Question

Aluminium chloride in acidified aqueous solution gives a complex ion 'X'. The hybridization of central atom in X is

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

Hint:

Al$^{3+}$ in aqueous solution forms the hydrated complex ion [Al(H$_2$O)$_6$]$^{3+}$.
H$_2$O acts as a ligand, donating electron pairs.
Coordination number of Al is 6, leading to an octahedral geometry.
For main group elements like Al forming octahedral complexes, the hybridization is typically $sp^3d^2$, utilizing $s, p,$ and outer $d$ orbitals (or $d$ orbitals of the same principal quantum shell if available and empty).
Al (Period 3) uses its $3s, 3p,$ and $3d$ orbitals for this hybridization.

Answer 1

The Correct Option is D

Aluminium chloride (AlCl$_3$) is a Lewis acid. In aqueous solution, it reacts with water. The Al$^{3+}$ ion is small and highly charged, leading to strong interactions with water molecules (hydration). In aqueous solution, aluminium ions typically exist as hydrated ions. The most common hydrated aluminium ion is hexaaquaaluminium(III), [Al(H$_2$O)$_6$]$^{3+}$. Water molecules (H$_2$O) act as Lewis bases (ligands), donating a pair of electrons to the Al$^{3+}$ ion to form coordinate covalent bonds. The complex ion 'X' formed is [Al(H$_2$O)$_6$]$^{3+}$. In this complex ion: The central atom is Aluminium (Al). The ligands are six water molecules (H$_2$O). The coordination number of Al is 6. The geometry of a 6-coordinate complex is typically octahedral. To determine the hybridization of the central Al atom in [Al(H$_2$O)$_6$]$^{3+}$: Electronic configuration of Al (atomic number 13): $1s^2 2s^2 2p^6 3s^2 3p^1$. Electronic configuration of Al$^{3+}$ ion: $1s^2 2s^2 2p^6$ (or [Ne]). For bonding with 6 H$_2$O ligands in an octahedral geometry, the Al$^{3+}$ ion needs 6 empty orbitals of suitable energy to accept electron pairs from the ligands. These orbitals are formed by hybridization. The available empty orbitals in Al$^{3+}$ for hybridization (from $n=3$ shell onwards) are $3s$, $3p_x, 3p_y, 3p_z$, and $3d$ orbitals ($3d_{x^2-y^2}, 3d_{z^2}$ for octahedral). For octahedral geometry (coordination number 6), the hybridization can be either $sp^3d^2$ (using outer $d$ orbitals) or $d^2sp^3$ (using inner $d$ orbitals). Since Al is a third-period element, its $n=3$ shell contains $3s, 3p,$ and $3d$ orbitals. Al$^{3+}$ has lost its $3s$ and $3p$ electrons. For bonding, it uses its empty $3s$, three $3p$, and two $3d$ orbitals. These are all from the same principal quantum shell ($n=3$). The hybridization for an octahedral complex involving $s, p,$ and $d$ orbitals of the same shell (or outer $d$) is $sp^3d^2$. Specifically, Al uses one $3s$, three $3p$, and two $3d$ orbitals ($3d_{x^2-y^2}$ and $3d_{z^2}$) to form six $sp^3d^2$ hybrid orbitals. These hybrid orbitals overlap with orbitals from the six H$_2$O ligands. Therefore, the hybridization of the central Al atom in [Al(H$_2$O)$_6$]$^{3+}$ is $sp^3d^2$. The acidified aqueous solution ensures that Al(OH)$_3$ does not precipitate; the Al$^{3+}$ remains in solution primarily as the hexaaqua complex. \[ \boxed{\text{sp}^3\text{d}^2} \]