23. Deduce the shape of the following molecules.(a) PCI5 (b) PF3 (c) SP6 (d) XeF4 (e) CH4 (f) NH3(Xe = 54, S = 16, P = 15, C = 6, N = 7,)
We s shall be happy to check your answers.
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/10%3A_Chemical_Bonding_II%3A_Molecular_Shapes%2C_Valance_Bond_Theory%2C_and_Molecular_Orbital_Theory/10.2%3A_VSEPR_Theory_-_The_Five_Basic_Shapes
You need to check SP6 to see if it was copied correctly... possibly SF6?
To deduce the shape of a molecule, you need to determine its electron geometry and then use the concept of VSEPR (Valence Shell Electron Pair Repulsion) theory to find its molecular geometry.
(a) PCI5:
Start by determining the electron geometry. Phosphorus (P) has five valence electrons, and each chlorine (Cl) atom contributes one electron, resulting in a total of 10 electrons for the 5 chlorine atoms. Since phosphorus has five bonding regions and no lone pairs, the electron geometry is trigonal bipyramidal.
To find the molecular geometry, consider the arrangement of bonding pairs and lone pairs. PCI5 has five chlorine atoms bonded to phosphorus, resulting in five bonding pairs. With no lone pairs, the molecular geometry will also be trigonal bipyramidal.
(b) PF3:
Phosphorus (P) has five valence electrons, and each fluorine (F) atom contributes one electron, resulting in a total of 8 electrons for the 3 fluorine atoms. Since phosphorus has four bonding regions and one lone pair, the electron geometry is tetrahedral.
To find the molecular geometry, consider the arrangement of bonding pairs and lone pairs. PF3 has three fluorine atoms bonded to phosphorus, resulting in three bonding pairs. With one lone pair, the molecular geometry will be trigonal pyramidal.
(c) SP6:
Since Sulfur (S) has six valence electrons, the maximum number of bonds it can form is six. Since there are no other atoms involved, the molecule SP6 does not exist. An S atom cannot form six bonds with adjacent P atoms due to its valence electron limitation.
(d) XeF4:
Xenon (Xe) has eight valence electrons, and each fluorine (F) atom contributes one electron, resulting in a total of 4 electrons for the 4 fluorine atoms. Since xenon has six bonding regions and two lone pairs, the electron geometry is octahedral.
To find the molecular geometry, consider the arrangement of bonding pairs and lone pairs. XeF4 has four fluorine atoms bonded to xenon, resulting in four bonding pairs. With two lone pairs, the molecular geometry will be square planar.
(e) CH4:
Carbon (C) has four valence electrons, and each hydrogen (H) atom contributes one electron, resulting in a total of 4 electrons for the 4 hydrogen atoms. Since carbon has four bonding regions and no lone pairs, the electron geometry is tetrahedral.
To find the molecular geometry, consider the arrangement of bonding pairs. CH4 has four hydrogen atoms bonded to carbon, resulting in four bonding pairs. With no lone pairs, the molecular geometry will also be tetrahedral.
(f) NH3:
Nitrogen (N) has five valence electrons, and each hydrogen (H) atom contributes one electron, resulting in a total of 3 electrons for the 3 hydrogen atoms. Since nitrogen has four bonding regions and one lone pair, the electron geometry is tetrahedral.
To find the molecular geometry, consider the arrangement of bonding pairs and lone pairs. NH3 has three hydrogen atoms bonded to nitrogen, resulting in three bonding pairs. With one lone pair, the molecular geometry will be trigonal pyramidal.