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The Bonds in Water

The oxygen atom in water (H2O) forms two covalent bonds. Because oxygen s electronic configuration shows that it has two unpaired valence electrons, oxygen does not need to promote an electron to form the two covalent bonds required to complete its octet. [Pg.38]

The experimentally observed bond angle in H2O is 104.5°. The bond angle indicates that oxygen, like carbon and nitrogen, uses hybrid orbitals when it forms covalent bonds. [Pg.38]

Also like carbon and nitrogen, the one s and three p orbitals hybridize to form four degenerate sj orbitals  [Pg.39]

Each of the two O—H bonds is formed by the overlap of an sp orbital of oxygen with the 5 orbital of a hydrogen. A lone pair occupies each of the two remaining [Pg.39]

The bond angle in water (104.5°) is even smaller than the bond angles in NH3 (107.3°) because oxygen has two relatively diffuse lone pairs, whereas nitrogen has only one. [Pg.39]


A single photon of visible light does not have enough energy to break the bonds in water four photons are required in this photolytic cleavage reaction. [Pg.738]

Let us now consider a delocalized description of the bonding in water. First we investigate the transformation properties of the various atomic orbitals, because we want to form as general a set of molecular orbitals as possible. [Pg.75]

Figure 6-10 Molecular-orbital description of the bonding in water. Figure 6-10 Molecular-orbital description of the bonding in water.
When using these values to characterize a bond, do not consider the subscripts in the chemical formula. That is, for H20, subtract the electronegativity of one H (2.1) from that of one 0 (3.5) to get a difference of 1.4. The bonds in water are therefore said to be polar covalent bonds. The smaller value should always be subtracted from the larger so that the difference is always positive. [Pg.124]

How can you explain the wide variety of properties that covalent compounds have Covalent compounds maybe solids, liquids, or gases at different temperatures. Some covalent compounds dissolve in water, and some do not. In fact, water itself is a covalent compound Examine Figures 3.22 and 3.23. Why are the bonds in water different from the bonds in dinitrogen monoxide Both of these compounds are made up of... [Pg.85]

How does the bonding in water molecules account for the fact that water is an exellent solvent ... [Pg.325]

The idea can be extended to materials formed from nonmetals, too. For instance, the formula H2O can be explained in terms of atomic shells. There are some subtleties regarding the difference between the bonding in sodium chloride and the bonding in water that we will explore later when we delve more into chemical bonding, but the rearrangement of electrons to fill shells is still our basic premise. [Pg.67]

The H-C-H bond angle in methane is 109.5°. The H-O-H bond angle of water is close to this number but the H-S-H bond angle of H2S is near 90°. What does this tell us about the bonding in water and H2S Draw a diagram of the molecular orbitals in H2S. [Pg.9]

Compounds with polar covalent bonds have different properties from compounds with pure covalent bonds. You saw that purely covalent compounds tend to have low melting points and boiling points. Carbon disulfide, CS2, as shown in Figure 9.7, is a triatomic molecule, with a AEN equal to zero. Carbon disulfide bods at 46°C. Water is also a triatomic molecule, but the bonding in water is polar covalent. Even though water is a much lighter molecule than carbon disulfide, its boiling point is 100°C. [Pg.311]

In principle, the bonding in water could also be described by formal sp hybridization on oxygen. However, the energetic cost is now too large (see Exercise 1-17). Nevertheless, for the sake of simplicity, we can extend the picture for ammonia to water as in Figure 1-20, with an HOH bond angle of 104.5°. [Pg.35]


See other pages where The Bonds in Water is mentioned: [Pg.344]    [Pg.55]    [Pg.115]    [Pg.111]    [Pg.110]    [Pg.501]    [Pg.515]    [Pg.145]    [Pg.452]    [Pg.170]    [Pg.38]    [Pg.39]    [Pg.220]    [Pg.180]   


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