NEET · Chemistry · STD 12 - p -Block elements - ll
Match the compounds given in \(column\, I\) with the hybridisation and shape given in \(column\, II\) and mark the correct option. \(\begin{array}{|l|l|} \hline Column\,\,I & Column\,\,II \\ \hline (A)\,XeF_6 & (i)\,\,distortedoctahedral \\ \hline (B)\,XeO_3 & (ii)\,\,square\,\,planar \\ \hline (C)\,XeOF_4 & (iii)\,\,pyramidal \\ \hline (D)\,XeF_4 & (iv)\,\,square\,\,pyramidal \\ \hline \end{array}\)
- A \(A-(iv),B-(iii),C-(i),D-(ii)\)
- B \(A-(iv),B-(i),C-(ii),D-(iii)\)
- C \(A-(i),B-(iii),C-(iv),D-(ii)\)
- D \(A-(i),B-(ii),C-(iv),D-(iii)\)
Answer & Solution
Correct Answer
(C) \(A-(i),B-(iii),C-(iv),D-(ii)\)
Step-by-step Solution
Detailed explanation
In \(XeF _6\), the central \(Xe\) atom undergoes \(sp ^3 d ^3\) hybridisation and the molecular geometry is distorted octahedral with \(1\) lone pair and \(6\) bond pairs of electrons. In \(XeO _3\), the central \(Xe\) atom undergoes \(sp ^3\) hybridisation and the molecular geometry is pyramidal with \(1\) lone pair and \(3\) bonding domains of electrons.
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