Determine whether each of the following molecules is polar or nonpolar. SiCl2F2 CO2 XeO4 XeF2
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Question “Determine whether each of the following molecules is polar or nonpolar. SiCl2F2 CO2 XeO4 XeF2”
Determine whether each of the following molecules is polar or
nonpolar.
SiCl2F2
CO2
XeO4
XeF2
Answer
This problem is based upon the polarity of a molecular.
A molecule with a negative charge density at one end and a positive charge density at the other is known as polar molecule. 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 nonpolarity, 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 = {n_{{\rm{Si}}}} + 2{n_{{\rm{Cl}}}} + 2{n_{\rm{F}}}Substitute 4 to $math_tag_1
\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( {\rm{7}} \right) + 2\left( 7 \right)\\\\ = 4 + 14 + 14\\\\ = 32\\\end{array}
(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 {\rm{SiC}}{{\rm{l}}_2}{{\rm{F}}_2}
(a)
The Si atom is composed of four atoms. Two {\rm{Si}} - {\rm{Cl}} atoms are attached to Si atom.
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 = {n_{\rm{C}}} + 2{n_{\rm{O}}}Substitute 4 to $math_tag_6
\begin{array}{c}\\n = {\rm{4}} + {\rm{2}}\left( 6 \right)\\\\ = 4 + 12\\\\ = 16\\\end{array}
(b)
The total number of valence particles is 16.
The Lewis structure can be described as follows:
(b)
The $math_tag_8 shape
Two oxygen atoms are less electronegative than one carbon atom.
Draw arrows that point towards oxygen
The two arrows that represent $math_tag_9
The molecule {\rm{C}}{{\rm{O}}_2} 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 = {n_{{\rm{Xe}}}} + 4{n_{\rm{O}}}Substitute 8 to $math_tag_12
\begin{array}{c}\\n = 8 + 4\left( 6 \right)\\\\ = 8 + 24\\\\ = 32\\\end{array}
(c)
The total number of valence particles is 32.
This is the Lewis structure:
(c)
The geometric shape $math_tag_14
O is also more electronegative than {\rm{Xe}}
To show the bond dipole, draw the arrows towards O.
The bond dipole for four {\rm{Xe}} - {\rm{O}}
The molecule $math_tag_14 is thus a molecule.
(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 = {n_{{\rm{Xe}}}} + 2{n_{\rm{F}}}Substitute 8 to $math_tag_12
\begin{array}{c}\\n = 8 + 2\left( 7 \right)\\\\ = 8 + 14\\\\ = 22\\\end{array}
(d)
22 total valence electrons are available
as the central atom and arrange fluorine electrons around it. The Lewis structure is then drawn:
(d)
The geometric shape $math_tag_21
F is also more electronegative than the Xe.
To show the bond dipole, draw the arrows towards F.
The bond dipole for two {\rm{Xe}} - {\rm{F}}
The molecule {\rm{Xe}}{{\rm{F}}_2}Ans is thus: Part a
Part b of the molecule {\rm{SiC}}{{\rm{l}}_2}{{\rm{F}}_2}
Part c: The molecule {\rm{C}}{{\rm{O}}_2}
Part d of the molecule {\rm{Xe}}{{\rm{O}}_4}
The molecule $math_tag_21
Conclusion
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