Coordination Compounds Test

Result

Coordination Compounds Test - 46

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

A complex compound of cobalt has molecular formula containing five $$NH_{3}$$ molecules, one $$NO_{2}$$ group and two $$Cl$$ atoms for one $$Co$$ atom. One mole of this compound produces three moles of ions in aqueous solution. On reacting with an excess of $$AgNO_{3}$$ solution, two moles of $$AgCl$$ get precipitated. The ionic formula of the compound is:

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

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

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

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

Solution

The ionic formula of the compound is $$[Co (NH_{3})_{5} (NO_{2})] Cl_{2}$$.

$$[Co (NH_{3})_{5} (NO_{2})] Cl_{2} \rightarrow [Co (NH_{3})_{5} (NO_{2})]^+ + 2Cl^{-}$$

Total 3 ions are produced in aqueous solution.

Hence, the correct option is $$\text{C}$$
Question 2
1 / -0

The total number of possible coordination isomers of the complex compound $$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$$ are:

A
3

B
6

C
2

D
4

Solution

The total number of possible isomers of the complex compound $$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$$ are 4.

$$\left [ Cu\left ( NH_{3} \right )_{4} \right ]\left [ PtCl_{4} \right ]$$

$$\left [ Cu\left ( NH_{3} \right )_{3}Cl \right ]\left [ Pt\left ( NH_{3} \right )Cl_{3} \right ]$$

$$\left [ Pt\left ( NH_{3} \right )_{3}Cl \right ]\left [ Cu\left ( NH_{3} \right ) Cl_{3} \right ]$$

$$\left [ Pt(NH_3)_{4} \right ]\left [ Cu\left (Cl \right )_{4} \right ]$$

Option D is correct.
Question 3
1 / -0

Which of the following statements is incorrect?

A
Complex $$[Co(NH_3)_4(H_2O)Cl]Br_2$$ can show both hydrate as well as ionization isomerism

B
Complex $$[Co(NH_3)_5(H_2O)](NO_3)_3$$ can show hydrate isomerism

C
Complex $$[Pt(NH_3)_4][PtCl_6]$$ cannot show coordination isomerism

D
$$[Co(NH_3)_4(NO_2)Cl]Cl$$ can show both ionization as well as linkage isomerism

Solution

Option (C) is correct.
(A) : $$[Co(NH_3)_4(H_2O)Cl]Br_2$$ and $$[Co(NH_3)_4Br_2]Cl.H_2O$$; ionization as well as hydrate isomers.
(B) : $$[Co(NH_3)_5(H_2O)](NO_3)_3$$ and $$[Co(NH_3)_5(NO_3)](NO_3)_2H_2O$$; hydrate isomers.
(C) : $$[Pt(NH_3)_4]^{2+} [PtCl_6]^{2-}$$ and $$[Pt(NH_3)_4Cl_2]^{2+} [PtCl_4]^{2-}$$; coordination isomers.
(D) : $$[Co(NH_3)_4(NO_2)Cl]Cl$$ ; $$[Co(NH_3)_4(ONO)Cl]Cl\!^-$$; linkage isomer and $$[Co(NH_3)_4Cl_2]{NO_{2}}^{-}$$; ionization isomer.
Question 4
1 / -0

Which pair is correctly matched?

A
$$[Fe(OH)_3] Fe^{3+} : Cl^{-}$$

B
$$[As_{2}S_{3} ]: As^{3+}$$

C
$$[\mathrm{S}\mathrm{n}\mathrm{O}_{2}]:\mathrm{S}\mathrm{n}\mathrm{O}_{3}^{2-}$$ in acidic medium

D
$$[\mathrm{A}\mathrm{g}\mathrm{I}]:\mathrm{I}^{-}$$ in excess of $$\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}_3$$
Question 5
1 / -0

Match column-I (chemical composition of precipitates) with column-II (colour of the precipitate) and select the correct answer using the codes given below in the columns :
(a) $$Co [Hg (SCN)_4]$$ (p) black
(b) $$Fe_3[Fe(CN)_6]_2$$ (q) rosy red
(c) $$Ni(dmg)_2$$ (r) prussian blue
(d) PbS (s) deep blue

A
a-s , b-p , c-q , d-r

B
a-s , b-r , c-q , d-p

C
a-s , b-p , c-r , d-q

D
a-p , b-r , c-q , d-s

E
a-p , b-q , c-r , d-s

Solution

$$Co[Hg(SCN)_4] $$- Deep Blue
$$Fe_3[Fe(CN_)6]_2$$- Prussian Blue
$$Ni(dmg)_2$$- Rosy red
$$PbS$$- Black
Question 6
1 / -0

In the analysis of an organic compound, it was found that it contains 7 carbon atoms, there are two C = C bonds and one $$C\equiv C$$ bond. This compound is a hydrocarbon. Hydrocarbons have only carbon and hydrogen elements. On structural analysis it was found that it is covalent in nature and expected structure is given above. The ratio between the pure and hybrid orbitals is?

A
$$7:10$$

B
$$8:7$$

C
$$5:9$$

D
$$1:2$$

Solution

In $$\displaystyle sp, sp^2 $$ and $$\displaystyle sp^3 $$ hybridization of $$C$$, the number of hybrid orbitals formed are $$2,3$$ and $$4 $$ respectively and the number of pure orbitals are $$2,1$$ and $$0$$ respectively.
In this compound, $$2$$ $$\displaystyle sp $$ hybridized, $$4$$ $$\displaystyle sp^2$$ hybridized and $$1$$ $$\displaystyle sp^3 $$ hybridized $$C$$ atom is present. The number of pure orbitals in these atoms are $$\displaystyle 4+4+0=8$$
Also there are $$6\: H$$ atoms which contribute $$6$$ pure orbitals. Hence,the total number of pure orbitals are $$\displaystyle 8+6=14 $$
The number of hybrid orbitals are $$\displaystyle 4+12+4+0=20 $$
Hence, the ratio of pure orbitals to hybrid orbitals is $$\displaystyle 14: 20 $$ or $$\displaystyle 7:10$$
Question 7
1 / -0

Which of the following orbitals can not undergo hybridisation amongst themselves?
$$(I)\ 3d, 4s$$
$$(II)\ 3d, 4d$$
$$(III)\ 3d, 4s$$ & $$4p $$
$$(IV) \ 3s, 3p$$ & $$4s$$

A
$$II$$

B
$$II$$ and $$III$$

C
$$I, II$$ and $$IV$$

D
$$II$$ and $$IV$$

Solution

$$(II) \ 3d$$ and $$4d$$ can not undergo hybridization without involving $$4s$$.
$$(IV) \ 3s, 3p$$ and $$4s$$ can not undergo hybridization without involving $$3d$$.
Question 8
1 / -0

Which of the following statements is incorrect?

A
Complex $$[Co(NH_3)_4(H_2O)Cl]Br_2$$ can show both hydrate as well as ionization isomerism

B
Complex $$[Co(NH_3)_5(H_2O)](NO_3)_3$$ can show hydrate isomerism

C
Complex $$[Pt(NH_3)_4][PtCl_6]$$ cannot show coordinate isomerism

D
$$[Co(NH_3)_4(NO_2)Cl]Cl$$ can both ionization as well as linkage isomerism

Solution

Complex $$[Pt(NH_3)_4][PtCl_6]$$ can show coordinate isomerism. This is because in both complex cation and complex anion, platinum metal is present in different oxidation states. Coordination isomerism is possible when two different metals are present or if same metal is present in different oxidation state.
Question 9
1 / -0

For the dark green compound, $$ CrCl_3. 6H_2O$$, which of the following is correct?

A
all the $$Cl$$ shows primary valency (PV).

B
two $$Cl$$ shows primary valency (PV) and one $$Cl$$ secondary valency (SV).

C
two $$Cl$$ shows primary valency (PV) and one $$Cl$$ (PV) as well as secondary valency (SV).

D
all the Cl show secondary valency (SV).

Solution

All the Cl shows primary valency (PV). All $$\displaystyle H_2O$$ show secondary valency (SV). The complex can be written as $$\displaystyle [Cr(H_2O)_6]Cl_3$$. $$\bf{Note:}$$ Primary valency is the number of negative ions which are equivalent to the charge on the metal ion. Secondary valency is the number of ligands that are attached or coordinated to the metal ion.
Question 10
1 / -0

Directions For Questions

Double salts are addition compounds which lose their identity in aqueous solution whereas complexes which are also addition compounds do not lose their identity in aqueous solution. The coordination compounds show isomerism and find applications in photography, qualitative analysis, metallurgy, water purification and in the treatment of various diseases.

...view full instructions

Which of the following statement is true for the complex, $$[Co(NH_3)_4Br_2]NO_2?$$

A
It shows ionisation, linkages and geometrical isomerism.

B
It does not show optical isomerism because its cis and trans forms each have at least one plane of symmetry.

C
Its ionisation isomers cannot be differentiated by silver nitrate solution.

D
(A) and (B) both.

Solution

The complex $$[Co(NH_3)_4Br_2]NO_2$$ shows geometrical isomerism but not optical isomerism as it contains at least one plane of symmetry.
The linkage isomers are $$ [Co(NH_3)_4BrNO_2]Br and [Co(NH_3)_4BrONO]Br$$The ionization isomers are $$[Co(NH_3)_4Br_2]NO2$$ and $$[Co(NH_3)_4BrNO_2]Br$$ . The action of $$AgNO_3$$ solution on $$[Co(NH_3)_4BrNO_2]Br$$ gives pale yellow precipitate.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Coordination Compounds Test - 46

Result

Coordination Compounds Test - 46

Score

-

Rank

-

SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

3

6

2

4

Solution

Question 3

Complex $$[Co(NH_3)_4(H_2O)Cl]Br_2$$ can show both hydrate as well as ionization isomerism

Complex $$[Co(NH_3)_5(H_2O)](NO_3)_3$$ can show hydrate isomerism

Complex $$[Pt(NH_3)_4][PtCl_6]$$ cannot show coordination isomerism

$$[Co(NH_3)_4(NO_2)Cl]Cl$$ can show both ionization as well as linkage isomerism

Solution

Question 4

$$[Fe(OH)_3] Fe^{3+} : Cl^{-}$$

$$[As_{2}S_{3} ]: As^{3+}$$

$$[\mathrm{S}\mathrm{n}\mathrm{O}_{2}]:\mathrm{S}\mathrm{n}\mathrm{O}_{3}^{2-}$$ in acidic medium

$$[\mathrm{A}\mathrm{g}\mathrm{I}]:\mathrm{I}^{-}$$ in excess of $$\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}_3$$

Question 5

a-s , b-p , c-q , d-r

a-s , b-r , c-q , d-p

a-s , b-p , c-r , d-q

a-p , b-r , c-q , d-s

a-p , b-q , c-r , d-s

Solution

Question 6

$$7:10$$

$$8:7$$

$$5:9$$

$$1:2$$

Solution

Question 7

$$II$$

$$II$$ and $$III$$

$$I, II$$ and $$IV$$

$$II$$ and $$IV$$

Solution

Question 8

Complex $$[Co(NH_3)_4(H_2O)Cl]Br_2$$ can show both hydrate as well as ionization isomerism

Complex $$[Co(NH_3)_5(H_2O)](NO_3)_3$$ can show hydrate isomerism

Complex $$[Pt(NH_3)_4][PtCl_6]$$ cannot show coordinate isomerism

$$[Co(NH_3)_4(NO_2)Cl]Cl$$ can both ionization as well as linkage isomerism

Solution

Question 9

all the $$Cl$$ shows primary valency (PV).

two $$Cl$$ shows primary valency (PV) and one $$Cl$$ secondary valency (SV).

two $$Cl$$ shows primary valency (PV) and one $$Cl$$ (PV) as well as secondary valency (SV).

all the Cl show secondary valency (SV).

Solution

Question 10

It shows ionisation, linkages and geometrical isomerism.

It does not show optical isomerism because its cis and trans forms each have at least one plane of symmetry.

Its ionisation isomers cannot be differentiated by silver nitrate solution.

(A) and (B) both.

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.