In beryllium chloride (BeCl_2), how many unhybridized p orbitals remain on the central beryllium atom after hybridization?

• A. 0
• B. 3
• C. 2
• D. 1

Answer: (C)

Verified step by step guidance

1. Identify the electron configuration of the central atom, beryllium (Be). The ground state electron configuration of Be is \$1s^2 2s^2$.
2. Determine the valence orbitals involved in bonding. For Be, the valence shell is the second shell, which includes the \$2s\( and \)2p$ orbitals.
3. Consider the hybridization that occurs in BeCl\(_2\). Since Be forms two bonds with Cl atoms and has a linear geometry, it undergoes \(sp\) hybridization, which mixes one \$2s\( and one \)2p\( orbital to form two equivalent \)sp$ hybrid orbitals.
4. Count the number of unhybridized \(p\) orbitals remaining after hybridization. Since one \$2p\( orbital is used in hybridization, the remaining \)2p$ orbitals that are not involved remain unhybridized.
5. Conclude that after \(sp\) hybridization, there are two unhybridized \(p\) orbitals left on the central Be atom.