Question:

The hybridized state of $Al^{3+}$ in the complex ion formed when $AlCl_3$ is treated with aqueous acid is

Updated On: Jun 7, 2024
  • $sp^3$
  • $dsp^2$
  • $sp^3d^2$
  • $sp^2d$
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The Correct Option is C

Solution and Explanation

$AlCl _{3}+$ aqueous acid $\rightarrow\left[ Al \left( H _{2} O \right)_{6}\right]^{3+}$ $Al ^{3+}$ ions give a complex ion in which $Al ^{3+}$ ion form six coordinate covalent bonds with six $H _{2} O$ molecules. Hence, it shows $s p^{3} d^{2}$ hybridisation. (Due to availability of vacant $3 d$ -orbitals.) $A l=3 s^{2} 3 p^{1}$ $A l^{3 T}=3 s^{0}$
Four electron pairs denoted by $6H_{2}O$ molecules
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Concepts Used:

Group 13 Elements

Group 13 is commonly known as the Boron Family. The boron family comprises:

  • Boron (B)
  • Aluminum (Al)
  • Gallium (Ga)
  • Indium (In)
  • Thallium (Tl)

Element 113 (Nihonium) gets the name of ununtrium Uut. Each one of the elements has three electrons in the external shell of their nuclear structure is one of the important and mutual properties of the group.

The general electronic configuration for the group 13 elements is ns2 np1.

Density of Boron Family:

Melting point and Boiling Point:

Electronegativity:

Atomic and Ionic Radii:

Compounds of Group 13 Elements:

  • Oxides
  • Halides
  • Borates
  • Boron Hydrides
  • Diborane
  • Borazine