Identify the series showing isolobal analogy.

A. CH₃, [Fe(CO)₅]⁺

B. \(CH^+_3\), [Cr(CO)₅]^-

C. \(CH^+_3\), Ni(CO)₃

D. CH⁺, CpCo

Concept:

The fragments are said to be isolobal if they have the same bonding properties and electronic structure.

Two fragments or molecule is said to be isolobal if their frontier orbitals are

• The same in number,
• Possesses the same symmetries,
• Have the same occupation of electrons,
• The orbitals are similar in radial extent.

We symbolize this relationship by,

Electron equivalent groups miss the same number of electrons necessary for the central atom to reach the stable electronic configuration. Electronic equivalent groups are always isolobal.

Explanation:-

A. CH3, [Fe(CO)5]+

For CH3:

Total number of electrons = 4+3 =7 (for 8 electrons require 1 more electron)

and for [Fe(CO)5]+

Total number of electrons = 8+5×2-1 =17 (for 18 electrons require 1 more electron)

Thus, CH3 ↔ [Fe(CO)5]+ (Isolobal)

B. \(CH^+_3\), [Cr(CO)5]-

For \(CH^+_3\):

Total number of electrons = 4+3 - 1 =6 (for 8 electrons require 2 more electron)

and for [Fe(CO)5]+

Total number of electrons = 8+5×2-1 =17 (for 18 electrons require 1 more electron)

Thus, \(CH^+_3\) and [Fe(CO)5]+ are not isolobal.

C. \(CH^+_3\), Ni(CO)3

For \(CH^+_3\):

Total number of electrons = 4+3 - 1 =6 (for 8 electrons require 2 more electron)

For Ni(CO)₃:

Total number of electrons = 10+3×2 = 16 (for 18 electrons require 2 more electron)

Thus, \(CH^+_3\) ↔ Ni(CO)3 (Isolobal)

D. CH+, CpCo

For CH⁺:

Total number of electrons = 4 + 1 - 1 = 4 (for 8 electrons require 4 more electron)

For CpCo:

Total number of electrons = 5+9 = 14 (for 18 electrons require 4 more electron)

Thus, CH+↔CpCo

Conclusion:-

Hence, only A, C, and D only shows isolobal analogy.