To determine hybridization in coordination compounds:
1. Find the oxidation state of the central metal.
2. Write the electronic configuration of the metal ion.
3. Identify the nature of the ligand (strong field or weak field) to determine if pairing occurs.
4. Determine the number of vacant orbitals required for the coordination number.
5. Assign the hybridization based on the orbitals used (e.g., \( \text{sp}^3 \), \( \text{dsp}^2 \), \( \text{sp}^3\text{d}^2 \), \( \text{d}^2\text{sp}^3 \)).
Remember that strong field ligands (like \( \text{CN}^- \), \( \text{CO} \), \( \text{NH}_3 \) for \( \text{Co}^{3+} \)) cause pairing, leading to inner orbital complexes, while weak field ligands (like \( \text{F}^- \), \( \text{Cl}^- \), \( \text{H}_2\text{O} \)) do not, leading to outer orbital complexes.
To determine the hybridization of the central metal ion in each complex, we need to consider the oxidation state of the metal, its d-electron configuration, and the nature of the ligands (strong field or weak field).
A) \( \text{[CoF}_6\text{]}^{3-} \)
\begin{itemize}
\item Central metal: Co
\item Oxidation state: Let oxidation state of Co be $x$. $x + 6(-1) = -3 \Rightarrow x = +3$. So, \( \text{Co}^{3+} \).
\item Electronic configuration of \( \text{Co} \): \( \text{[Ar]} 3\text{d}^7 4\text{s}^2 \)
\item Electronic configuration of \( \text{Co}^{3+} \): \( \text{[Ar]} 3\text{d}^6 \)
\item Ligand: \( \text{F}^- \) (Fluoride) is a weak field ligand.
\item Since \( \text{F}^- \) is a weak field ligand, it does not cause pairing of electrons in the 3d orbitals.
\item The 6 ligands will occupy 4s, 4p, and 4d orbitals.
\item Hybridization: \( \text{sp}^3\text{d}^2 \) (outer orbital complex)
\item This matches with IV) \( \text{sp}^3\text{d}^2 \).
\end{itemize}
B) \( \text{[NiCl}_4\text{]}^{2-} \)
\begin{itemize}
\item Central metal: Ni
\item Oxidation state: Let oxidation state of Ni be $x$. $x + 4(-1) = -2 \Rightarrow x = +2$. So, \( \text{Ni}^{2+} \).
\item Electronic configuration of \( \text{Ni} \): \( \text{[Ar]} 3\text{d}^8 4\text{s}^2 \)
\item Electronic configuration of \( \text{Ni}^{2+} \): \( \text{[Ar]} 3\text{d}^8 \)
\item Ligand: \( \text{Cl}^- \) (Chloride) is a weak field ligand.
\item Since \( \text{Cl}^- \) is a weak field ligand, it does not cause pairing of electrons.
\item For a coordination number of 4, with weak field ligands, tetrahedral geometry and \( \text{sp}^3 \) hybridization are common.
\item Available orbitals for hybridization: 3d (full), 4s, 4p.
\item Hybridization: \( \text{sp}^3 \)
\item This matches with III) \( \text{sp}^3 \).
\end{itemize}
C) \( \text{[Ni(CN)_4\text{]}^{2-} \)}
\begin{itemize}
\item Central metal: Ni
\item Oxidation state: Let oxidation state of Ni be $x$. $x + 4(-1) = -2 \Rightarrow x = +2$. So, \( \text{Ni}^{2+} \).
\item Electronic configuration of \( \text{Ni} \): \( \text{[Ar]} 3\text{d}^8 4\text{s}^2 \)
\item Electronic configuration of \( \text{Ni}^{2+} \): \( \text{[Ar]} 3\text{d}^8 \)
\item Ligand: \( \text{CN}^- \) (Cyanide) is a strong field ligand.
\item Since \( \text{CN}^- \) is a strong field ligand, it causes pairing of electrons in the 3d orbitals. For \( \text{3d}^8 \), the two unpaired electrons will pair up, leaving one 3d orbital vacant.
\item Available orbitals for hybridization: one 3d, 4s, two 4p.
\item Hybridization: \( \text{dsp}^2 \) (inner orbital complex, square planar geometry)
\item This matches with I) \( \text{dsp}^2 \).
\end{itemize}
D) \( \text{[Co(NH}_3)_6\text{]^{3+} \)}
\begin{itemize}
\item Central metal: Co
\item Oxidation state: Let oxidation state of Co be $x$. $x + 6(0) = +3 \Rightarrow x = +3$. So, \( \text{Co}^{3+} \).
\item Electronic configuration of \( \text{Co} \): \( \text{[Ar]} 3\text{d}^7 4\text{s}^2 \)
\item Electronic configuration of \( \text{Co}^{3+} \): \( \text{[Ar]} 3\text{d}^6 \)
\item Ligand: \( \text{NH}_3 \) (Ammonia) is a strong field ligand for \( \text{Co}^{3+} \).
\item Since \( \text{NH}_3 \) is a strong field ligand, it causes pairing of electrons in the 3d orbitals. For \( \text{3d}^6 \), the electrons will pair up, leaving two 3d orbitals vacant.
\item Available orbitals for hybridization: two 3d, 4s, three 4p.
\item Hybridization: \( \text{d}^2\text{sp}^3 \) (inner orbital complex, octahedral geometry)
\item This matches with II) \( \text{d}^2\text{sp}^3 \).
\end{itemize}
Step 6: Form the Correct Match
A - IV
B - III
C - I
D - II Step 7: Analyze Options
\begin{itemize}
\item Option (1): A-IV, B-I, C-III, D-II. Incorrect.
\item Option (2): A-II, B-I, C-III, D-IV. Incorrect.
\item Option (3): A-IV, B-III, C-I, D-II. Correct, as it matches our derived hybridizations.
\item Option (4): A-II, B-III, C-I, D-IV. Incorrect.
\end{itemize}
