Coordination Compounds Test

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Coordination Compounds Test - 8

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Weekly Quiz Competition

Question 1
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Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Q.
Among the following complexes the one which shows zero crystal field stabilisation energy is

A
[Co(H₂ O)₆ ]²⁺

B
[Co(H₂ O)₆ ]³⁺

C
[Mn(H₂ O)₆ ]³⁺

D
[Fe(H₂ O)₆ ]³⁺

Solution

In [Fe(H₂ O)₆ ]³⁺ , iron has 3d⁵ configuration. Since, H₂ O is weak ligand the distribution is .
Hence, CFSE = - 3 x 0.4 + 2 x 0.6 = 0.0
Question 2
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The wavelength of light absorbed is highest in

A
[Co(NH₃ )₅ CI]²⁺

B
[Co(NH₃ )₅ H₂ O]³⁺

C
[Co(NH₃ )₆ ]³⁺

D
[Co(en)₃ ]³⁺

Solution

Among all the coordination aforesaid in complex [Co(NH₃ )₅ Cl]²⁺ unpaired electrons are present with the central element cobalt so it required less energy for excitation. Thus, the wavlength of light absorbed will be the highest.
Question 3
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The CFSE for octahedral [CoCI₆ ]^4- is 18000 cm^-1 . Then the CFSE for tetrahedral [CoCI₄ ]^2- will be

A
18000cm^-1

B
16000cm^-1

C
8000cm^-1

D
20000cm^-1

Solution

CFSE for [CoCI₄ ]^2- will be
i.e. = 8000cm^-1
Question 4
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The crystal field splitting energy (Δ_o ) of

I. [CoBr₆ ]^3- II. [CoF₆ ]^3-
III. [Co(NCS)₆ ]^3- IV. [Co(CN)₆ ]^3- is in the order of

A
I < II < III < IV

B
II < IV < I < III

C
Ill < I < IV < II

D
IV < III < II < I

Solution

Crystal field splitting energy depends on strength of ligand. Strong ligands have more value of CFSE (Δ_o).
Hence, Br^- < F^- < NCS < CN^-
Question 5
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A magnetic moment of 1.73 BM will be shown by one among the following

A
[Cu(NH₃ )₄ ]²⁺

B
[Ni(CN)₄ ]^{2 –}

C
TiCl₄

D
[CoCl₆ ]^{4 –}

Solution

The correct answer is Option A.
Electronic configuration of Cu²⁺ ion in [Cu(NH₃ )₄ ]²⁺ .
Cu²⁺ ion =[Ar]3d⁹ 4s⁰ .
∴Cu²⁺ ion has one unpaired electron.
Magnetic moment of [Cu(NH₃ )₄ ]²⁺ (μ) = BM
where, n = no. of unpaired electrons

Whereas Ni²⁺ in [Ni(CN)₄ ]^{2 −} , Ti⁴⁺ in TiCl₄ and Co²⁺ ion [COCl₆ ]^{4 −} has 2,0 and 3 unpaired electrons respectively.
Question 6
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The correct order o f ligand field strength is

A
H₂O < Cl^- < CO < NH₃

B
CO < NH₃ < Cl^- < H₂O

C
H₂O < CO < NH₃ < Cl^-

D
Cl^- < H₂O < NH₃ < CO
Question 7
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The crystal field splitting energies (CFSE) of high spin and low spin d⁶ metal complexes in octahedral complex in terms of Δ_o respectively are

A
-0.4 and -2.4

B
-2.4 and -0.4

C
-0.4 and 0.0

D
-2.4 and 0.0

Solution

For high spin d⁶ , CFSE = - 4 x 0.4 + 2 x 0.6 = -0.4
For low spin d⁶ , CFSE= - 6 x 0.4 = -2.4
Question 8
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The Δ_t of the following complexes follows the order
I. [CoCI₄ ]^2-
II. [CoBr₄ ]^2-
III. [Co(NCS)₄ ]^2-

A
Ill >I >II

B
I >II >III

C
II >I >III

D
Ill >II >I

Solution

Correct Answer :- d
Explanation : Bonding Energy :
[CoCI₄ ]² -2084.67
[CoBr₄ ]² -3139.39
[Co(NCS)₄ ]^2- - 8559.48
Correct order is : III >II >I
Question 9
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In which of the following coordination entities, the magnitude Δ_o (CFSE in octahedral field) will be maximum ?

A
[Co(H₂ O)₆ ]³⁺

B
[Co(NH₃ )₆ ]³⁺

C
[Co(CN)₆ ]^3-

D
[Co(C₂ O₄ )^3-

Solution

Increasing order of ligand field strength
l^- < Br^- < SCN^- < CI^- < S^2- < F^- < OH^- < H₂O < NCS^- < edta^4- < NH₃ < en < CN^- < CO
Question 10
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The increasing order of wavelength of absorption for the complex ions

I. [Cr(NH₃ )₆ ]³⁺

II. [CrCI₆ ]^3-

III. [Cr(H₂ O)₆ ]³⁺
IV. [Cr(CN)₆ ]^3-

is :

A
IV < I < III < II

B
IV < III < II < I

C
IV < II < III < I

D
II < III < I < IV

Solution

Stronger ligand causes greater splitting between two energy levels.

The energy of light absorbed is more and the wavelength of light absorbed is less for the stronger ligand.

Hence order of wavelength absorption will be reverse of the order of the strength of ligands.

Hence A is correct.

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Coordination Compounds Test - 8

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Question 1

[Co(H2 O)6 ]2+

[Co(H2 O)6 ]3+

[Mn(H2 O)6 ]3+

[Fe(H2 O)6 ]3+

Solution

Question 2

[Co(NH3 )5 CI]2+

[Co(NH3 )5 H2 O]3+

[Co(NH3 )6 ]3+

[Co(en)3 ]3+

Solution

Question 3

18000cm-1

16000cm-1

8000cm-1

20000cm-1

Solution

Question 4

I < II < III < IV

II < IV < I < III

Ill < I < IV < II

IV < III < II < I

Solution

Question 5

[Cu(NH3 )4 ]2+

[Ni(CN)4 ]2 –

TiCl4

[CoCl6 ]4 –

Solution

Question 6

H2O < Cl- < CO < NH3

CO < NH3 < Cl- < H2O

H2O < CO < NH3 < Cl-

Cl- < H2O < NH3 < CO

Question 7

-0.4 and -2.4

-2.4 and -0.4

-0.4 and 0.0

-2.4 and 0.0

Solution

Question 8

Ill >I >II

I >II >III

II >I >III

Ill >II >I

Solution

Question 9

[Co(H2 O)6 ]3+

[Co(NH3 )6 ]3+

[Co(CN)6 ]3-

[Co(C2 O4 )3-

Solution

Question 10

IV < I < III < II

IV < III < II < I

IV < II < III < I

II < III < I < IV

Solution

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[Co(H₂ O)₆ ]²⁺

[Co(H₂ O)₆ ]³⁺

[Mn(H₂ O)₆ ]³⁺

[Fe(H₂ O)₆ ]³⁺

[Co(NH₃ )₅ CI]²⁺

[Co(NH₃ )₅ H₂ O]³⁺

[Co(NH₃ )₆ ]³⁺

[Co(en)₃ ]³⁺

18000cm^-1

16000cm^-1

8000cm^-1

20000cm^-1

[Cu(NH₃ )₄ ]²⁺

[Ni(CN)₄ ]^{2 –}

TiCl₄

[CoCl₆ ]^{4 –}

H₂O < Cl^- < CO < NH₃

CO < NH₃ < Cl^- < H₂O

H₂O < CO < NH₃ < Cl^-

Cl^- < H₂O < NH₃ < CO

[Co(H₂ O)₆ ]³⁺

[Co(NH₃ )₆ ]³⁺

[Co(CN)₆ ]^3-

[Co(C₂ O₄ )^3-