Question

The hybridization of metals involved in the following complexes, respectively are [Mn(CN)$_6$]$^{3-}$, [CoF$_6$]$^{3-}$

• A. sp$^3$d$^2$, sp$^3$d$^2$
• B. sp$^3$d$^2$, d$^2$sp$^3$
• C. d$^2$sp$^3$, d$^2$sp$^3$
• D. d$^2$sp$^3$, sp$^3$d$^2$

Hint:

Determine oxidation state of central metal and its $d$-electron count.
Identify ligands as strong field or weak field.
Strong field ligands (e.g., CN$^-$, CO, NH$_3$, en) cause electron pairing $\rightarrow$ low spin complex $\rightarrow$ may use inner $(n-1)d$ orbitals for $d^2sp^3$ hybridization if available.
Weak field ligands (e.g., F$^-$, Cl$^-$, H$_2$O) cause less/no pairing $\rightarrow$ high spin complex $\rightarrow$ may use outer $nd$ orbitals for $sp^3d^2$ hybridization.
Mn$^{3+}$ is $3d^4$. CN$^-$ (strong field) $\rightarrow$ low spin $(t_{2g})^4 (e_g)^0$. Empty $e_g$ ($3d$) orbitals available for $d^2sp^3$.
Co$^{3+}$ is $3d^6$. F$^-$ (weak field) $\rightarrow$ high spin $(t_{2g})^4 (e_g)^2$. Inner $e_g$ ($3d$) orbitals are occupied. Uses outer $4d$ orbitals for $sp^3d^2$.

Answer 1

The Correct Option is D

Both complexes have a coordination number of 6, suggesting an octahedral geometry. For octahedral geometry, the hybridization can be $d^2sp^3$ (inner orbital complex) or $sp^3d^2$ (outer orbital complex). This depends on whether inner ($n-1)d$ orbitals or outer $nd$ orbitals are used for hybridization. Ligand strength plays a key role. 1. [Mn(CN)$_6$]$^{3-$:} \begin{itemize} \item Central metal ion: Manganese (Mn). \item Oxidation state of Mn: Let it be $x$. CN$^-$ is a ligand with -1 charge. $x + 6(-1) = -3 \Rightarrow x - 6 = -3 \Rightarrow x = +3$. So, Mn is in the +3 oxidation state (Mn$^{3+}$). \item Electronic configuration of Mn (Z=25): [Ar] $3d^5 4s^2$. \item Electronic configuration of Mn$^{3+}$: [Ar] $3d^4$ (loses two $4s$ and one $3d$ electron). \item Ligand: CN$^-$ (cyanide ion). CN$^-$ is a strong field ligand. Strong field ligands cause pairing of electrons in the $d$ orbitals if possible, leading to low spin complexes and often using inner $d$ orbitals for hybridization ($d^2sp^3$). \item For Mn$^{3+}$ ($3d^4$) in an octahedral field with strong ligands (CN$^-$): The four $3d$ electrons will occupy the $t_{2g}$ orbitals. Pairing will occur. $t_{2g}$ orbitals can hold up to 6 electrons. $d^4$ low spin: $(t_{2g})^4 (e_g)^0$. Three electrons in separate $t_{2g}$ orbitals, then the fourth pairs up in one of them. Or if pairing energy is less than $\Delta_o$, then configuration could be two paired and two empty $t_{2g}$ if using for $d^2sp^3$. For $d^2sp^3$ hybridization, two $(n-1)d$ orbitals (here, $3d$), one $ns$ (here, $4s$), and three $np$ (here, $4p$) orbitals are used. The $3d$ orbitals split into $t_{2g}$ (lower energy) and $e_g$ (higher energy) sets in an octahedral field. For Mn$^{3+}$ ($3d^4$) with a strong field ligand (CN$^-$), it will be a low spin complex. The electrons fill $t_{2g}$ first: $t_{2g}^4 e_g^0$. For example, $(d_{xy})^2 (d_{yz})^1 (d_{zx})^1$. The two $e_g$ orbitals ($d_{z^2}, d_{x^2-y^2}$) are empty and available from the $(n-1)d$ set. These can be used for $d^2sp^3$ hybridization if they are the correct ones. Actually, for $d^2sp^3$, the specific $d$ orbitals used are $d_{x^2-y^2}$ and $d_{z^2}$ (which are the $e_g$ set). In Mn$^{3+}$ ($3d^4$), the $d$ electrons are arranged in $t_{2g}$ and $e_g$ orbitals. Low spin $d^4$: $(t_{2g})^4 (e_g)^0$. Here, the two $e_g$ orbitals are empty. So, these two $3d$ orbitals ($d_{x^2-y^2}, d_{z^2}$) can participate in $d^2sp^3$ hybridization along with $4s$ and $4p$ orbitals. Thus, hybridization for [Mn(CN)$_6$]$^{3-}$ is $d^2sp^3$ (inner orbital complex). \end{itemize} 2. [CoF$_6$]$^{3-$:} \begin{itemize} \item Central metal ion: Cobalt (Co). \item Oxidation state of Co: Let it be $y$. F$^-$ is a ligand with -1 charge. $y + 6(-1) = -3 \Rightarrow y - 6 = -3 \Rightarrow y = +3$. So, Co is in the +3 oxidation state (Co$^{3+}$). \item Electronic configuration of Co (Z=27): [Ar] $3d^7 4s^2$. \item Electronic configuration of Co$^{3+}$: [Ar] $3d^6$ (loses two $4s$ and one $3d$ electron). \item Ligand: F$^-$ (fluoride ion). F$^-$ is a weak field ligand. Weak field ligands generally do not cause pairing of electrons in the $d$ orbitals, leading to high spin complexes and often using outer $d$ orbitals for hybridization ($sp^3d^2$). \item For Co$^{3+}$ ($3d^6$) in an octahedral field with weak ligands (F$^-$), it will be a high spin complex. The six $3d$ electrons are distributed as $(t_{2g})^4 (e_g)^2$ according to Hund's rule (fill $t_{2g}$ with 3, then $e_g$ with 2, then pair in $t_{2g}$ for the 6th electron). $t_{2g}: d_{xy}^1 d_{yz}^1 d_{zx}^1 \rightarrow d_{xy}^2 d_{yz}^1 d_{zx}^1$ (one pairing in $t_{2g}$) $e_g: d_{z^2}^1 d_{x^2-y^2}^1$ Total $3d^6$ high spin: $(t_{2g})^4 (e_g)^2$. Both $e_g$ orbitals ($d_{z^2}, d_{x^2-y^2}$) from the $3d$ set are occupied by single electrons. They are not empty to participate in $d^2sp^3$ hybridization. Therefore, the complex uses outer $4d$ orbitals for hybridization. The hybridization will be $sp^3d^2$, using one $4s$, three $4p$, and two $4d$ orbitals. Thus, hybridization for [CoF$_6$]$^{3-}$ is $sp^3d^2$ (outer orbital complex). \end{itemize} So, the hybridizations are $d^2sp^3$ for [Mn(CN)$_6$]$^{3-}$ and $sp^3d^2$ for [CoF$_6$]$^{3-}$. This matches option (d). \[ \boxed{d^2sp^3, sp^3d^2} \]