Coordination Compounds Test

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Coordination Compounds Test - 9

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Weekly Quiz Competition

Question 1
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Direction : 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.

Which of the following metal ions cannot form both high spin and low spin octahedral complexes?

A
Ti³⁺

B
Co²⁺

C
Cu²⁺

D
Option A,C are Correct

Solution

Ti³⁺ has 3d¹ configuration and Cu²⁺ has 3d⁹ configuration.
Question 2
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Which is/are the correct statements?

A
CFSE of [Co(NH₃ )₆ ]³⁺ <[Rh(NH₃ )₆ ]³⁺ <[lr(NH₃ )₆ ]³⁺

B
CFSE of [Co(NH₃ )₆ ]³⁺ <[Co(en)₃ ]³⁺

C
[Ni(H₂ O)₆ ]²⁺ complex is green while [Ni(en)₃ ]²⁺ is blue

D
All are correct

Solution

If Δ_o < P high spin complexes are favoured.
Question 3
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CFSE is zero for the complexes with

A
d⁵ weak field

B
d¹⁰ weak field

C
d¹⁰ strong field

D
All are correct

Solution

For d⁵ in weak field, CFSE = -3 x 0.4 + 2 x 0.6 = 0
For d¹⁰ in weak and strong ligand field, CFSE = -6 x 0.4 + 4 x 0.6 = 0
Question 4
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CFSE value depends on

A
nature of ligands

B
charge on the metal ion

C
whether the metal is 3d, 4d or 5d series

D
All are correct

Solution

Geometry of complex, charge on metal ion, nature of ligands and series of d orbital are the factors on which CFSE value depends.
Question 5
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Octahedral complex of Cr (III) will be

A
sp³ d² in case of weak field ligand

B
d² sp³ always

C
always coloured complexes

D
Option B,C are Correct

Solution

Cr (III) ion have 3d³ configuration.
Cr (III) ion

Flence, it will form always d² sp³ hybridisation and show colour due to the presence of 3 unpaired electrons.
Question 6
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Comprehension Type

Direction : This section contains a paragraph, describing theory, experiments, data, etc. Two questions related to the paragraph have.

Passage

In octahedral complexes due to repulsion between the ligands and d-orbitals, there is splitting of d-orbitals into two sets, i.e. two orbitals of higher energy called eg and three orbitals of lower energy called t_2g . The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δ_o . For any given metal cation, the magnitude of Δ_o depends on the nature of ligands.

Q.

The CFSE for d⁷ configuration for strong ligand field is

A
-1.6 Δ_o

B
-2.0 Δ_o

C
-2.4 Δ_o

D
-1.8 Δ_o

Solution

The total crystal field stabilisation energy is given by CFSE_(octahedral) = are the number of electron occupying the t_2g and e_g orbitatls respectively.
For d⁷ in strong field,

CFSE = -0.4 x 6 + 0.6 x 1 = -2.4 + 0.6 = -1.8 Δ_o
Question 7
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In octahedral complexes due to repulsion between the ligands and d-orbitals, there is splitting of d-orbitals into two sets, i.e. two orbitals of higher energy called eg and three orbitals of lower energy called t_2g . The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δ_o . For any given metal cation, the magnitude of Δ_o depends on the nature of ligands.

Q.

In the following complexes of manganese, the distribution of electrons in d-orbitals of manganese

i. [Mn(H₂ O)₆ ]²⁺

ii. [Mn(CN)₆ ]^4-

A

B

C

D

Solution

In [Mn (H₂ O)₆ ]²⁺ , manganese have d⁵ configuration. H₂ O is a weak field ligand. Hence, it will form high spin complex.

In [Mn(CN)₆ ]^4- , manganese have d⁵ configuration. CN^- ion is a strong field ligand. It will form low spin complex by pairing of electron in t_2g orbitals.
Question 8
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Matching List Type

Direction : 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 below.

A
a

B
b

C
c

D
d

Solution

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

A
a

B
b

C
c

D
d

Solution

(i) →(q), (ii) →(p,s), (iii) →(r,t), (iv) →(s)
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.

Number of unpaired electrons in d⁶ , low spin octahedral complex.

A
0

B
1

C
2

D
3

Solution

In d⁶ low spin octahedral complex.

Hence, zero number of unpaired electrons.
Question 11
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The number of ligands which have strong crystal field splitting than

H₂ O among SCN^- , NCS^- , EDTA^4- , , , Br^- , PPh₃ , F^-

A
0

B
4

C
1

D
2

Solution

NCS^- edta^4- , and PPh3 are strong field ligand than H₂ O.
Question 12
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The total number of unpaired electrons in the two complexes [Cr(H₂ O)₆ ]²⁺ and [Cr(CN)₆ ]^4- having octahedral geometry are

A
0

B
1

C
6

D
9

Solution

In [Cr(H₂ O)₆ ]²⁺ chromium have d⁴ configuration and H₂ O is a weak field ligand.
d⁴ (weak field)

In [Cr(CN)₆ ]^4- chromium have d⁴ configuration and CN^- ion is a strong field ligand. Hence, CN^- ion causing pairing of electron. d⁴ (strong field)

Hence, total number of unpaired electrons in both the complex are 6.
Question 13
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Number of unpaired electrons in t_2g and e_g orbitals in weak octahedral ligand fields with d⁷ configuration.

A
3

B
4

C
5

D
6

Solution

d⁷ (weak octahedral field)

Hence, 3 unpaired electrons.
Question 14
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In tetrahedral complexes having weak field ligand, pairing starts with ...... electron.

A
1

B
2

C
3

D
6

Solution

The pairing starts with six electrons in the tetrahedral complexes which have weak field ligand.
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 : The value of CFSE for the octahedral complex [Co(NH₃ )₆ ]³⁺ is - 2.4 Δ_o .

Statement II : The ammonia molecule is strong field ligand.

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 is correct but Statement I is incorrect

Solution

In [Co(NH₃ )₆ ]³⁺ Co, have d⁶ configuration and NH₃ is strong field ligand.

CFSE_(OCt) = -0.4 x 6 + 0.6 x 0 = -2.4 Δ_o