11th Class Chemistry Chemical Bonding and Molecular Structure

  • question_answer 11) Discuss the shapes of the following molecules using VSEPR model: \[BeC{{l}_{2}},BC{{l}_{3}},SiC{{l}_{4}},As{{F}_{3}},{{H}_{2}}S,P{{H}_{3}}\]

    Answer:

    \[BeC{{l}_{2}}=Cl_{\bullet }^{\bullet }BeCl\] Be is surrounded by two bond pairs and no lone pair. Hence, its shape is linear. \[BC{{l}_{3}}=Cl\underset{\bullet }{\mathop{_{\bullet }^{\bullet }B_{\bullet }^{\bullet }}}\,Cl\] is surrounded by three bond pairs and no lone pair. Hence, its shape is trigonal planar. \[SiC{{l}_{4}}=Cl\,\underset{\begin{smallmatrix} \bullet \,\,\bullet \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ \bullet \,\,\bullet \end{smallmatrix}}{\mathop{_{\bullet }^{\bullet }Si\,_{\bullet }^{\bullet }}}}\,Cl\] Bond pairs = 4, lone pairs = 0. Hence, shape is tetrahedral. \[As{{F}_{3}}=F\,\underset{\begin{smallmatrix} \bullet \,\,\bullet \\ F \end{smallmatrix}}{\overset{\bullet \,\,\bullet }{\mathop{_{\bullet }^{\bullet }As\,_{\bullet }^{\bullet }}}}\,F\] Bond pairs = 3, lone pair = 1. Hence, electron group geometry is tetrahedral but actual shape is pyramidal as one position is occupied by a lone pair.   \[{{H}_{2}}S=H\,\underset{\begin{smallmatrix} \bullet \,\,\bullet \\ H \end{smallmatrix}}{\overset{\bullet \,\,\bullet }{\mathop{_{\bullet }^{\bullet }S\,_{\bullet }^{\bullet }}}}\,\] Bond pairs = 2, lone pairs = 2. Hence, electron group geometry is tetrahedral bat actual shape is V-shaped as two positions are occupied by lone pairs.   \[P{{H}_{3}}=H\,\underset{\begin{smallmatrix} \bullet \,\,\bullet \\ H \end{smallmatrix}}{\overset{\bullet \,\,\bullet }{\mathop{_{\bullet }^{\bullet }P\,_{\bullet }^{\bullet }}}}\,H\] Bond pairs = 3, lone pair = 1. Hence, electron group geometry is tetrahedral but actual shape is pyramidal as one position is ocupied by a lone pair.


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