A -> d, B -> b, C -> c, D -> a

Solution

Question 8

$$7 : 12$$

$$14 : 13$$

$$6 : 5$$

$$5 : 6$$

Solution

Question 9

Ti$$^{+3}$$

V$$^{+2}$$

Co$$^{+2}$$

Cr$$^{+2}$$

Solution

Question 10

A-4, B-3, C-2, D-1

A-1, B-4, C-3, D-2

A-4, B-1, C-2, D-3

A-4, B-2, C-3, D-1

Solution

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