Final Answer:
The hybridization of the Zn atom in the compound (CH₃)₂Zn is sp³.
Step-by-step explanation:
In (CH₃)₂Zn, the zinc atom exhibits a coordination number of 4, forming four sigma bonds with the surrounding atoms. Each methyl group (CH₃) contains three sigma bonds - one sigma bond with carbon and three sigma bonds with hydrogen atoms. Considering each methyl group contributes one sigma bond to the central Zn atom, and there are two methyl groups in the compound, this totals to four sigma bonds around Zn. According to the theory of hybridization, the number of sigma bonds and lone pairs around an atom determines its hybridization state.
In sp³ hybridization, one s orbital and three p orbitals hybridize to form four sp³ hybrid orbitals. These orbitals are directed towards the corners of a tetrahedron, maximizing the bond angles between them. For Zn in (CH₃)₂Zn, its four sp³ hybrid orbitals overlap with the orbitals of carbon and hydrogen atoms, resulting in the formation of four sigma bonds, consistent with the observed coordination number and molecular structure.
The arrangement of the four sp³ hybrid orbitals around the Zn atom in (CH₃)₂Zn allows for the formation of stable sigma bonds with the carbon and hydrogen atoms. This tetrahedral geometry maximizes the separation between electron pairs, minimizing repulsion and stabilizing the molecule. Therefore, the sp³ hybridization of the Zn atom in (CH₃)₂Zn facilitates the formation of these bonds and influences the compound's overall structure and stability.