Question 9.3: Classify the bond formed between each pair of atoms as coval...
Classify the bond formed between each pair of atoms as covalent, polar covalent, or ionic.
(a) Sr and F (b) N and Cl (c) N and O
(a) From Figure 9.7, find the electronegativities of Sr (1.0) and of F (4.0). The electronegativity difference is ∆EN = 4.0 – 1.0 = 3.0. Using Table 9.1, classify this bond as ionic.
(b) From Figure 9.7, find the electronegativities of N (3.0) and of Cl (3.0). The electronegativity difference (∆EN) is ∆EN = 3.0 – 3.0 = 0. Using Table 9.1, classify this bond as covalent.
(c) From Figure 9.7, find the electronegativities of N (3.0) and of O (3.5). The electronegativity difference (∆EN) is ∆EN = 3.5 – 3.0 = 0.5. Using Table 9.1, classify this bond as polar covalent.
| TABLE 9.1 The effect of electronegativity Difference on Bond Type | ||
| Electronegativity Difference (∆EN) | Bond Type | Example |
| Small (0-0.4) | Covalent | Cl_{2} |
| Intermediate (0.4-2.0) | Polar covalent | HCl |
| Large (2.0+) | Ionic | NaC |
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