Coordination Compounds Test

Result

Coordination Compounds Test - 13

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Question 1
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One or More than One Options Correct Type

Direction (Q. Nos. 1 - 5) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.

Q.

The formation of complex which involves d-orbitals of outer shell are called

A
low spin complex

B
high spin complex

C
outer orbital complex

D
Option B,C are Correct

Solution

The complex which involves d-orbitals of outer shell are called high spin complex as well as outer orbital complex.
Question 2
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sp³ -hybridisation is found in

A
[ZnCI₄ ]^2-

B
[CuCI₄ ]^2-

C
[Ni(CO)₄ ]

D
All are correct

Solution

[ZnCI₄ ]^2- , [CuCI₄ ]^2- and [Ni(CO)₄ ] have sp³ hybridisation and all three contains weak field ligands.
Question 3
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Regarding the complex [Co(H₂ O)₆ ]^2- , correct statement(s) is/are

A
Paramagnetic complex

B
Outer orbital complex

C
Magnetic moment is 3.873 BM

D
All are correct

Solution

Since, F^- ion is weak ligand it forms outer orbital complex arid it has 3d⁷ configuration with 3 unpaired electrons.
Question 4
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For which the EAN value is equal to the atomic number of a noble gas?

A
K₂ [HgI₄ ]

B
[Cdl₄ ]^2-

C
Co₂ (CO)₈

D
All are correct

Solution

K₂ [Hgl₄ ] In this EAN = 80 - 2 + 8 = 86 (Equal to noble gas, radon).
[Pd(NH₃ )CI₂ ] In this EAN = 46 - 2 + 8 = 52 (Not equal to noble gas, xenon)
[Cdl₄ ]^2- In this EAN = 48 - 2 + 8 = 54 (Equal to noble gas, xenon)
Co₂ (CO)₈ In this EAN = Electrons from Co-atom (27) + electrons from 4 CO molecules (8) + one electron from Co —Co bond (1) = 36 (Equal to noble gas krypton)
Question 5
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[Fe(H₂ O)₆ ]²⁺ and [Fe(CN)₆ ]^4- differ in

A
magnetic moment

B
colour

C
hybridisation

D
All are correct

Solution

[Fe(H₂ O)₆ ]²⁺ and [Fe(CN)₆ ]^4- have different hybridisation, magnetic moment and colour also.
Question 6
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Comprehension Type

Direction (Q. Nos. 6 and 7) This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Passage

When crystals of CuSO₄ . 4NH₃ are dissolved in water, there is hardly any evidence for the presence of Cu²⁺ ions or ammonia molecules. A new ion [Cu(NH₃ )₄ ]²⁺ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)₂ .4KCNdoes not give tests of Fe²⁺ and CN^- ions but give test for new ion [Fe(CN)₆ ]^4- called ferrocyanide ion.

Q.

The hybridisation and geometry and magnetic property of [Cu(NH₃ )₄ ]²⁺ ion are

A
sp³ , tetrahedral, diamagnetic

B
sp³ , tetrahedral, paramagnetic

C
dsp² , square planar, diamagnetic

D
dsp² , square planar, paramagnetic

Solution

Cu(Z = 29) copper configuration (3d¹⁰ 4s¹ )

Hence, [Cu(NH₃ )₄ ]²⁺ has square planar geometry and 1 unpaired electron in Ad orbital which is transferred from 3d orbital with magnetic moment 1.73 BM.
Question 7
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When crystals of CuSO₄ . 4NH₃ are dissolved in water, there is hardly any evidence for the presence of Cu²⁺ ions or ammonia molecules. A new ion [Cu(NH₃ )₄ ]²⁺ is furnished in which ammonia molecules are directly linked with the metal ion. Similarly, aqueous solution of Fe(CN)₂ .4KCNdoes not give tests of Fe²⁺ and CN^- ions but give test for new ion [Fe(CN)₆ ]^4- called ferrocyanide ion.

Q.

The hybridisation, geometry and magnetic property of [Fe(CN)₆ ]^4- are

A
d² sp³ , octahedral, diamagneti

B
d² sp³ , octahedral, paramagnetic

C
sp³ d² , octahedral, diamagnetic

D
sp³ d² , octahedral, paramagnetic

Solution

It is inner orbital complex, diamagnetic.
Question 8
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Matching List Type

Direction (Q. Nos. 8 and 9) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d) out of which one is correct.

Q.

Match the Column I with Column II and mark the correct option from the codes given bleow.

A
a

B
b

C
c

D
d

Solution

(i) →(p.s), (ii) →(p,s), (iii) →(q,t), (iv) →(q.r)
Question 9
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Match the Column I with Column il and mark the correct option from the codes given bleow.

A
a

B
b

C
c

D
d
Solution
A →(iii) (b) B →(i) (c) C →(iv) (d) D →(ii)

a) Six empty orbitals (two d, one s and three p) are available, so whether the ligand is weak (H2 O) the six orbitals hybridize to give six equivalent d2sp3 hybrid orbitals. Hence, in [Cr(H2 O)6 ]3+ ions chromium is in a state of d2sp3 hybridization.

b) CN is a strong ligand it makes the unpaired electrons of cobalt to pair up and occupies the space.

The ligands occupy one d orbital, one s orbital and 2 p orbitals. Thus the hybridisation is “dsp2 ” with “square planar ” geometry.

It has 1 unpaired electron.

c) In [Ni(NH3)6]2+, Ni is in +2 state and has configuration 3d8. In presence of NH3, the 3d electrons do not pair up. The hybridization is sp3d2 forming an outer orbital complex.

It has been found that the complex has two unpaired electrons .
Question 10
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One Integer Value Correct Type

Direction : This section contains 5 questions. When worked out will result in an integer from 0 to 9 (both inclusive).

Q.

How many of the following species obey Sidgwick EAN rule?

K₃ [Fe(CN)₆ ], [Ru(CO)₅ ], [Cr(NH₃ )₆ ] ³⁺ , [Co(NH₃ )₆ ]³⁺ , [Ni(NH₃ )₆ ]²⁺ , [Fe(CO)₅ ],[W(CO)₆ ]

A
4

B
3

C
2

D
1

Solution
Question 11
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How many of the following are diamagnetic?

[Zn(OH)₄ ]^2- ,[Ni(NH₃ )₆ ]²⁺ , K₄ [Fe(CN)₆ ], K₃ [Fe(CN)₆ ], [Cu(NH₃ )₄ ]²⁺ ,[PdBr₄ ]^2- , [Ni(CO)₄ ], [CoF₆ ] ^3-

A
9

B
4

C
3

D
1

Solution

Diamagnetic are
[Zn(OH)₄ ]^2- , K₄ [Fe(CN)₆ ], [PdBr₄ ]^2- ,[Ni(CO)₄ ]
Question 12
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The number of unpaired electrons in the complex [Mn(acac)₃ ] is

(Atomic number of Mn= 25)

A
1

B
2

C
4

D
0

Solution

In this, Mn has +3 oxidation state. Mn³⁺ configuration = [Ar]3d⁴ . This forms outer complex. Hence, unpaired electrons are 4.
Question 13
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The magnetic moment of [Ru(H₂ O)₆ ]²⁺ corresponds to the presence of ...... unpaired electrons.

A
9

B
8

C
7

D
0

Solution

R[44] = [Kr]4d⁷ 5s¹ (in ground state)
In Ru[2+] =>4d⁶
=>(t₂ g)⁶ (eg)⁰

^{Hence, no unpaired electron. Therefore, correct answer is zero.}
Question 14
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The number of unpaired electrons in [Fe(dipy)₃ ]²⁺ are

A
0

B
3

C
5

D
1

Solution

In [Fe(dipy)₃ ]²⁺ complex ion, Fe has 3d⁶ configuration. Due to pairing, this has no unpaired electrons.
Question 15
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Statement Type

Direction : This section is based on Statement I and Statement II. Select the correct answer from the codes given below.

Q.

Statement I : Both [Ni(CN)₄ ]^2- and [NiCI₄ ]^2- have same shape and same magnetic behaviour.

Statement II : Both are square planar and diamagnetic

A
Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

B
Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II and Statement I is incorrect

Solution

Statement I [Ni(CN)₄ ]^2- is square planar and diamagnetic whereas [NiCI₄ ]^2- is tetrahedral and paramagnetic .
Statement II [Ni(CN)₄ ]^2- is square planar while [NiCI₄ ]^2- is tetrahedral.