Question

Which of the following complexes has unpaired electrons?

• A. [Fe(CN)$_{6}$]$^{4-}$
• B. [Co(NH$_{3}$)$_{6}$]$^{3+}$
• C. [Mn(H$_{2}$O)$_{6}$]$^{2+}$
• D. [Zn(CN)$_{4}$]$^{2-}$

Hint:

Remember the general spectrochemical series: $I^-<Br^-<SCN^-<Cl^-<F^-<OH^-<H_2O<NH_3<en<CN^-<CO$. Ligands on the right are strong-field, and those on the left are weak-field. Also, any complex with a $d^{10}$ central ion (like Zn$^{2+}$, Cd$^{2+}$) will always be diamagnetic.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to determine which of the given coordination complexes is paramagnetic, meaning it has one or more unpaired electrons in its d-orbitals. This requires analyzing the oxidation state of the central metal ion, its d-electron count, and the effect of the ligand (strong-field or weak-field) on electron pairing.
Step 2: Key Formula or Approach:
1. Determine the oxidation state of the central metal ion.
2. Write the electronic configuration of the metal ion (d-electron count).
3. Identify the ligand as strong-field (causes pairing, low-spin complex) or weak-field (no pairing, high-spin complex) based on the spectrochemical series.
4. Fill the d-orbitals according to Crystal Field Theory (splitting into $t_{2g}$ and $e_g$ levels for octahedral complexes) and check for unpaired electrons.
Step 3: Detailed Explanation:
\begin{itemize} \item (A) [Fe(CN)$_{6}$]$^{4-$}:
Let the oxidation state of Fe be x. $x + 6(-1) = -4 \implies x = +2$.
Fe(II) has an electronic configuration of [Ar] $3d^6$.
CN$^{-}$ is a strong-field ligand, causing electron pairing. In the octahedral field, the six electrons will fill the lower energy $t_{2g}$ orbitals first.
Configuration: $t_{2g}^{6}e_g^{0}$. All electrons are paired. Number of unpaired electrons = 0. (Diamagnetic)
\item (B) [Co(NH$_{3}$)$_{6}$]$^{3+$}:
Let the oxidation state of Co be x. $x + 6(0) = +3 \implies x = +3$.
Co(III) has an electronic configuration of [Ar] $3d^6$.
NH$_{3}$ is a strong-field ligand, especially with Co$^{3+}$, causing electron pairing.
Configuration: $t_{2g}^{6}e_g^{0}$. All electrons are paired. Number of unpaired electrons = 0. (Diamagnetic)
\item (C) [Mn(H$_{2}$O)$_{6}$]$^{2+$}:
Let the oxidation state of Mn be x. $x + 6(0) = +2 \implies x = +2$.
Mn(II) has an electronic configuration of [Ar] $3d^5$.
H$_{2}$O is a weak-field ligand, so it does not cause pairing (high-spin complex). The electrons will occupy all d-orbitals singly before pairing up.
Configuration: $t_{2g}^{3}e_g^{2}$. All five electrons are unpaired. Number of unpaired electrons = 5. (Paramagnetic)
\item (D) [Zn(CN)$_{4}$]$^{2-$}:
Let the oxidation state of Zn be x. $x + 4(-1) = -2 \implies x = +2$.
Zn(II) has an electronic configuration of [Ar] $3d^{10}$.
The d-subshell is completely filled. Regardless of the ligand or geometry, all electrons are paired. Number of unpaired electrons = 0. (Diamagnetic)
\end{itemize} Step 4: Final Answer:
The complex [Mn(H$_{2}$O)$_{6}$]$^{2+}$ has unpaired electrons.