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Question 57 of 66

Q.(i) What is the hybridization state of central atom of XeF4? Draw its structure. [1]

(ii) Complete the following reaction: PCl5 + D2O → ? [1] OR Out of H2O and H2S which one has higher 'bond angle' and why? [2]
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 2mImportance★★★★★
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XeF4XeF_4: sp3d2sp^3d^2 hybridisation, square planar shape. PCl5PCl_5 hydrolyses with D2OD_2O just as with H2OH_2O, giving POCl3+DClPOCl_3 + DCl. H2OH_2O's bond angle (104.5°) exceeds H2SH_2S's (~92°) because O is more electronegative and uses more s-character.

(i) XeF4XeF_4: Xenon has 8 valence electrons; in XeF4XeF_4, 4 are used to form 4 Xe-F bonds and the remaining 4 electrons form 2 lone pairs, giving a total of 6 electron domains around Xe. Six electron domains require sp3d2sp^3d^2 hybridisation (octahedral electron-pair geometry). To minimise lone pair-lone pair repulsion, the two lone pairs occupy mutually trans (axial) positions, leaving the four F atoms in a plane around Xe — so the molecular shape is square planar.

(ii) PCl5+D2OPCl_5 + D_2O: PCl5PCl_5 reacts with water (or heavy water) by hydrolysis, replacing Cl atoms with OD (or OH) groups and releasing DCl (or HCl):

PCl5+D2O→POCl3+2DClPCl_5 + D_2O \rightarrow POCl_3 + 2DCl

(analogous to the normal reaction PCl5+H2O→POCl3+2HClPCl_5 + H_2O \rightarrow POCl_3 + 2HCl, with deuterium simply replacing hydrogen).

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