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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