Chemistry Coordination Compounds
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New answer posted
a month agoContributor-Level 9
The oxidation states of iron in these compounds will be
A = +2
B = +4
C = 0
The sum of oxidation states will be = 6.
New answer posted
a month agoContributor-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.
New answer posted
a month agoContributor-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?
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
Δ? > P
Pairing of electron will take place
Number of unpaired electron = 0
µ = 0.00
New answer posted
a month agoContributor-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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