Chemistry Coordination Compounds

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4 months ago

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R
Raj Pandey

Contributor-Level 9

With weak field ligands Mn (II) will be of high spin and with strong field ligands it will be of low spin.
Ni (II) tetrahedral complexes will be generally of high spin due to sp³ hybridisation. Mn (II) is of light pink color in aqueous solution.

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Raj Pandey

Contributor-Level 9

Since spin only magnetic moment is 4.90BM so number of unpaired electrons must be 4, so If the complex is octahedral, then it has to be high spin complex with configuration t? g? e²g¹ in that case
CFSE = 4* (-0.4Δ? ) + 2*0.6Δ? = -0.4Δ?
If the complex is tetrahedral then its electronic configuration will be e? ², t? g² and CFSE will be = 3* (-0.6Δ? ) + 3* (0.4Δ? ) = -0.6Δ?

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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Vishal Baghel

Contributor-Level 10

Δ? > P
Pairing of electron will take place


Number of unpaired electron = 0
µ = 0.00

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Vishal Baghel

Contributor-Level 10

(A) Trans

(B) Cis

Optically inactive due to presence of plane of symmetry

Optically active no plane of symmetry

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R
Raj Pandey

Contributor-Level 9

C o N H 3 6 C l 3 + 3 A g N O 3 ? 3 A g C l

  Mole of   C o N H 3 6 C l 3 1 =   Mole of   A g N O 3 3

0.3 267.46 = 0.125 * V * 10 - 3 3

V = 26.92 m L

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R
Raj Pandey

Contributor-Level 9

M a 3 b 3  - type complex compound show fac-and mer-isomer.

C o N H 3 3 N O 2 3

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A
alok kumar singh

Contributor-Level 10

In strong ligand field C o 3 + will have t 2 g 6 e g 0 of configuration and Δ t = 4 9 Δ 0

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A
alok kumar singh

Contributor-Level 10

M A 2 B 2 shows geometrical and not optical isomerism

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