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TS EAMCET · Chemistry · Chemical Bonding and Molecular Structure

\(\begin{aligned} & \mathrm{Xe}(\mathrm{g})+\mathrm{F}_2(\mathrm{~g}) \frac{573 \mathrm{~K}}{60-70 \mathrm{bar}} \mathrm{X}(\mathrm{s}) \ & (1: 20)\end{aligned}\) \(\begin{aligned} & \mathrm{X}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Y}+\mathrm{HF} \ & \mathrm{X}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{Z}+\mathrm{HF}\end{aligned}\) The correct statements regarding \(Y\) and \(Z\) are \(\mathrm{Y}\) has square pyramidal geometry \(\mathrm{Y}\) has linear geometry \(\mathrm{Z}\) has \(3 \sigma, 3 \pi\) bonds and 1 lone pair of electrons on the central atóm \(\mathrm{Z}\) has tetrahedral geometry

  1. A I & III only
  2. B II & III only
  3. C III & IV only
  4. D I & IV only
Verified Solution

Answer & Solution

Correct Answer

(A) I & III only

Step-by-step Solution

Detailed explanation

The complete reaction are:- The structures of \(\mathrm{XeOF}_4\) and \(\mathrm{XeO}_3\) are:- and \(3 \pi\) bonds with a lone pair of electrons on the central atom. Therefore, (I) and (III) are correct.