MCQ
Aluminium $(III)$ chloride forms a dimer because
- ✓Higher coordination number can be achieved by aluminium
- BAluminium has high ionization energy
- CAluminium belongs to $III$ group
- DIt cannot form a trimer
✓
Answer
Correct option: A.
Higher coordination number can be achieved by aluminium
a
$AlCl _3$ can form dimer and exists as $Al _2 Cl _6$, aluminium has vacant $d$-orbitals which can accommodate electron from chlorine atom. $AlCl _3$ is an electron deficient compound in $Al$ (octet incomplete) thus behaves as Lewis acid and $Al$ completes it by taking electron pair from $Cl$-atom as shown in figure.
$AlCl _3$ can form dimer and exists as $Al _2 Cl _6$, aluminium has vacant $d$-orbitals which can accommodate electron from chlorine atom. $AlCl _3$ is an electron deficient compound in $Al$ (octet incomplete) thus behaves as Lewis acid and $Al$ completes it by taking electron pair from $Cl$-atom as shown in figure.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A reaction, $A+B \rightarrow C+D+q$ is found to have a positive entropy change. The reaction will be
→Find out stability order of following compounds
Given that
→$\Lambda_{ m }^{\alpha}=133.4\,\left( AgNO _{3}\right) ; \Lambda_{ m }^{\alpha}=149.9( KCl )$
$\Lambda_{ m }^{\alpha}=144.9\, S\, cm ^{2} \,mol ^{-1}\left( KNO _{3}\right)$
the molar conductivity at infinite dilution for $AgCl$ is $.......\, S \,cm ^{2}\, mol ^{-1}$
The correct order of the ionic radii of
→$O ^{2-}, N ^{3-}, F ^{-}, Mg ^{2+}, Na ^{+}$ and $Al ^{3+}$ is
Which of the following alcohols cannot be prepared from hydration of an alkene?
$2-$ chloropentane on halogenation with chlorine gives $2,3,$ dichloropentane. What will be the structure of free radical species form in the reaction ?
→Acetoacetic ester behaves as
The number of oxygen atoms bonded to one phosphorous atom in $P_4O_6$ is
→For the allotropic change represented by equation $C(diamond) \to C(graphite)$; the enthalpy change is $\Delta H = - 1.89\,kJ$. If $6\,g$ of diamond and $6\,g$ of graphite are separately burnt to yield carbon dioxide, the heat liberated in the first case is
→In the final answer of the expression $\frac{{(29.2 - 20.2)\,(1.79 \times {{10}^5})}}{{1.37}}$. The number of significant figures is
→