Coordination Compounds Test

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Coordination Compounds Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following complexes will have six isomers?

A
$$[Co(en)(NH_3)_2Cl_2]Cl$$

B
$$[Cr(H_2O)_4Cl_2]Cl$$

C
$$[Co(ox)_3]^{3-}$$

D
$$[Co(en)_2Br_2]Cl$$

Solution

The isomers of $$[Co(en)(NH_3)_2Cl_2]Cl$$,
(i) and (ii) have cis-optical isomer also so the total isomers are $$4$$ geometrical and $$2$$ optical isomers.
Question 2
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The hybridisation of each carbon in the following compound is:
$$\underset {i}{CH_{3}} - \underset {ii}{\overset {\overset {O}{||}}{C}} - \underset {iii}{CH_{2}} - \underset {iv}{CN}$$

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

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

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

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

Solution

$$\underset {i}{CH_{3}} - \underset {ii}{\overset {\overset {O}{||}}{C}} - \underset {iii}{CH_{2}} - \underset {iv}{CN}$$
In (i), all single bonds occur therefore $$sp^{3}$$ hybridization.
In (ii), one double bond occurs therefore $$sp^{2}$$ hybridization.
In (iiI), all single bond occurs therefore $$sp^{3}$$ hybridization.
In (iv), one triple bond occurs, therefore $$sp$$ hybridization.
Question 3
1 / -0

Out of given reaction which show change in hybridization of central atom__________.

A
$${ H }_{ 2 }{ BO }_{ 3 }$$ dissolve in water

B
$${ H }_{ 2 }{ SO }_{ 4 }$$ dissolve in water

C
$${ N }_{ 2 }{ O }_{ 5(g) }\longrightarrow { N }_{ 2 }{ O }_{ 5(s) }$$

D
$$P{ Br }_{ 5(g) }\longrightarrow P{ Br }_{ 5(s) }$$

E
$${ C }_{ 2 }{ H }_{ 6 }\xrightarrow [ bond\ cleavage\ of\ C-C\ bond ]{ Homolytic } $$

Solution

Correct Option $$- A$$
Reason:-
$$H_{3}BO_{3}\longrightarrow HBO_{2} + H_{2}O$$
$$H_{3}BO_{3}$$ has $$sp^{3}$$ hybridisation at the central atom and $$HBO_{2}$$ has $$sp^{2}$$ hybridisation at the central atom.
Question 4
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Which of the following primary and secondary valencies are not correctly marked against the compound?

A
$$[Cr(NH_3)_6]Cl_3,\, p= 3, \,s= 6$$

B
$$K_2[PtCl_4],\, p=2,\,s= 4$$

C
$$[Pt(NH_3)_2Cl_2],\, p= 0, \, s=4$$

D
$$[Cu(NH_3)_4]SO_4,\, p=4, \, s= 4$$

Solution

Primary valency is satisfied by anions or cations. Secondary valency refers to coordination number.

In $$[Cr(NH_3)_6]Cl_3$$ primary valency is 3 and secondary valency is 6. Since chromium is coordinated to 6 ammonia ligands, secondary valency is 6. Since chlorine ions have $$-1$$ charge and there are 3 of them, primary valency is 3.

In $$K_2[PtCl_4]$$ primary valency is 2 and secondary valency is 4. Since platinum is coordinated to 4 chloride ligands, secondary valency is 4. Since potassium ions have $$+1$$ charge and there are 2 of them, primary valency is 2.

In $$[Pt(NH_3)_2Cl_2]$$ primary valency is 0 and secondary valency is 4. Since platinum is coordinated to 2 ammonia and 2 chloride ligands, secondary valency is 4. Since no ions are attached to the complex, primary valency is 0.

In $$[Cu(NH_3)_4]SO_4$$ primary valency is 2 and secondary valency is 4. Since, copper is co-ordinated to 4 ammonia ligands, secondary valence is 4. Since sulphate ions has $$-2$$ charge, primary valence is 2.
Question 5
1 / -0

The magnitude of CFSE (Crystal Field Splitting Energy, $$\Delta_0$$) can be related to the configuration of d - orbitals in a coordination entity as?

A
If $$\Delta_0$$ < P, the configuration is $$t_{2g}^{3}e_{g}^{1}$$ = weak filed ligand and high spin complex.

B
If $$\Delta_0$$ > P, the configuration is $$t_{2g}^{3}e_{g}^{1}$$ = strong filed ligand and high spin complex.

C
If $$\Delta_0$$ > P, the configuration is $$t_{2g}^{4}e_{g}^{0}$$ = strong filed ligand and high spin complex.

D
If $$\Delta_0=$$ P, the configuration is $$t_{2g}^{4}e_{g}^{0}$$ = strong filed ligand and high spin complex.

Solution

$$I.$$ In case of weak field ligand Crystal Field Splitting Energy, $$\Delta _o< P \Rightarrow$$ So, pairing will not take place.

For - $$d^4$$, the configuration is $$t^3_{2g} e^1_g$$ and high spin complex will form

$$II.$$ In case of strong field ligand (low spin) Crystal Field Splitting Energy, $$\Delta _o> P \Rightarrow$$ Pairing will take place
For - $$d^4$$, the configuration is $$t^4_{2g} e^0_g$$ and low spin complex will form
Question 6
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Aqueous solution of nickel sulphate on treating with pyridine and then adding a solution of sodium nitrate gives dark blue crystals of:

A
$$[Ni(py)_4]SO_4$$

B
$$[Ni(py)_2(NO_2)_2]$$

C
$$[Ni(py)_4(NO_2)_2]$$

D
$$[Ni(py)_3(NO_2)]_2SO_4$$

Solution

$$NiSO_4+Py\longrightarrow [Ni[Py]_4]^{2+}SO_4^{2-}$$
$$[Ni[Py]_4]^{2+}SO_4^{2-}+NaNO_3\rightarrow [Ni(Py)_4(NO_2)_2]$$(Blue crystal)
Question 7
1 / -0

The terminal and bridged $$CO$$ ligands in the compound $$[Co_2(CO)_8]$$ are respectively:

A
0, 2

B
6, 1

C
5, 2

D
6, 2

Solution

The structure of $$[Co(CO)_8]$$ is as shown. There are 6 terminal and 2 bridged CO ligands in the compound $$[Co_2(CO)_8]$$.
Question 8
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The two isomers X and Y with the formula $$Cr(H_2O)_5ClBr_2$$ were taken for experiment on depression in freezing point. It was found that one mole of X gave depression corresponding to $$2$$ moles of particles and one mole of Y gave depression due to $$3$$ moles of particles. The structural formulae of X and Y respectively are:

A
$$[Cr(H_2O)_5Cl]Br_2$$; $$[Cr(H_2O)_4Br_2]Cl\cdot H_2O$$

B
$$[Cr(H_2O)_5Cl)]Br_2$$; $$[Cr(H_2O)_3ClBr]\cdot 2H_2O$$

C
$$[Cr(H_2O)_5Br]BrCl$$; $$[Cr(H_2O)_4ClBr]Br\cdot H_2O$$

D
$$[Cr(H_2O)_4Br_2]Cl\cdot H_2O$$; $$[Cr(H_2O)_5Cl]Br_2$$

Solution

$$Cr(H_2O)_5ClBr_2\longrightarrow$$ 2 moles of particle.
It compoubnd is $$[Cr(H_2O)_4(Br_2)]Cl.H_2O$$
It gives $$[Cr(H_2O)_4Br_2]^{+},C\bar l\Rightarrow $$ Two particles.
So, $$X=[Cr(H_2O)_4Br_2]Cl.H_2O$$
If $$Y=[Cr(H_2O)_5Cl]Br_2$$
It gives $$[Cr(H_2O)_5Cl^{2+},2B\bar r\Rightarrow $$ 3 particles.
Question 9
1 / -0

Which of the following complex species is not expected to exhibit optical isomerism?

A
$$[Co(en)(NH_3)_2Cl_2]^+$$

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

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

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

Solution

The complex species $$ [Co(NH_3)_3Cl_3]$$ is not expected to exhibit optical isomerism. It shows geometrical isomerism in the form of facial (cis)-isomerism (trans) isomerism due to presence of plane of symmetry.
Question 10
1 / -0

Isostructural species are those which have the same shape and hybridisation. Among the given species, identify the isostructural pairs.

A
[$$NF_3$$ and $$BF_3$$]

B
[$$BF_4^-$$ and $$NH_4^+$$]

C
[$$BCl_3$$ and $$BrCl_3$$]

D
[$$NH_3$$ and $$NO^-_3$$]

Solution

From the structural point of the view, we see that,
$$NF_{3}$$ is pyramidal whereas $$BF_{3}$$ is planar triangular in structure.
$$BF^{-}_{4}$$ and $$NH^{+}_{4}$$ ions both are tetrahedral in structure.
$$BCl_{3}$$ is triangular planar whereas $$BrCl_{3}$$ is pyramidal in structure.
$$NH_{3}$$ is pyramidal whereas $$NO_{3}$$ is triangular planar structure.
So as we can see the only option B has both the elements same in structure.
The correct answer is option B.

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Coordination Compounds Test - 53

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Coordination Compounds Test - 53

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Question 1

$$[Co(en)(NH_3)_2Cl_2]Cl$$

$$[Cr(H_2O)_4Cl_2]Cl$$

$$[Co(ox)_3]^{3-}$$

$$[Co(en)_2Br_2]Cl$$

Solution

Question 2

$$sp^{3}\ sp^{2}\ sp^{3}\ sp$$

$$sp^{3}\ sp^{3}\ sp^{2}\ sp$$

$$sp^{3}\ sp\ sp^{3}\ sp^{2}$$

$$sp^{3}\ sp^{2}\ sp\ sp^{3}$$

Solution

Question 3

$${ H }_{ 2 }{ BO }_{ 3 }$$ dissolve in water

$${ H }_{ 2 }{ SO }_{ 4 }$$ dissolve in water

$${ N }_{ 2 }{ O }_{ 5(g) }\longrightarrow { N }_{ 2 }{ O }_{ 5(s) }$$

$$P{ Br }_{ 5(g) }\longrightarrow P{ Br }_{ 5(s) }$$

$${ C }_{ 2 }{ H }_{ 6 }\xrightarrow [ bond\ cleavage\ of\ C-C\ bond ]{ Homolytic } $$

Solution

Question 4

$$[Cr(NH_3)_6]Cl_3,\, p= 3, \,s= 6$$

$$K_2[PtCl_4],\, p=2,\,s= 4$$

$$[Pt(NH_3)_2Cl_2],\, p= 0, \, s=4$$

$$[Cu(NH_3)_4]SO_4,\, p=4, \, s= 4$$

Solution

Question 5

If $$\Delta_0$$ < P, the configuration is $$t_{2g}^{3}e_{g}^{1}$$ = weak filed ligand and high spin complex.

If $$\Delta_0$$ > P, the configuration is $$t_{2g}^{3}e_{g}^{1}$$ = strong filed ligand and high spin complex.

If $$\Delta_0$$ > P, the configuration is $$t_{2g}^{4}e_{g}^{0}$$ = strong filed ligand and high spin complex.

If $$\Delta_0=$$ P, the configuration is $$t_{2g}^{4}e_{g}^{0}$$ = strong filed ligand and high spin complex.

Solution

Question 6

$$[Ni(py)_4]SO_4$$

$$[Ni(py)_2(NO_2)_2]$$

$$[Ni(py)_4(NO_2)_2]$$

$$[Ni(py)_3(NO_2)]_2SO_4$$

Solution

Question 7

0, 2

6, 1

5, 2

6, 2

Solution

Question 8

$$[Cr(H_2O)_5Cl]Br_2$$; $$[Cr(H_2O)_4Br_2]Cl\cdot H_2O$$

$$[Cr(H_2O)_5Cl)]Br_2$$; $$[Cr(H_2O)_3ClBr]\cdot 2H_2O$$

$$[Cr(H_2O)_5Br]BrCl$$; $$[Cr(H_2O)_4ClBr]Br\cdot H_2O$$

$$[Cr(H_2O)_4Br_2]Cl\cdot H_2O$$; $$[Cr(H_2O)_5Cl]Br_2$$

Solution

Question 9

$$[Co(en)(NH_3)_2Cl_2]^+$$

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

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

$$ [Co(NH_3)_3Cl_3]$$

Solution

Question 10

[$$NF_3$$ and $$BF_3$$]

[$$BF_4^-$$ and $$NH_4^+$$]

[$$BCl_3$$ and $$BrCl_3$$]

[$$NH_3$$ and $$NO^-_3$$]

Solution

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