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Hint: While solving a question related to the shape of a molecule, we need to find out the valence electron pairs in the molecule and then the hybridization, number of bond pairs, number of lone pairs and the repulsion between the electron pairs. The VSEPR theory will help us determine the structure with minimized repulsion.
Complete answer:
To solve this question, let us analyze the shape of each molecule.
Water $\left( {{H_2}O} \right)$ is a molecule consisting of one oxygen atom bonded to two different hydrogen atoms. The oxygen atom is the central atom with two polar bonds to the two hydrogen atoms. It has two bond pairs and two lone pairs. Since there is lone pair-lone pair repulsion, the bond angle reduces from a standard $109.5^\circ $ of $s{p^3}$hybridized molecules to around $105^\circ $ . This results in a bent shape of water molecules.
${\text{B}}{{\text{H}}_3}$ is Borane, which has a central boron atom surrounded by three different hydrogen atoms. There are three bond pairs and no lone pair in this molecule. The molecule is $s{p^2}$ hybridized and hence has trigonal planar shape with bond angles of $120^\circ .$
${\text{C}}{{\text{H}}_4}$ is Methane, which has a central carbon atom bonded to four hydrogen atoms. There are four bond pairs and no lone pair in this molecule. The molecule is $s{p^3}$ hybridized and hence has tetrahedral shape with bond angles of $109.5^\circ .$
${\text{C}}{{\text{O}}_2}$ is Carbon dioxide, where a central carbon atom is doubly bonded to two different oxygen atoms. It is $sp$ hybridized, with the molecule having four bond pairs (two $\sigma $ and two $\pi $ bonds) and four lone pairs (two on each oxygen atom). The molecule has bond angle $108^\circ $ and has linear shape.
Hence, the required answer is d) ${\text{C}}{{\text{O}}_2}.$
Note:
The linear shape of carbon dioxide molecules is due to its hybridization as well as due to $p\pi - p\pi $ bonding between the carbon atom and the oxygen atoms. ${\text{C}}{{\text{O}}_2}$ molecule has two regions of electron density, and the carbon atom has no lone pair of electrons on it.
Complete answer:
To solve this question, let us analyze the shape of each molecule.
Water $\left( {{H_2}O} \right)$ is a molecule consisting of one oxygen atom bonded to two different hydrogen atoms. The oxygen atom is the central atom with two polar bonds to the two hydrogen atoms. It has two bond pairs and two lone pairs. Since there is lone pair-lone pair repulsion, the bond angle reduces from a standard $109.5^\circ $ of $s{p^3}$hybridized molecules to around $105^\circ $ . This results in a bent shape of water molecules.
${\text{B}}{{\text{H}}_3}$ is Borane, which has a central boron atom surrounded by three different hydrogen atoms. There are three bond pairs and no lone pair in this molecule. The molecule is $s{p^2}$ hybridized and hence has trigonal planar shape with bond angles of $120^\circ .$
${\text{C}}{{\text{H}}_4}$ is Methane, which has a central carbon atom bonded to four hydrogen atoms. There are four bond pairs and no lone pair in this molecule. The molecule is $s{p^3}$ hybridized and hence has tetrahedral shape with bond angles of $109.5^\circ .$
${\text{C}}{{\text{O}}_2}$ is Carbon dioxide, where a central carbon atom is doubly bonded to two different oxygen atoms. It is $sp$ hybridized, with the molecule having four bond pairs (two $\sigma $ and two $\pi $ bonds) and four lone pairs (two on each oxygen atom). The molecule has bond angle $108^\circ $ and has linear shape.
Hence, the required answer is d) ${\text{C}}{{\text{O}}_2}.$
Note:
The linear shape of carbon dioxide molecules is due to its hybridization as well as due to $p\pi - p\pi $ bonding between the carbon atom and the oxygen atoms. ${\text{C}}{{\text{O}}_2}$ molecule has two regions of electron density, and the carbon atom has no lone pair of electrons on it.
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