[tex]{\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] and [tex]{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}[/tex] do not have the same geometry. Both [tex]{\text{C}}{{\text{O}}_2}[/tex] and [tex]{{\text{H}}_2}{\text{O}}[/tex] has two side atoms but the number of bond pair and lone pair are different. [tex]{\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] has only two bond pair with no lone pair while [tex]{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}[/tex] has two bond pair with two lone pair.
Further Explanation:
Lewis structure of [tex]{\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}[/tex] :
The total number of valence electrons of [tex]{\text{C}}{{\text{O}}_2}[/tex] is calculated as,
Total valence electrons = [(1) (Valence electrons of C) + (2) (Valence electrons of O)]
[tex]\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}}\right)&=\left[ {\left( {\text{1}} \right)\left({\text{4}}\right)+\left( {\text{2}}\right)\left({\text{6}}\right)}\right]\\ &=16\\\end{aligned}[/tex]
In [tex]{\text{C}}{{\text{O}}_2}[/tex] , the total number of valence electrons is 16. Here, carbon atoms form two double bond with the two oxygen atoms, therefore, 8 pair of electrons are used in the formation of two double bonds with the oxygen atoms. Rest 8 electrons are distributed to complete the octet of oxygen atoms.
According to the Lewis structure of [tex]{\text{C}}{{\text{O}}_2}[/tex] central atom carbon has only two bond pair with no lone pair thus, [tex]{\text{C}}{{\text{O}}_2}[/tex] has [tex]{\text{A}}{{\text{B}}_2}[/tex] electron group arrangement. Therefore, [tex]{\text{C}}{{\text{O}}_2}[/tex] has linear geometry and linear shape.
Lewis structure of [tex]{{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}[/tex] :
The total number of valence electrons of [tex]{{\text{H}}_2}{\text{O}}[/tex] is calculated as,
Total valence electrons = [(1) (Valence electrons of O) + (2) (Valence electrons of H)]
[tex]\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}} \right)&=\left[{\left( {\text{1}}\right)\left({\text{6}}\right)+\left({\text{2}}\right)\left({\text{1}}\right)}\right]\\&=8\\\end{aligned}[/tex]
In [tex]{{\text{H}}_2}{\text{O}}[/tex] , the total number of valence electrons is 8. Here, oxygen forms single bond with the hydrogen atom, therefore, 2 pair of electrons are used in the formation of two single bonds with hydrogen atom. Rest 4 electrons are used to complete the octet of oxygen atom.
According to the Lewis structure of [tex]{{\text{H}}_2}{\text{O}}[/tex] central atom oxygen has two bond pair with two lone pair and [tex]{{\text{H}}_2}{\text{O}}[/tex] has [tex]{\text{A}}{{\text{B}}_2}{{\text{E}}_2}[/tex] electron group arrangement. Therefore, [tex]{{\text{H}}_2}{\text{O}}[/tex] has a tetrahedral geometry and has bent shape.
Therefore, [tex]{\text{C}}{{\text{O}}_2}[/tex] and [tex]{{\text{H}}_2}{\text{O}}[/tex] do not have the same geometry.
Learn more:
1. Molecular shape around each of the central atoms in the amino acid glycine: https://brainly.com/question/4341225
2. How many molecules will be present on completion of reaction?: https://brainly.com/question/4414828
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Molecular structure and chemical bonding
Keywords: Shape, molecular shape, tetrahedral, bent, bent shape, lewis structure, valence, valence electron, H2O, CO2, lone pair, bond pair, charge, geometry, linear.
