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Question

The ease of formation of the adduct, $NH_3 \cdot BX_3$ (where, $X = F, Cl, Br$) follows the order

The correct answer is
$BF_3 < BCl_3 < BBr_3$

Understanding Adduct Formation: $NH_3 \cdot BX_3$

The formation of an adduct between a Lewis acid ($BX_3$) and a Lewis base ($NH_3$) depends on the Lewis acidity of the boron halide ($BX_3$). The ease of formation indicates the stability of the adduct.

Lewis Acidity and Back-Bonding in $BX_3$

Boron trifluoride ($BF_3$), boron trichloride ($BCl_3$), and boron tribromide ($BBr_3$) act as Lewis acids due to the electron-deficient boron atom. The strength of their Lewis acidity is influenced by:

  • Electronegativity: The electronegativity order is $F > Cl > Br$. Higher electronegativity of the halogen withdraws electron density from boron, increasing its Lewis acidity.
  • Back-bonding ($\pi$-bonding): Halogens possess lone pairs that can overlap with the empty p-orbital of boron, forming a $\pi$-bond. This 'back-bonding' reduces the electron deficiency of boron, thereby decreasing its Lewis acidity. The effectiveness of back-bonding depends on the orbital overlap.

The overlap efficiency follows the order: 2p(B)-2p(F) > 2p(B)-3p(Cl) > 2p(B)-4p(Br). Therefore, back-bonding is strongest in $BF_3$ and weakest in $BBr_3$.

Order of Lewis Acidity and Adduct Formation

The extent of back-bonding decreases the Lewis acidity. Consequently, the Lewis acidity order is:

$ BBr_3 > BCl_3 > BF_3 $

Since $NH_3$ is a Lewis base, the ease of adduct formation ($NH_3 \cdot BX_3$) directly corresponds to the Lewis acidity of $BX_3$. A stronger Lewis acid forms the adduct more readily.

Thus, the ease of formation follows the order:

$ BBr_3 > BCl_3 > BF_3 $

This can also be written as:

$ BF_3 < BCl_3 < BBr_3 $

Conclusion

The ease of formation of the adduct $NH_3 \cdot BX_3$ follows the order $BF_3 < BCl_3 < BBr_3$, aligning with the increasing Lewis acidity of the boron halides from $BF_3$ to $BBr_3$ due to reduced back-bonding.

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Important Questions from p Block Elements

  1. Borax on treatment with NaOH and $H_2O_2$ forms X. The compound X on reaction with PhCN at $60$ °C in methanol-water mixture gives Y as the major product. 
    X and Y, respectively, are

  2. The reaction of $P_2O_5$ with $HNO_3$ and $HClO_4$, respectively, gives
  3. Ammonolysis of $S_2Cl_2$ in an inert solvent gives
  4. In cyclophosphazenes, $(NPX_2)_3$ (X = F, Cl, Br and Me), the strength of P-N $\pi$-bond varies with X in the order
  5. Among the following, the group of molecules that undergoes rapid hydrolysis is
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