Coordination Compounds Test

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Coordination Compounds Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Among the following complexes (K-P) :
$$K_3[Fe(CN_6]-K $$ and $$ [Co(NH_3)_6]Cl_3-L$$ ;
$$Na_3[Co(oxalate)_3]-M $$ and $$ [Ni(H_2O)_6]Cl_2-N$$ ;
$$K_2[Pt(CN)_4]-O$$ and $$[Zn(H_2O)_6](NO_3)_2 - P$$;
the diamagnetic complexes are:

A
K,L,M,N

B
K,M,O,P

C
L,M,O,P

D
all

Solution

$$K_3[Fe(CN_6]-K $$ = $$ 3d^5$$ it is paramagnetic

$$ [Co(NH_3)_6]Cl_3-L$$ = $$3d^6$$ is diamagnetic

$$Na_3[Co(oxalate)_3]-M $$ = $$3d^6$$ is diamagnetic

$$ [Ni(H_2O)_6]Cl_2-N$$ = $$3d^8$$ is paramagnetic

$$K_2[Pt(CN)_4]-O$$ = $$d^3 $$ it is diamagnetic

and $$[Zn(H_2O)_6](NO_3)_2 - P$$;= $$ d^6$$ it is diamagnetic
Question 2
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Three arrangements are shown for the complex, $$[Co(en)(NH_3)_2Br_2]^+$$. Which one is the wrong statement?

A
I and II are geometrical isomers.

B
II and III are optical isomers.

C
I and III are optical isomers.

D
II and III are geometrical isomers.

Solution

I and II are geometrical isomers. I is cis as two $$Br$$ atoms are adjacent to each other. II is trans as two $$Br$$ atoms are opposite to each other.
II and III are geometrical isomers. II is trans as two $$Br$$ atoms are opposite to each other. III is cis as two $$Br$$ atoms are adjacent to each other.
I and III are optical isomers. They are non-superimposable mirror images of each other.
Question 3
1 / -0

Copper sulphate dissolves in ammonia due to the formation of:

A
$$Cu_2O$$

B
$$[Cu(NH_3)_4]SO_4$$

C
$$[Cu(NH_3)_4]OH$$

D
$$[Cu(H_2O)_4]SO_4$$

Solution

Copper sulphate dissolves in ammonia due to the formation of $$[Cu(NH_3)_4]SO_4$$.
$$CuSO_4 + 4NH_3 \rightarrow [Cu(NH_3)_4]SO_4$$
$$[Cu(NH_3)_4]SO_4$$ contains complex cation $$[Cu(NH_3)_4]^{2+}$$.
Question 4
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Which one of the following has largest number of isomers?
(R = alkyl group,en = ethylenediamine)

A
$$[Ru(NH_3)_4Cl_2]^+$$

B
$$[Co(NH_3)_5Cl]^{2+}$$

C
$$[Ir(PR_3)_2H(CO)]^{2+}$$

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

Solution

$$[Co(en)_2Cl_2]^+ $$ show geometrical as well as optical isomerism.
$$[Ru(NH_3)_4Cl_2]^+$$ and $$[Ir(PR_3)_2H(CO)]^{2+}$$ shows geometrical isomerism only.
No isomerism is shown by $$[Co(NH_3)_5Cl]^{2+}$$.
Question 5
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Both geometrical and optical isomerism are shown by:

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

B
$$[Co(NH_3)_5Cl]^{2+}$$

C
$$[Co(NH_3)_4Cl_2]^{+}$$

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

Solution

Both geometrical and optical isomerism is shown by $$[Co(en)_2Cl_2]^+$$.

No isomerism is shown by $$[Co(NH_3)_5Cl]^{2+}$$.

$$[Co(NH_3)_4Cl_2]^{+}$$ shows geometrical isomerism.

$$[Cr(ox)_3]^{3-}$$ shows optical isomerism.
Question 6
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Which of the following statements is true about hybridisation?

A
The hybridised orbitals have different energies for each orbital.

B
The number of hybrid orbitals is equal to the number of atomic orbitals that are hybridised.

C
Hybrid orbitals form multiple bonds.

D
The orbitals with different energies undergo hybridisation.

Solution

In hybridization, the number of hybrid orbitals is equal to the number of atomic orbitals that are hybridized.
Let's take an example:- $$CH_4$$
Carbon electronic configuration is $$2S^2$$$$2P^2$$ in ground state, in excited state one electron of $$s$$ orbital move into $$p$$ orbital and thus we get $$4$$ unpaired electrons in atomic orbital which will take part in bonding,
The hybridisation of carbon in $$CH_4$$ is $$sp^3$$ , means $$4$$ hybrid orbitals. (one $$s$$ orbital and three $$p$$ orbitals)
So the correct option is [B]
Question 7
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According to Werner's coordination theory, there are _______ kinds of valency, _______ and ______. The primary valency of a central metal ion is satisfied with ________.

A
three, negative, positive, cations

B
different, negative, positive, anions

C
two, primary, secondary, anions

D
two, saturated, unsaturated, cations

Solution

According to Werner's coordination theory, there are two kinds of valency, primary and secondary. The primary valency of a central metal ion is satisfied with anions.
For example, in $$[Cu(NH_3)_4]SO_4$$ primary valency is 2 and secondary valency is 4.
Secondary valence refers to coordination number. Since copper is coordinated to 4 ammonia ligands, secondary valence is 4. Primary valence is satisfied by anions. Since sulphate ion has $$-2$$ charge, primary valence is 2.
Question 8
1 / -0

The compounds $$[Co(SO_4) (NH_3)_5]Br$$ and $$ [Co(SO_4)(NH_3)_5]Cl$$ represent:

A
linkage isomerism

B
ionisation isomerism

C
coordination isomerism

D
no isomerism

Solution

Isomers are compounds having the same chemical formula but a different structure.

The compounds $$[Co(SO_4) (NH_3)_5] Br$$ and $$ [Co(SO_4)(NH_3)_5]Cl$$ have different chemical formulas.

So they represent no isomerism.
Question 9
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Which of the following pairs of isomers is not correctly matched with its type of isomerism?

A
$$[Cr(NH_3)_6][Cr(CN)_6]$$ and $$[Cr(NH_3)_4(CN)_2][Cr(NH_3)_2(CN)_4]$$ - Coordination isomerism

B
$$Co[(NH_3)_5NO_2]Cl_2$$ and $$[Co(NH_3)_5ONO]Cl_2$$- Linkage isomerism

C
$$[Co(py)_2(H_2O)_2Cl_2]Cl$$ and $$[Co(py)_2(H_2O)Cl_3]H_2O$$ -Coordination isomerism

D
$$[Pt(NH_3)_4Br_2]Cl_2 $$ and $$[Pt(NH_3)_4Cl_2]Br_2$$ -Ionisation isomerism

Solution

Water molecules are exchanged inside and outside coordination sphere. Number of water molecules is different inside and outside coordination sphere. These are called hydrate isomers.
(i) $$[Co(py)_2(H_2O)_2Cl_2]Cl $$ has two water molecules inside coordination sphere and zero water molecules outside coordination sphere.
(ii) $$[Co(py)_2(H_2O)Cl_3]H_2O$$ has one water molecule inside coordination sphere and one water molecule outside coordination sphere.

Question 10

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Match the column I with column II and mark the appropriate choice.

	Column I (Complex)		Column II (Isomerism)
(A)	$$[Co(NH_3)_6] [Cr(CN)_6]$$	(i)	Geometrical isomerism
(B)	$$[Co(en)_2(NO_2)Cl]Br$$	(ii)	Optical isomerism
(C)	$$[Pt(en)_2Cl_2]$$	(iii)	Coordination isomerism
(D)	$$[Cr(CN)_2(NH_3)_4]^{2+}$$	(iv)	Linkage isomerism

A - (iv), B - (ii), C - (iii), D - (i)

A - (ii), B - (iii), C - (i), D - (iv)

A - (iii), B - (iv), C - (ii), D - (i)

A - (i), B - (iii), C - (iv), D - (ii)

Solution

$$[CO(NH_3)_6][Cr(CN)_6]\Longleftrightarrow [CO(CN)_6][Cr(NH_3)_6]$$(Co-ordinate Isomerism)

$$[CO(en)_2[NO_2]Cl]Br \Rightarrow [CO(en)_2(ONO)Cl]Br$$ (Linkage Isomerism)

$$[Pt(en)_2Cl_2]$$ (Optical Isomerism)(Refer to image 01)

$$[Cr(CN)_2(NH_3)_6]^{2+}$$ (Refer to image 02)

Geometric Isomerism (Refer to image 03)

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Coordination Compounds Test - 28

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Coordination Compounds Test - 28

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Question 1

K,L,M,N

K,M,O,P

L,M,O,P

all

Solution

Question 2

I and II are geometrical isomers.

II and III are optical isomers.

I and III are optical isomers.

II and III are geometrical isomers.

Solution

Question 3

$$Cu_2O$$

$$[Cu(NH_3)_4]SO_4$$

$$[Cu(NH_3)_4]OH$$

$$[Cu(H_2O)_4]SO_4$$

Solution

Question 4

$$[Ru(NH_3)_4Cl_2]^+$$

$$[Co(NH_3)_5Cl]^{2+}$$

$$[Ir(PR_3)_2H(CO)]^{2+}$$

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

Solution

Question 5

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

$$[Co(NH_3)_5Cl]^{2+}$$

$$[Co(NH_3)_4Cl_2]^{+}$$

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

Solution

Question 6

The hybridised orbitals have different energies for each orbital.

The number of hybrid orbitals is equal to the number of atomic orbitals that are hybridised.

Hybrid orbitals form multiple bonds.

The orbitals with different energies undergo hybridisation.

Solution

Question 7

three, negative, positive, cations

different, negative, positive, anions

two, primary, secondary, anions

two, saturated, unsaturated, cations

Solution

Question 8

linkage isomerism

ionisation isomerism

coordination isomerism

no isomerism

Solution

Question 9

$$[Cr(NH_3)_6][Cr(CN)_6]$$ and $$[Cr(NH_3)_4(CN)_2][Cr(NH_3)_2(CN)_4]$$ - Coordination isomerism

$$Co[(NH_3)_5NO_2]Cl_2$$ and $$[Co(NH_3)_5ONO]Cl_2$$- Linkage isomerism

$$[Co(py)_2(H_2O)_2Cl_2]Cl$$ and $$[Co(py)_2(H_2O)Cl_3]H_2O$$ -Coordination isomerism

$$[Pt(NH_3)_4Br_2]Cl_2 $$ and $$[Pt(NH_3)_4Cl_2]Br_2$$ -Ionisation isomerism

Solution

Question 10

A - (iv), B - (ii), C - (iii), D - (i)

A - (ii), B - (iii), C - (i), D - (iv)

A - (iii), B - (iv), C - (ii), D - (i)

A - (i), B - (iii), C - (iv), D - (ii)

Solution

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