$\mathrm{AlCl}_3$ is an electron deficient compound but $\mathrm{AlF}_3$ is not. This is because

$\mathrm{AlCl}_3$ is an electron deficient compound but $\mathrm{AlF}_3$ is not. This is because
  1. atomic size of $\mathrm{F}$ is smaller than $\mathrm{Cl}$ which makes $\mathrm{AlF}_3$ more covalent
  2. $\mathrm{AlCl}_3$ is a covalent compound while $\mathrm{AlF}_3$ is an ionic compound
  3. $\mathrm{AlCl}_3$ exists as dimer but $\mathrm{AlF}_3$ does not
  4. $\mathrm{Al}$ in $\mathrm{AlCl}_3$ is in $s p^3$ hybrid state but $\mathrm{Al}$ in $\mathrm{AlF}_3$ is in $s p^2$ hybrid state

Solution

According to Fajan's rule, larger is the size of anion, more is the covalent character. So, $\mathrm{AlCl}_3$ is a covalent, while $\mathrm{AlF}_3$ is ionic, so $\mathrm{AlCl}_3$ is electron deficient but $\mathrm{AlF}_3$ is not.

Asked in: AP EAMCET 2021 (23 Aug Shift 2)

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