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

 

(i) Overall charge balance:

X + 3 (-2) = -3 X = + 3

Oxidation state of Co is + 3.

As there are 3 oxolate ion and being bidentate, coordination no. Of complex is 6. So it is octahedral complex.

d orbital occupation: t_2g⁶eg⁰ (oxolate ion is weak field ligand, does not cause pairing of electron as the energy required for pairing of electron is more than CFSE).

(ii) Overall charge balance:

X + balance4 (-1) = -2 X = + 2

Oxidation state of Co is + 2.

As there are 4 fluoride ion, coordination no. Of complex is 4 i.e. Tetrahedral complex.

d orbital occupation: eg⁴t_2g³ (fluoride ion is weak field ligand, does not cause pairing of electron as the energy required for pairing of electron is more than CFSE).

(iii) Overall charge balance:

X + 2 (-1) + 2 (0) = + 1 X = + 3

Oxidation state of Cr is + 3

Coordination no. Of complex is 6. (-en is a bidentate ligand). Octahedral complex. d orbital complex: t_2g³

(iv) Overall charge balance:

X + 6 (0) = + 2 X = + 2

Oxidation state of Mn is + 2.

Coordination no. Of complex is 6. Octahedral complex.

d orbital occupation: t_2g³e ² (water is weak field ligand, does not cause pairing of the electron).

Compounds containing carbonyl

Ligands only are known as homoleptic carbonyl. Such types of compounds are formed by most of the transition metals. These metal carbonyls always have simple, well-defined structures. In metal carbonyls the metal - carbonyl bond possess both s and p.character. M-C-bond is sigma bond. It is formed by the donation of lone pair of electrons of the carbonyl carbon into the vacant orbital of the metal. The M-C pi bond is formed by the donation of a pair of electron from a filled d orbital of a metal into the vacant antibonding? orbital of carbon monoxide. Such type of metal to ligand bonding creates a synergic effect which strengthens the bond between CO and metal.