Here is a detailed explanation of the hybridization orbitals in the compound K4[Fe(CN)6]:
- **Hybridization Orbitals in K4[Fe(CN)6]**
In K4[Fe(CN)6], the central Fe atom is bonded to six CN ligands, forming an octahedral complex. The hybridization of the central Fe atom can be determined using the formula:
Hybridization = 1/2 (V + M - C + A)
where V is the number of valence electrons on the central atom, M is the number of monovalent atoms bonded to the central atom, C is the charge of the cation, and A is the charge of the anion.
- **Valence Electrons and Monovalent Atoms**
The Fe atom in K4[Fe(CN)6] has a total of 8 valence electrons (Fe = 8), and it is bonded to 6 CN ligands (CN = 6 monovalent atoms).
- **Charge of the Cation and Anion**
In this compound, the cation (K+) has a charge of +4, and the anion (Fe(CN)6)^4- has a charge of -4.
- **Calculating Hybridization**
Using the formula for hybridization, we can calculate:
Hybridization = 1/2 (8 + 6 - 4 + 4) = 7
- **Hybridization Orbitals**
Based on the calculated hybridization value of 7, the Fe atom in K4[Fe(CN)6] will exhibit sp3d3 hybridization. This means that the Fe atom will utilize 3d, 3s, and 3p orbitals to form six hybrid orbitals for bonding with the six CN ligands in an octahedral geometry.
- **Conclusion**
In conclusion, the hybridization orbitals in K4[Fe(CN)6] are sp3d3, involving the utilization of 3d, 3s, and 3p orbitals by the central Fe atom to form six hybrid orbitals for bonding with the surrounding CN ligands.