Explanation:

Introduction:
In the context of hybridization, the empty atomic orbitals of the central atom of a monomer can participate in the formation of hybrid orbitals. However, in certain dimers, the empty atomic orbital of the central atom of a monomer may not be involved in hybridization.

Analysis:
Let's analyze each option:

A) GaH6:
Gallium (Ga) has the electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1. In GaH6, each hydrogen atom donates one electron to form Ga-H bonds. The hybridization of gallium in GaH6 involves the 3s, 3p, and 3d orbitals. Therefore, the empty atomic orbital of the central atom of GaH6 is involved in hybridization.

B) Al2Br6:
Aluminum (Al) has the electronic configuration 1s2 2s2 2p6 3s2 3p1. In Al2Br6, each bromine atom donates one electron to form Al-Br bonds. The hybridization of aluminum in Al2Br6 involves the 3s and 3p orbitals. Therefore, the empty atomic orbital of the central atom of Al2Br6 is involved in hybridization.

C) Be2H4:
Beryllium (Be) has the electronic configuration 1s2 2s2. In Be2H4, each hydrogen atom donates one electron to form Be-H bonds. Beryllium does not undergo hybridization, as it only forms two bonds. Therefore, the empty atomic orbital of the central atom of Be2H4 is not involved in hybridization.

D) Cl2O7:
Chlorine (Cl) has the electronic configuration 1s2 2s2 2p6 3s2 3p5. In Cl2O7, each oxygen atom donates two electrons to form Cl-O bonds. The hybridization of chlorine in Cl2O7 involves the 3p orbitals. Therefore, the empty atomic orbital of the central atom of Cl2O7 is involved in hybridization.

Conclusion:
Based on the analysis, the correct answer is C) Be2H4. In Be2H4, the empty atomic orbital of the central atom (beryllium) does not involve in hybridization.

C) Be2H4