Transition Elements: Question 8

Worked solution

Step 1: Determine the formula and overall charge

Chromium is present as $\text{Cr}^{3+}$ (chromium(III)), directly bonded to two chloride ligands and four water ligands, giving the formula $[\text{CrCl}_2(\text{H}_2\text{O})_4]^{n}$, where $n$ is found by summing the charges of the metal ion and its ligands.

$\text{Cl}^-$ is a $1-$ ligand, so the two chloride ligands contribute $2\times(-1)=-2$. $\text{H}_2\text{O}$ is a neutral ligand, contributing $0$ regardless of how many are attached. So: $\text{overall charge} = (+3) + 2\times(-1) + 4\times0 = +3-2 = +1$

The complex ion is therefore $[\text{CrCl}_2(\text{H}_2\text{O})_4]^+$.

Step 2: Determine the coordination number and shape

Chromium forms six coordinate bonds in total. Two to the chloride ligands and four to the water ligands (the two types of ligand present do not determine the coordination number; every individual donor atom counts). So the coordination number is 6. A coordination number of 6 corresponds to an octahedral shape.

Step 3: Evaluate each option

• A: correctly gives the formula $[\text{CrCl}_2(\text{H}_2\text{O})_4]^+$, coordination number 6, and octahedral shape. This matches the reasoning above.
• B: incorrect charge sign. Combining $\text{Cr}^{3+}$ with two $\text{Cl}^-$ ligands gives an overall charge of $1+$, not $1-$; this option has made a sign error.
• C: correct formula and shape, but the coordination number is wrong. 2 is the number of distinct ligand types (chloride and water), not the total number of coordinate bonds (6).
• D: correct formula and coordination number, but the shape is wrong. A 6-coordinate complex is octahedral, not tetrahedral (tetrahedral geometry corresponds to a coordination number of 4).

Final answer

• The complex ion is $[\text{CrCl}_2(\text{H}_2\text{O})_4]^+$, with coordination number 6 and an octahedral shape, option A.