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AP EAMCET · Chemistry · Chemical Bonding and Molecular Structure

\(\mathrm{XeF}_4\) is square planar where as \(\mathrm{CCl}_4\) is tetrahedral because

  1. A in \(\mathrm{XeF}_4\), ' \(\mathrm{Xe}\) ' is \(s p^2\) hybridised and in \(\mathrm{CCl}_4\) ' \(\mathrm{C}\) ' is \(s p^3\) hybridised
  2. B in both \(\mathrm{XeF}_4\) and \(\mathrm{CCl}_4\) the central atom is \(s p^3\) hybridised
  3. C in \(\mathrm{XeF}_4\), ' \(\mathrm{Xe}\) ' is \(s p^3 d^2\) hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in \(\mathrm{CCl}_4\) 'C' is \(s p^3\) hybridised
  4. D \(X e\) is noble gas, whereas \(C\) is a non-metal
Verified Solution

Answer & Solution

Correct Answer

(C) in \(\mathrm{XeF}_4\), ' \(\mathrm{Xe}\) ' is \(s p^3 d^2\) hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in \(\mathrm{CCl}_4\) 'C' is \(s p^3\) hybridised

Step-by-step Solution

Detailed explanation

\(\mathrm{XeF}_4\) is \(s p^3 d^2\) hybridised due to \(4 \sigma+2 l p\) and shape is square planar. \(\mathrm{CCl}_4\) is \(s p^3\) hybridised due to \(4 \sigma\)-bond.