Coordination Compounds Test

Result

Coordination Compounds Test - 24

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The number of isomers for the compound with the molecular formula $$\mathrm{C}_2\mathrm{Br Cl F I}$$ is :

A
3

B
4

C
5

D
6

Solution

total 6 isomers are possible
Question 2
1 / -0

The coordination of $$Pt$$ in the complex ion $$[Pt (en)_2Cl_2]^{2+}$$ is:

A
3

B
4

C
5

D
6

Solution

The coordination of Pt in the complex ion $$\displaystyle [Pt (en)_2Cl_2]^{2+}$$ is $$6.$$
en (ethylene diamine) is bidentate ligand.

So, $$2\ en$$ ligands and $$2 \ Cl^-$$ ligands corresponds to six donor atoms.
Question 3
1 / -0

$$Al_4C_3$$ on hydrolysis yields

A
nitrogen

B
methane

C
hydrogen

D
all the above

Solution

The hydrolysis reaction of $$Al_4C_3$$

$$Al_4C_3 + 12 H_2O \rightarrow 4Al(OH)_3 + 3CH_4$$

So, $$Al_4C_3 $$ on hydrolysis gives methane

Hence option B is correct.
Question 4
1 / -0

Square planar complex of the type $$MAXBL$$(where A, B, X and L are unidentate ligands) shows following set of isomers.

A
Two cis and one trans

B
Two trans and one cis

C
Two cis and two trans

D
Three cis and one trans

Solution

Square planar complex of the type $$M{AXBL}$$ (where A, B, X, and L are unidentate ligands) shows two cis and one trans isomer as shown in the image.
Question 5
1 / -0

In $$TeCl_4$$, the central atom tellurium involves:

A
$$sp^3$$ hybridisation

B
$$sp^3d$$ hybridisation

C
$$dsp^2$$ hybridisation

D
$$sp^3d^2$$ hybridisation

Solution

Number of hybrid orbitals $$=\dfrac{1}{2}$$ (no. of electrons in valence shell of atom + no. of monovalent atoms - charge no cation + charge on the anion

Number of hybride orbitals $$=\dfrac{1}{2}(6+4+0+0)=5$$

Hence, $$TeCl_4$$ shows $$sp^3d$$ hybridisation.

Therefore, option B is correct.
Question 6
1 / -0

Which of the following $$0.1M$$ complex compound solutions will have the minimum electrical conductivity?

A
Hexammine platinum (IV) chloride

B
Chloropenta ammine platinum (IV) chloride

C
Dichloro tetrammine platinum (IV) chloride

D
Trichloro tetrammine platinum (IV) chloride

Solution

The formula of given complex are as follows:
(a) Hexammine platinum (IV) chloirde
$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 4 }$$
(b) Chloropentammine platinum (IV) chloirde
$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 3 }$$
(c) Dichlorotetrammine platinum (IV) chloirde
$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] { Cl }_{ 2 }\quad $$
(d) Trichlorotriammine platinum (IV) chloirde
$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] Cl$$
In aqueous solution the complexes ionise as
$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] { Cl }_{ 4 }\rightleftharpoons \underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] }^{ 4+ }+4{ Cl }^{ - }}_{5\, ions}$$

$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] { Cl }_{ 4 }\rightleftharpoons\underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 5 }Cl \right] }^{ 3+ }+3{ Cl }^{ - }}_{4\, ions}$$

$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] Cl\rightleftharpoons\underbrace{ { \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 4 }{ Cl }_{ 2 } \right] }^{ 2+ }+2{ Cl }^{ - }}_{3\, ions}$$

$$\left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] Cl\rightleftharpoons\underbrace{{ \left[ Pt{ \left( { NH }_{ 3 } \right) }_{ 3 }{ Cl }_{ 3 } \right] }^{ + }+{ Cl }^{ - }}_{2\, ions} $$

Trichlorotriammine platinum (IV) chloride gives the minimum number of ions in the solution. Hence, it has the minimum electrical conductivity
Question 7
1 / -0

Coordinate bond is formed in:

A
$$NaCl$$

B
$${ NH }_{ 4 }Cl$$

C
$${ Cl }_{ 2 }$$

D
$$B{ Cl }_{ 3 }$$

Solution

$$NH_4Cl$$ exists as $$NH_4^+Cl^-$$
in $$NH_4^+$$
only coordinate bond is present.
Question 8
1 / -0

When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

A
$$[Co(NH_3)_6]Cl_3$$

B
$$[Co(NH_3)_6Cl_3]$$

C
$$[Co(NH_3)_5Cl]Cl_2$$

D
$$[Co(NH_3)_4Cl_2]NO_3$$

Solution

$$4.305 \, g \, AgCl =\dfrac {4.305}{143.5} mol= 0.03\, mol$$

As 0.01 mole of the complex gives 0.03 mole of $$AgCI$$, this shows that there are three ionisable $$Cl$$.

Hence, formula is $$[Co(NH_3)_6]Cl_3$$.
Question 9
1 / -0

In spectrochemical series, chlorine is above water i.e., $$Cl> {H}_{2}O$$. This is due to:

A
Good $$\pi -$$ acceptor properties of $$Cl$$

B
Strong $$\sigma -$$ and good $$\pi -$$ acceptor properties of $$Cl$$

C
Good $$\pi -$$ donor properties of $$Cl$$

D
Larger size of $$Cl$$ than $${H}_{2}O$$

E
None of the option

Solution

None of the given options is correct.

$${H}_{2}O$$ is a stronger ligand than $$Cl$$ and the order of the ligands in spectrochemical series is determined experimentally.
Question 10
1 / -0

What is the electronic configuration of elements of $$III$$ rd group

A
$$1{ s }^{ 2 },2{ s }^{ 2 }2p{ s }^{ 3 }$$

B
$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 2 }3{ p }^{ 1 }$$

C
$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 }$$

D
$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 1 }\quad $$

Solution

$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 2 }3{ p }^{ 1 }$$
As it is p block element therefore the group is $$12+1=13$$ (III A group)

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 24

Result

Coordination Compounds Test - 24

Score

-

Rank

-

SHARING IS CARING

Question 1

3

4

5

6

Solution

Question 2

3

4

5

6

Solution

Question 3

nitrogen

methane

hydrogen

all the above

Solution

Question 4

Two cis and one trans

Two trans and one cis

Two cis and two trans

Three cis and one trans

Solution

Question 5

$$sp^3$$ hybridisation

$$sp^3d$$ hybridisation

$$dsp^2$$ hybridisation

$$sp^3d^2$$ hybridisation

Solution

Question 6

Hexammine platinum (IV) chloride

Chloropenta ammine platinum (IV) chloride

Dichloro tetrammine platinum (IV) chloride

Trichloro tetrammine platinum (IV) chloride

Solution

Question 7

$$NaCl$$

$${ NH }_{ 4 }Cl$$

$${ Cl }_{ 2 }$$

$$B{ Cl }_{ 3 }$$

Solution

Question 8

$$[Co(NH_3)_6]Cl_3$$

$$[Co(NH_3)_6Cl_3]$$

$$[Co(NH_3)_5Cl]Cl_2$$

$$[Co(NH_3)_4Cl_2]NO_3$$

Solution

Question 9

Good $$\pi -$$ acceptor properties of $$Cl$$

Strong $$\sigma -$$ and good $$\pi -$$ acceptor properties of $$Cl$$

Good $$\pi -$$ donor properties of $$Cl$$

Larger size of $$Cl$$ than $${H}_{2}O$$

None of the option

Solution

Question 10

$$1{ s }^{ 2 },2{ s }^{ 2 }2p{ s }^{ 3 }$$

$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 2 }3{ p }^{ 1 }$$

$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 }$$

$$1{ s }^{ 2 },2{ s }^{ 2 }2{ p }^{ 6 },3{ s }^{ 1 }\quad $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.