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Chapter 30: Problem 4

Discuss the structure of \(\mathrm{Re}_{2} \mathrm{Cl}_{8}\) and justify for its bond order using molecular orbital theory.

Short Answer

Expert verified
The compound \(\mathrm{Re}_{2} \mathrm{Cl}_{8}\) has a bond order of 6.

Step by step solution

01

Identify the Structure

The compound \(\mathrm{Re}_{2} \mathrm{Cl}_{8}\) has a cubic structure of chlorine atoms around rhenium metal atoms. Each Re atom is bonded to 4 Cl atoms.
02

Apply Molecular Orbital Theory

Molecular Orbital theory suggests that atomic orbitals merge to form molecular orbitals, which can be filled with electrons. For \(\mathrm{Re}_{2} \mathrm{Cl}_{8}\), since there are two Re atoms, the orbitals will be doubled. There are 5 atomic orbitals for Re d-orbitals so it will generate 10 molecular orbitals.
03

Calculate Bond Order

The bond order can be calculated using the formula: Bond Order = (No. of electrons in bonding orbitals - No. of electrons in anti-bonding orbitals)/2 . For \(\mathrm{Re}_{2} \mathrm{Cl}_{8}\), all 12 electrons from the \(5d\) orbital of each Re atom are in bonding orbitals with no electron in antibonding orbitals. So, the bond order = \(12-0)/2 = 6.

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