Question

04. Predict the shapes of the following molecules using the valence shell electron pair repulsion (VSEPR) theory. (1 mark each) BeCl, SiCly AsF5, H2S, HgBr2, PH3, GeF2 [NCERT] [KVS-2007]

JEE/Engineering Exams

Chemistry

Solution

136

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Be ( C l_{2} geqslant ) There is no lone pair on the contral atom. U atoms are bonded to Be by bond pairs. ( Rightarrow ) The shape must be a linear phape [ C l-B e-C ] Si ( C l y rightarrow ) Again there are no lone pairs on the contral afom and ce and bonded by 4 bond pairs. ( Rightarrow ) Shape is tetrahedral Asts ( rightarrow ) As has no lone paird and ( f ) are bonded by 5 bond Pairs ( Rightarrow ) Shape is trigenal bipyramidal. ( F ) ( F_{1} geqslant A_{s}^{prime}-F ) ( F^{N} frac{1}{F} ) ( mathrm{H}_{2} mathrm{S} ) - Contral atom has one lonc pair and two bond pairs. ( Rightarrow ) Shape is ( B e n t ) . HgBrg- There is no lone pair and Br aboms are bonded by 2 bond pairs. ( Rightarrow ) Shape is lírear. [ B gamma-4 g-B gamma ] ( P H_{3}- ) Central atom has one lone pair and 3 Broud pairs cure bonded by 2 bond pairs ( Rightarrow ) shape is Bent