Question

Do carbon dioxide and water have the same molecular geometry? if not, what attributes cause these molecules with similar formulas, ab2, to have different geometries?

Answer 1

No, they do not.
Carbon dioxide has a linear geometry because the lone pair and bond pair repulsion cancels out; however, water has a bent structure because only the oxygen atom possesses a lone pair which brings the bonding electron pairs closer.

Answer 2

${\mathbf{C}}{{\mathbf{O}}_{\mathbf{2}}}$ 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: brainly.com/question/4341225

2. How many molecules will be present on completion of reaction?: 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.