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Determine whether each of the following molecules is polar or nonpolar.
a, SiCl2F2
b, CO2
c, XeO4
d, XeF2
♦ Relevant knowledgePolarity is a physical property that a compound has. It is a result of the intermolecular interactions which occur among the molecules. The compound may be non-polar or polar.
This problem is based upon the polarity of a molecular.
A polar molecule is a molecule with negative charge density at one end and positive charge densities at the other. The shape of a molecule can reveal the polarity of that molecule. Due to the difference in electronegativity between the atoms, a polar molecule is an asymmetrical shape that contains one or more polar bond.
To determine if a molecule has polarity or nonpolar properties, first determine its Lewis structure.
Next, determine the shape of the molecule. The symmetric shapes of trigonal pyramidal, linear, and tetrahedral shapes, respectively, are bent and trigonal pyramidal.
A polar molecule does not have a symmetrical shape.
Nonpolar molecules have symmetric shapes. It is also symmetric in that all the atoms around its central atom are identical. The molecule could be polar if the atoms attached are different to the central atom. Based on the differences in electronegativities, identify if the bonds are polar and nonpolar. If there are no polar bond, the molecule is considered nonpolar.
If there are polar bonds draw arrows that point towards the more electronegative element.
a) There are 4-7, 7 and 7 valence electrons respectively in Si, Cl, and F.
The following is how the total number of valence electrons can be calculated:
n = nSi + 2nCl + 2nF
Substitute 4 to nSi
n = 4+2( 7 )+2( 7 )
= 4 + 14 + 14
= 32
(a) There are 32 total valence electrons. These electrons are organized in such a manner that each atom’s valency is complete.
The Lewis structure for SiCl2F2
(a) There are four atoms attached at the Si atom. There are two Si–Cl
To represent the polar bonds, draw arrows.
The arrows don’t balance one another.
Therefore, the molecule can be referred to as polar.
(b) There are 6 and 4 valence electrons, respectively, in C and O.
The following is how the total number of valence electrons can be calculated:
n = nC+2nO
Substitute 4 to nC
n = 4 + 2( 6 ) = 4 + 12 = 16
(b) The total number of valence particles is 16.
The Lewis structure can be described as follows:
(b) The form of CO2
Two oxygen atoms are less electronegative than one carbon atom.
Draw arrows that point towards oxygen
The two arrows that signify C–O
The molecule CO2 is thus derived.
(c) There are 8 and 6 valence electrons respectively in Xe, O.
The following is how the total number of valence electrons can be calculated:
n = nXe + 4nO
Substitute 8 to nXe
n = 8 + 4( 6 )= 8 + 24 = 32
(c) The total number of valence particles is 32.
This is the Lewis structure:
(c) The geometric shape XeO4
O is also more electronegative than Xe.
To show the bond dipole, draw the arrows towards O.
The bond dipole for four Xe–O
The molecule XeO4 is thus derived
(d) There are 8 and 7 valence electrons respectively in Xe, F.
The following is how the total number of valence electrons can be calculated:
n =nXe+2nF
Substitute 8 to nXe
n = 8 + 2( 7 ) = 8 + 14 = 22
(d) 22 total valence electrons are available
This will place Xe as the central atom and arrange fluorine electrons around it. The Lewis structure is then drawn:
(d) The geometric shape XeF2
And F is more electron-negative than Xe.
To show the bond dipole, draw the arrows towards F.
The bond dipole for two Xe–F bonds are equal
The molecule XeF2 iss nonpolar
Ans is thus: Part a
The molecule SiCl2F2 is polar
Part b
The molecule CO2 is nonpolar
Part c
The molecule XeO4 is nonpolar
Part d
The molecule XeF2 is nonpolar
This problem is based upon the polarity of a molecular.
A polar molecule is a molecule with negative charge density at one end and positive charge densities at the other. The shape of a molecule can reveal the polarity of that molecule. Due to the difference in electronegativity between the atoms, a polar molecule is an asymmetrical shape that contains one or more polar bond.
To determine if a molecule has polarity or nonpolar properties, first determine its Lewis structure.
Next, determine the shape of the molecule. The symmetric shapes of trigonal pyramidal, linear, and tetrahedral shapes, respectively, are bent and trigonal pyramidal.
A polar molecule does not have a symmetrical shape.
Nonpolar molecules have symmetric shapes. It is also symmetric in that all the atoms around its central atom are identical. The molecule could be polar if the atoms attached are different to the central atom. Based on the differences in electronegativities, identify if the bonds are polar and nonpolar. If there are no polar bond, the molecule is considered nonpolar.
If there are polar bonds draw arrows that point towards the more electronegative element.
(a)
There are 4-7, 7 and 7 valence electrons respectively in Si, Cl, and F.
The following is how the total number of valence electrons can be calculated:
[katex]n = {n_{{\rm{Si}}}} + 2{n_{{\rm{Cl}}}} + 2{n_{\rm{F}}}[/katex]
Substitute 4 to [katex]{n_{{\rm{Si}}}}[/katex]
[katex]\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( {\rm{7}} \right) + 2\left( 7 \right)\\\\ = 4 + 14 + 14\\\\ = 32\\\end{array}[/katex]
(a)
There are 32 total valence electrons. These electrons are organized in such a manner that each atom’s valency is complete.
The Lewis structure for [katex]{SiCl_2F_2}[/katex]
(a)
There are four atoms attached at the Si atom. There are two [katex]{\rm{Si}} – {\rm{Cl}}[/katex]
To represent the polar bonds, draw arrows.
The arrows don’t balance one another.
Therefore, the molecule can be referred to as polar.
(b)
There are 6 and 4 valence electrons, respectively, in C and O.
The following is how the total number of valence electrons can be calculated:
[katex]n = {n_{\rm{C}}} + 2{n_{\rm{O}}}[/katex]
Substitute 4 to [katex]{n_{\rm{C}}}[/katex]
[katex]\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( 6 \right)\\\\ = 4 + 12\\\\ = 16\\\end{array}[/katex]
(b)
The total number of valence particles is 16.
The Lewis structure can be described as follows:
(b)
The form of [katex]{\rm{C}}{{\rm{O}}_2}[/katex]
Two oxygen atoms are less electronegative than one carbon atom.
Draw arrows that point towards oxygen
The two arrows that signify [katex]{\rm{C}} – {\rm{O}}[/katex]
The molecule [katex]{\rm{C}}{{\rm{O}}_2}[/katex] is thus derived.
(c)
There are 8 and 6 valence electrons respectively in Xe, O.
The following is how the total number of valence electrons can be calculated:
[katex]n = {n_{{\rm{Xe}}}} + 4{n_{\rm{O}}}[/katex]
Substitute 8 to [katex]{n_{{\rm{Xe}}}}[/katex]
[katex]\begin{array}{c}\\n = 8 + 4\left( 6 \right)\\\\ = 8 + 24\\\\ = 32\\\end{array}[/katex]
(c)
The total number of valence particles is 32.
This is the Lewis structure:
(c)
The geometric shape [katex]{\rm{Xe}}{{\rm{O}}_4}[/katex]
O is also more electronegative than [katex]{\rm{Xe}}[/katex].
To show the bond dipole, draw the arrows towards O.
The bond dipole for four [katex]{\rm{Xe}} – {\rm{O}}[/katex]
The molecule [katex]{\rm{Xe}}{{\rm{O}}_4}[/katex] is thus derived.
(d)
There are 8 and 7 valence electrons respectively in Xe, F.
The following is how the total number of valence electrons can be calculated:
[katex]n = {n_{{\rm{Xe}}}} + 2{n_{\rm{F}}}[/katex]
Substitute 8 to [katex]{n_{{\rm{Xe}}}}[/katex]
[katex]\begin{array}{c}\\n = 8 + 2\left( 7 \right)\\\\ = 8 + 14\\\\ = 22\\\end{array}[/katex]
(d)
22 total valence electrons are available
This will place
as the central atom and arrange fluorine electrons around it. The Lewis structure is then drawn:
(d)
The geometric shape [katex]{\rm{Xe}}{{\rm{F}}_2}[/katex]
And F is more electron-negative than Xe.
To show the bond dipole, draw the arrows towards F.
The bond dipole for two [katex]{\rm{Xe}} – {\rm{F}}[/katex]
The molecule [katex]{\rm{Xe}}{{\rm{F}}_2}[/katex]Ans is thus: Part a
The molecule [katex]{\rm{SiC}}{{\rm{l}}_2}{{\rm{F}}_2}[/katex]Part B
The molecule [katex]{\rm{C}}{{\rm{O}}_2}[/katex]Part C
The molecule [katex]{\rm{Xe}}{{\rm{O}}_4}[/katex]Part D
The molecule [katex]{\rm{Xe}}{{\rm{F}}_2}[/katex]