Coordination Compounds

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: When light wavelengths from a specific part of the spectrum are absorbed by a substance, the result is a complimentary colour. When a complex absorbs a wavelength of light, it reflects a complementary colour. If a violent colour is absorbed, for example, yellow is conveyed. The CFSE value, often known as the colour definer for any complex, comes next. To determine the wavelength value in order to determine which colour absorbs the most energy.

Δe=$\frac{\mathrm{h}\mathrm{c}}{\mathrm{\lambda }}$

As λ has shorter wavelength.

Low spin complexes absorb shorter wavelengths, while high spin complexes absorb longer ones. The colour theory, complex, and wavelength are all determined using this method.

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans:

[Mn (CN)₆]³⁻

Electronic configuration is Mn³⁺= [Ar]3 d⁴ hence box electronic structure

(i) Type of hybridisation d²sp³

(ii) Inner orbital complex

(iii) paramagnetic, due to presence of three unpaired electrons.

(iv) Spin only magnetic moment is calculated using the formula : n=2 in this case, we get spin only magnetic moment in BMas = $\sqrt[]{2(2+2}$ = $\sqrt[]{8}$  = 2.87BM

[Co (NH₃)₆]³⁺

Electronic configuration of Co³⁺= [Ar]3 d⁶

(i) Hyb As shown in the above box electronic structure the type of hybridisation is . d²sp³

(ii) Inner orbital complex

(iii) Diamagnetic

(iv) Zero ( Since no unpaired electrons are present, as depicted above)

[Cr (H₂O)₆]³⁺

(i) type of hybridisation. d²sp³

(ii) Inner orbital complex

(iii) paramagnetic

(iv) 3.87BM

[Fe (Cl)₆]⁴⁻

electronic configuration is of Fe²⁺= [Ar]3 d⁶

(i) type of hybridisation will be sp³d²

(ii) Outer orbital complex

(iii) Paramagnetic

(iv) 4.9BM

This is a Long Answer Type Questions  as classified in NCERT Exemplar

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment =   μ= $\sqrt[]{n(n+2}$   =  $\sqrt[]{4(4+2}$

   =  $\sqrt[]{24}$   = 4.9 BM 

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3(3+2}$ = $\sqrt[]{15}$ = 3.87BM

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Energy level diagram:

Number of unpaired electrons =  μ= $\sqrt[]{n(n+2}=\mathrm{}\sqrt[]{5(5+2}$ =  μ= $\sqrt[]{35}$ = 5.95BM [Fe(H₂O)₆]²⁺

Fe²⁺=[Ar]3 d⁶

Energy level diagram:

Magnetic moment

Since, number of unpaired electrons (n)=4, therefore magnetic moment is-

$\sqrt[]{4(4+2}$ = $\sqrt[]{24}$ = 4.9BM

[Fe(CN)₆]⁴⁻

Electronic configuration: Fe²⁺=[Ar]3 d⁶

Energy level diagram

Since CN⁻ is strong field ligand, it forces all the electrons to get paired up Magnetic moment.

In the absence of unpaired electrons it behaves as diamagnetic ( ie. no magnetic behaviour ).

So magnetic moment = zero