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Test: Crystal Field Theory & Coloured Complexes (January 2) - NEET MCQ


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10 Questions MCQ Test - Test: Crystal Field Theory & Coloured Complexes (January 2)

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Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 1

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

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 1

In [Fe(H2O)6]3+, iron has 3d5 configuration. Since, H2O is weak ligand the distribution is .
Hence, CFSE = - 3 x 0.4 + 2 x 0.6 = 0.0

Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 2

The crystal field splitting energy (Δo) of

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

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 2

Crystal field splitting energy depends on strength of ligand. Strong ligands have more value of CFSE (Δo).
Hence, Br- < F- < NCS < CN-

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Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 3

The increasing order of wavelength of absorption for the complex ions

I. [Cr(NH3)6]3+

II. [CrCI6]3-

III. [Cr(H2O)6]3+
IV. [Cr(CN)6]3-

is :

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 3

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.

*Multiple options can be correct
Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 4

CFSE is zero for the complexes with

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 4

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

Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 5

Comprehension Type

Direction (Q. Nos. 16 and 17) 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 t2g. The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δo. For any given metal cation, the magnitude of Δo depends on the nature of ligands.

Q. 

The CFSE for d7 configuration for strong ligand field is

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 5

The total crystal field stabilisation energy is given by CFSE(octahedral) = are the number of electron occupying the t2g and eg orbitatls respectively.
For d7 in strong field,


CFSE = -0.4 x 6 + 0.6 x 1 = -2.4 + 0.6 = -1.8Δo 

Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 6

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 t2g. The difference of energy between the two sets of d-orbitals is called crystal field stabilisation energy denoted by Δo. For any given metal cation, the magnitude of Δo depends on the nature of ligands.

Q. 

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

i. [Mn(H2O)6]2+

ii. [Mn(CN)6]4-

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 6

In [Mn (H2O)6]2+, manganese have d5 configuration. H2O 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 t2g orbitals.

*Answer can only contain numeric values
Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 7

The number of ligands which have strong crystal field splitting than

H2O among SCN-, NCS-, EDTA4- , , , Br-, PPh3, F-


Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 7

NCS- edta4- , and PPh3 are strong field ligand than H2O.

*Answer can only contain numeric values
Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 8

The total number of unpaired electrons in the two complexes [Cr(H2O)6]2+ and [Cr(CN)6]4- having octahedral geometry are


Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 8

In [Cr(H2O)6]2+ chromium have d4 configuration and H2O 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.

*Answer can only contain numeric values
Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 9

Number of unpaired electrons in t2g and eg orbitals in weak octahedral ligand fields with d7 configuration.


Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 9

d7 (weak octahedral field)

Hence, 3 unpaired electrons.

Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 10

The crystal field splitting energies (CFSE) of high spin and low spin d6 metal complexes in octahedral complex in terms of Δo respectively are

Detailed Solution for Test: Crystal Field Theory & Coloured Complexes (January 2) - Question 10

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

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