Question

explain geometry of water molecule on the basis of hybridisation​

Answer 1

Answer:

Formation of water molecules on the basis of $sp^{3}$ hybridisation

Explanation:

• The water molecule (H2O) consists of two hydrogen atoms and one oxygen atom.
• Oxygen's (Z = 8) ground state electrical arrangement is $1s^{2} 2s^{2} 2p_{x} ^{2} 2p_{y} ^{1} 2p_{z} ^{1}$
• The observed oxygen valency in the H2O molecule, which is 2, can be explained by the ground state electronic arrangement..
• The oxygen atom's 2s, 2px, 2py, and 2pz orbitals combine and are recast to create four sp3 hybrid orbitals with comparable energy.
• These orbitals are aligned in space tetrahedrally.
• Lone electron pairs are present in two of the sp hybrid orbitals.
• The axial overlap of two of the O atom's half-filled sp3 hybrid orbitals with two different hydrogen atoms' half-filled 1s orbitals results in the formation of two O-H (sp3 -s) sigma covalent bonds.
• Two of oxygen's sp hybrid orbitals contain two lone pairs of electrons, and the bonding pair and lone pair of electrons repel one another.
• The HO-H bond angle decreases from the standard tetrahedral angle of 109°28′ to 104°35′ as a consequence.
• The H2O molecule has an angular or V-shaped form.

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Answer 2

Answer:

Hybridization Type: sp3

Bond Angle: 104.5o

Geometry: Angular or V-shaped

Explanation:

Lone pairs occupy two hybrid orbitals, while two are utilised to bond with hydrogen atoms. H2O has an angular geometry because lone pairs do not contribute to a molecule's geometry. Among the four pairings, two are oxygen lone pairs and two are bound to hydrogen, forming the shape of a v.

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