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Bonds as a Continuum CHEMISTRY 11 MS. MCGRATH. Bond Types We have discussed each bond type (covalent and ionic bonds) as a separate and unique entity.

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Presentation on theme: "Bonds as a Continuum CHEMISTRY 11 MS. MCGRATH. Bond Types We have discussed each bond type (covalent and ionic bonds) as a separate and unique entity."— Presentation transcript:

1 Bonds as a Continuum CHEMISTRY 11 MS. MCGRATH

2 Bond Types We have discussed each bond type (covalent and ionic bonds) as a separate and unique entity for the other types of bonds In reality, bonds exist on a spectrum This spectrum has completely ionic at one end and completely covalent at the other end We will investigate this spectrum using electronegativities

3 Electronegativity of Elements On most periodic tables, electronegativity for almost all elements are stated Electronegativity refers to the electron affinity of each atom or how strongly held the valance electrons are to the nucleus Find the atoms with the largest and smallest electronegativities and which have a zero value

4 Electronegativities Fluorine has the largest electronegativity with a 3.98 ◦This means that fluorine does not easily give up electrons but that it will take electrons from other elements Francium has the lowest electronegativity with a 0.7 ◦This means that francium will give away its electron but will not attract electrons from other atoms

5 Electronegativities The Noble Gases do not have an electronegativity This is because the Noble Gases do not want to gain or lose electrons due to their stability

6 Bond Type and Electronegativity When considering a bond between two atoms, instead of following previous rules for ionic or covalent, we will now consider difference of electronegativity To determine the bond type, subtract the smaller electronegativity from the larger

7 Bond type We will use the following values to determine bond type: ◦Ionic: > 1.7 ◦Polar Covalent: 0.4 - 1.7 ◦Slightly Polar Covalent: < 0.4 ◦Non Polar Covalent: 0

8 Water If we consider water, we can see the ΔEN = ΔEN O - ΔEN H ΔEN = 3.44 – 2.20 ΔEN = 1.24 which means that it is polar covalent. This means that a water molecule will have a slightly positive end (the hydrogen atoms) and a slightly negative end (the oxygen atom)

9 Practice Problems Complete #7 – 9, page 178


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