Coordination Compounds Test

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Coordination Compounds Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following will give $$Fe^{3+}$$ ions in the solution?

A
$$[Fe(CN)_{6}]^{3-}$$

B
$$[Fe(CN)_{6}]^{4-}$$

C
$$NH_{4}(SO_{4})_2.FeSO_{4}.6H_{2}O$$

D
$$Fe_{2}(SO_{4})_{3}$$

Solution

$$[Fe(CN)_{6}]^{3-}$$ and $$[Fe(CN)]_{6}^{4-}$$ are complex ions.

They retain their identity in the solution.

Hence, they will not give $$Fe^{3+}$$ ions in the solution.

$$(NH_{4})_{2}SO_{4}.FeSO_{4}.6H_{2}O$$ is a double salt.

In the solution, it breaks into individual ions, but it will give $$Fe^{2+}$$ ions in the solution.

Similarly, $$Fe_{2}(SO_{4})_{3}$$ also breaks into individual ions.

Hence, they will give $$Fe^{3+}$$ ions in the solution.

Thus, option D is the correct answer.
Question 2
1 / -0

Co-ordination compounds are mostly formed by:

A
s-block elements

B
p-block elements

C
d-block elements

D
f-block elements

Solution

since 'd' block elements have higher oxidation states and variable oxidation states and their tendency to form co-ordinate bonds due to presence of unpaired electrons, they form co-ordination complexes
Question 3
1 / -0

The chemical formula of iron(III) hexacyanoferrate(II) is:

A
$$Fe\left [ Fe\left ( CN \right )_{6} \right ]$$

B
$$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]$$

C
$$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]_{4}$$

D
$$Fe_{4}\left [ Fe\left ( CN \right )_{6} \right ]_{3}$$

Solution

We proceed from left to right, iron (III) means $$Fe^{3+}$$, hexacyano means $$6CN^{-}$$ ions are present and ferrate (II) means $$Fe^{2+}$$ is the central metal atom.

So, the complex ion is $$[Fe(CN)_{6}]^{4-}$$ and the counter ion is $$Fe^{3+}$$. So, after balancing the charge, the formula becomes $$Fe_{4}[Fe(CN)_{6}]_{3}$$.

Option D is correct.
Question 4
1 / -0

IUPAC name of $$Na_{3}\left [ Co\left ( ONO \right )_{6} \right ]$$ is:

A
Sodium cobaltinitrite

B
Sodium hexanitritocobaltate (III)

C
Sodium hexanitrocobalt (III)

D
Sodium hexanitritocobaltate (II)

Solution

The above complex $${ Na }_{ 3 }[Co(ONO)_{ 6 }]$$ has two ions: $${ Na }^{ + }$$ and a complex ion which is negatively charged $$[Co(ONO)_{ 6 }]^{ - }$$

According to the rules, name of the positive ion comes first (i.e., sodium), do not mention the number of sodium as it is not present in a square bracket.

After naming the positive ion, next will be the negative ion, Now in this example,the negative ion is a complex compound, which contain's 'metal' and 'ligand'. While naming a complex, name of ligand comes first, also mention the number of ligand by using prefix (di, tri, tetra. etc..) and if a bidentate ligand is present then mention the atom which is attached to central metal (i.e., hexanitrito $${ - }O{ - })$$ followed by name of metal with the ending 'ate' (i.e., cobaltate, always use the word 'ate' if the complex is negatively charge) and than oxidation number of metal.

To show the oxidation state, we use Roman numerals inside parenthesis.

Option B is correct.
Question 5
1 / -0

Example for a coordination compound is ____________.

A
$$KCl.MgCl_{2}.6H_{2}O$$

B
$$K_{2}SO_{4}Al_{2}(SO_{3})24H_{2}O$$

C
$$CoCl_{3}.6NH_{3}$$

D
$$FeSO_{4}.(NH_{4})_{2}SO_{4}.6H_{2}O$$

Solution

Carnallite $$KCl.MgCl_{2}.6H_{2}O$$
Alum $$K_{2}SO_{4}Al_{2}(SO_{4})_3.24H_{2}O$$
Ferrous ammonium sulphate $$(NH_4)_2Fe(SO_4)_26H_2O$$.
All are double salts.
Only $$[Co(NH_3)_6].Cl_3$$ is complex compound.
Question 6
1 / -0

The primary valence of the metal ion is satisfied by :

A
neutral molecules

B
positive ions

C
negative ions

D
all

Solution

The primary or principal valency; this is the ionisable valency.

A metal always gives electron and becomes positively charged that can be neutralized by bonding with negative ion only. In a coordination compound, the number of negative ions needed to satisfy the charge on the central metal ion is it's primary valency.
Question 7
1 / -0

Which is a coordination compund ?

A
Ferrous ammonium sulphate

B
Carnallite

C
Potassium ferrocyanide

D
Gypsum

Solution

Only 'Potassium ferrocyanide' (i.e. $$[Fe(CN)_{ 5 }]$$ ) shows properties of coordination compound.

When it gets dissolve they do not form simple ion like $$Fe^{ +2 }$$ or $$CN^{ - }$$ , but instead their complex ions remain intact.

whereas carnallite (i.e. $$KMg{ Cl }_{ 3 }$$ ), Gypsum (i.e. $$Ca{ SO }_{ 4 }$$) and Ferrous ammonium sulphate (i.e. $$({ NH }_{ 4 })_ { 2 }Fe (SO_4)_2$$ ) they dissovle and give simple ions.

Option C is correct.
Question 8
1 / -0

Which of the following statement is correct?

A
Energy difference should be more between orbitals which undergo hybridization.

B
Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.

C
Hybrid orbitals arrange around the center of the atom unsymmetrically.

D
Hybrid orbitals can form $$\pi$$ bonds.

Solution

Number of hybrid orbitals formed is equal to number of atomic orbitals involved in hybridisation. $$\pi$$ bonds will be formed always by pure orbitals (p or d orbitals). Hybrid orbitals arrange around the centre of atom gymmetrically depending on number of bond pairs and lone pairs.
Question 9
1 / -0

Which of the following has the highest molar conductivity?

A
$$[Co(NH_{3})_{6}]Cl_{3}$$

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

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

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

Solution

Hint: Molar conductivity is dependent on the number of ions produced in the solution.
Explanation:
The dissociation of $$[Co{(N{H_3})_6}]C{l_3}$$ is as follows;
$$[Co{(N{H_3})_6}]C{l_3} \rightleftharpoons {[Co{(N{H_3})_6}]^{3 + }} + 3C{l^ - }$$
The number of ions produced here is $$4$$. So the molar conductivity is the highest.
Final Answer: The molar conductivity of $$[Co{(N{H_3})_6}]C{l_3}$$ is the highest. Hence the correct answer is option $$A$$.
Question 10
1 / -0

The primary and secondary valency of $$Co$$ in the octahedral complex $$CoCl_{3}.5NH_{3}$$ are __________.

A
6 and 3

B
3 and 6

C
4 and 3

D
3 and 5

Solution

The primary valency of $$Co$$ is 3 and the secondary valency of $$Co$$ is 6.
The primary valency represents the number of ions in the ionization sphere and the secondary valency represents the number of ligands.
In this example, one chlorine atom serves dual role.

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Coordination Compounds Test - 14

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Coordination Compounds Test - 14

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Question 1

$$[Fe(CN)_{6}]^{3-}$$

$$[Fe(CN)_{6}]^{4-}$$

$$NH_{4}(SO_{4})_2.FeSO_{4}.6H_{2}O$$

$$Fe_{2}(SO_{4})_{3}$$

Solution

Question 2

s-block elements

p-block elements

d-block elements

f-block elements

Solution

Question 3

$$Fe\left [ Fe\left ( CN \right )_{6} \right ]$$

$$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]$$

$$Fe_{3}\left [ Fe\left ( CN \right )_{6} \right ]_{4}$$

$$Fe_{4}\left [ Fe\left ( CN \right )_{6} \right ]_{3}$$

Solution

Question 4

Sodium cobaltinitrite

Sodium hexanitritocobaltate (III)

Sodium hexanitrocobalt (III)

Sodium hexanitritocobaltate (II)

Solution

Question 5

$$KCl.MgCl_{2}.6H_{2}O$$

$$K_{2}SO_{4}Al_{2}(SO_{3})24H_{2}O$$

$$CoCl_{3}.6NH_{3}$$

$$FeSO_{4}.(NH_{4})_{2}SO_{4}.6H_{2}O$$

Solution

Question 6

neutral molecules

positive ions

negative ions

all

Solution

Question 7

Ferrous ammonium sulphate

Carnallite

Potassium ferrocyanide

Gypsum

Solution

Question 8

Energy difference should be more between orbitals which undergo hybridization.

Number of hybrid orbitals formed should be same as the number of atomic orbitals involved in hybridization.

Hybrid orbitals arrange around the center of the atom unsymmetrically.

Hybrid orbitals can form $$\pi$$ bonds.

Solution

Question 9

$$[Co(NH_{3})_{6}]Cl_{3}$$

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

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

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

Solution

Question 10

6 and 3

3 and 6

4 and 3

3 and 5

Solution

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