Question

Account for the following observations:

A $AlC{l}_3$ is Lewis acid.

B Though Fluorine is more electronegative than chlorine yet $B{F}_3$ is a weaker Lewis acid than $BC{l}_3$.

C $Pb{O}_2$ is stronger oxidising agent than $Sn{O}_2$

D The $+1$ oxidation state of thallium is more stable than the $+3$ state.

Answer 1

A. Aluminium has $3$ electrons in its valence shell $\left[Ne\right]3s^{2}3p^1$. In $AlC{l}_3$, the central atom $Al$ has only $6$ electrons. So, it is electron deficient and therefore it has a tendency to accept the lone pair of electrons to achieve stable electronic configuration and thus, behave as a Lewis acid




B. The boron atom in $B{F}_3$ is $s{p}^2$ hybridized, with a vacant $2p$ orbital. Fluorine contains $3$ lone pairs of electron with $2$ lone pairs on $2p$ orbital. These lone pairs on $F$ can overlap with the empty $2p$ orbital on $B$, thereby forming a coordinate bond.

This increases the electron density on the Boron atom and hence the Boron atom does not tend to accept lone pair and thus behave as less acidic. This donation of lone pair of electrons by fluorine is called back bonding.


So, $B{F}_3$ is an electron deficient species, it tries to complete the octet of Boron atom by forming $p\pi -p\pi$ back bonding with fluorine.

While, In $BC{l}_3$, the $3p$ orbitals of $Cl$ are bigger than the $2p$ orbital on $B$, so orbital overlap is less efficient, and back bonding is not that effective

Hence, stronger back bonding in $B{F}_3$ makes it a weaker Lewis acid, even though fluorine is more electronegative than chlorine.


$\text{C. Oxidising agent}$

• Oxidising agent oxidises the other species by itself getting reduced.
• In $Pb{O}_2$, $Pb$ is in $+4$ oxidation state.
• Due to the inert pair effect in $Pb$, the stability of $+4$ oxidation is less and $Pb$ forms more stable compounds in $+2$ state.
• On the other hand $Sn$ is more stable in $+4$ state than $+2$ oxidation state.
• In $SnO$, $Sn$ is in $+2$ oxidation state, so it tries to oxidize itself from $+2$ to $+4$ and thus acts as a good reducing agent.
• Therefore, one can say that $Pb{O}_2$ is a stronger oxidising agent than $Sn{O}_2$.

$\text{D. Stability of Oxidation State of Thallium}$

• Thallium shows both Oxidation states of $+1$ and $+3$.
• But due to the inert pair effect, $Tl$ forms more stable compound in $+1$ state.
• Due to poor shielding effect of $4f$ and $5d$ electrons in $Tl$ atom, effective nuclear charge increases on valence electrons and its effect is more upon $6s$ valence electrons as it has uniform shape around the nucleus and less effect on $6p$ electrons due to its diffused dumbbell shape.
• This high attraction of nucleus on $6s$ electrons makes them unavailable for bonding and only $6p$ electrons are available to form bond in $Tl$ atom.
• Thus, $Tl$ tries to remain in $+1$ oxidation state via losing only $2p$ electron. Hence, $+1$ oxidation state is more stable than $+3$ oxidation state due to the inert pair effect.