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Exercises · 11.4

Q.Consider the compounds, BCl3 and CCl4. How will they behave with water? Justify.

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Step 1 - BCl3 hydrolyses

Boron is electron deficient (only 6 electrons, empty 2p2p orbital). A water molecule's oxygen lone pair can attack this vacant orbital, forming an intermediate adduct that rapidly loses HCl in successive steps:

BCl3+3H2O→B(OH)3+3HCl\text{BCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{B(OH)}_3 + 3\text{HCl}

This hydrolysis is fast and essentially complete.

Step 2 - CCl4 does not hydrolyse

Carbon in CCl4\text{CCl}_4 is sp3sp^3 hybridised with a complete octet (4 bonds, no lone pair, no empty orbital) and, being a period-2 element, has no accessible d orbitals to expand its coordination number beyond 4. A water molecule therefore has no vacant orbital on carbon to attack, and there is also steric shielding by the four Cl atoms - so CCl4\text{CCl}_4 does not hydrolyse:

CCl4+H2O→No reaction\text{CCl}_4 + \text{H}_2\text{O} \rightarrow \text{No reaction}

Step 3 - Contrast with SiCl4 …

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