Question

discuss the nature of bonding of [Co(C2O4)3]^3- on the basis of valence bond theory​

Answer 1

Answer:

Fe(CN)

6

]

4−

Co(III) with 3d

6

outer electronic configuration is d

2

sp

3

hybridised. A diamagnetic octahedral complex is formed. It is inner orbital, low spin or spin paired complex.

(ii) [FeF

6

]

3−

Fe(III) with 3d

5

outer electronic configuration is sp

3

d

2

hybridised. A paramagnetic octahedral complex is formed. It is outer orbital, high spin or spin free complex.

(iii) [Co(C

2

O

4

)

3

]

3−

Co(III) with 3d

6

outer electronic configuration is d

2

sp

3

hybridised. A diamagnetic octahedral complex is formed. It is inner orbital, low spin or spin paired complex.

(iv) [CoF

6

]

3−

Co(III) with 3d

6

outer electronic configuration is sp

3

d

2

hybridised. A paramagnetic octahedral complex is

formed. It is outer orbital, high spin or spin free complex.

Answer 2

Valence Bond Theory states that:

• A bond will be formed only if an orbital of one atom comes to occupy the same space of an orbital of another atom.

• Bonds will be formed such that maximum number of electrons involved will be 2.

• The strength of the bond will be directly proportional to the orbital overlap between the two atoms.

Considering these rules:

${Co}^{ + 3} \rightarrow \boxed{ \uparrow \downarrow |\uparrow | \uparrow | \uparrow | \uparrow } \: \: \: \boxed{ \overset{}{ - } } \: \: \: \boxed{ \: \: | \: \: \: | \: \: }$

In presence of oxalate ;

${Co}^{ + 3} \rightarrow \boxed{ \uparrow \downarrow |\uparrow \downarrow | \uparrow \downarrow| \: \: \: | \: \: \: } \: \: \: \boxed{ \overset{}{ - } } \: \: \: \boxed{ \: \: | \: \: \: | \: \: }$

So, the required hybridisation is :

$\boxed{ \sf{hybridisarion = {d}^{2} s {p}^{3} }}$

So, the type of bonding is coordinate bonding characterised by electron donation by oxalate chelation into the empty orbitals of Cobalt ion.