\({\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}\) and \({{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\) do not have the same geometry. Both \({\text{C}}{{\text{O}}_2}\) and \({{\text{H}}_2}{\text{O}}\) has two side atoms but the number of bond pair and lone pair are different. \({\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}\) has only two bond pair with no lone pair while \({{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\) has two bond pair with two lone pair.
Further Explanation:
Lewis structure of \({\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}\) :
The total number of valence electrons of \({\text{C}}{{\text{O}}_2}\) is calculated as,
Total valence electrons = [(1) (Valence electrons of C) + (2) (Valence electrons of O)]
\(\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}}\right)&=\left[ {\left( {\text{1}} \right)\left({\text{4}}\right)+\left( {\text{2}}\right)\left({\text{6}}\right)}\right]\\ &=16\\\end{aligned}\)
In \({\text{C}}{{\text{O}}_2}\) , 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 \({\text{C}}{{\text{O}}_2}\) central atom carbon has only two bond pair with no lone pair thus, \({\text{C}}{{\text{O}}_2}\) has \({\text{A}}{{\text{B}}_2}\) electron group arrangement. Therefore, \({\text{C}}{{\text{O}}_2}\) has linear geometry and linear shape.
Lewis structure of \({{\mathbf{H}}_{\mathbf{2}}}{\mathbf{O}}\) :
The total number of valence electrons of \({{\text{H}}_2}{\text{O}}\) is calculated as,
Total valence electrons = [(1) (Valence electrons of O) + (2) (Valence electrons of H)]
\(\begin{aligned}{\text{Total valence electrons}}\left({{\text{TVE}}} \right)&=\left[{\left( {\text{1}}\right)\left({\text{6}}\right)+\left({\text{2}}\right)\left({\text{1}}\right)}\right]\\&=8\\\end{aligned}\)
In \({{\text{H}}_2}{\text{O}}\) , 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 \({{\text{H}}_2}{\text{O}}\) central atom oxygen has two bond pair with two lone pair and \({{\text{H}}_2}{\text{O}}\) has \({\text{A}}{{\text{B}}_2}{{\text{E}}_2}\) electron group arrangement. Therefore, \({{\text{H}}_2}{\text{O}}\) has a tetrahedral geometry and has bent shape.
Therefore, \({\text{C}}{{\text{O}}_2}\) and \({{\text{H}}_2}{\text{O}}\) 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.
