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

$BeC{{l}_{2}}=Cl_{\bullet }^{\bullet }BeCl$ 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.
You will be redirected in 3 sec 