{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

chapter08_part1

# chapter08_part1 - 336 Chapter 8 Covalent Bonding...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 8: Covalent Bonding 336 End-of-Chapter Solutions for Chapter 8 Summary Problem Answer: (a) O–H < N–H < C–H < C=O < C–O < C–N < C–C (b) C–N < C–O < C–C < C–H < O–H < C=O (c) C–C < C–H < C–O < O–H (d) see structure below (e) see structures below (d) see structure below (e) N has +1 formal charge, O has –1 formal charge, net charge is 0. Strategy and Explanation : (a) Order a given set of bonds by increasing length. The lengths of covalent bonds are given in Table 8.1. O–H (94 pm) < N–H (98 pm) < C–H (110 pm) < C=O (122 pm) < C–O (143 pm) < C–N (147 pm) < C–C (154 pm) This order makes sense because H atom, from Period 1, is very small compared to the others. The smallest of the two Period 2 elements is O, since it further to the right in the same period. N is the next smallest atom, then C. Double bonds are shorter than single bonds between atoms of the same type. (b) Order a given set of bonds by increasing bond strength. The strengths of covalent bonds increase as the bond enthalpy increases. Bond enthalpies are given in Table 8.2. C–N (285 kJ/mol) < C–O (336 kJ/mol) < C–C (356 kJ/mol) < C–H (416 kJ/mol) < O–H (467 kJ/mol) < C=O (695 kJ/mol) This order makes sense because significantly shorter bonds, such as C–H and O–H, will be stronger. Bonds involve multiple bonds are stronger than ones with single bonds between atoms of the same type. (c) Order a given set of bonds by increasing polarity. The polarity of covalent bonds increase as the electronegativity difference between the atoms increases. Electronegativities for atoms are given in Figure 8.6. Look up electronegativity values for the atoms: EN C = 2.5, EN O = 3.5, EN N = 3.0, EN H = 2.1, EN I = 2.5 Calculate the electronegativity differences. Δ EN C–N = EN N – EN C = 3.0 – 2.5 = 0.5 Δ EN C–O = EN O – EN C = 3.5 – 2.5 = 1.0 Δ EN C–H = EN H – EN C = 2.1 – 2.5 = 0.4 Δ EN O–H = EN H – EN O = 2.1 – 3.5 = 1.4 Δ EN C–C = EN C – EN C = 2.5 – 2.5 = 0.0 C–C ( Δ EN = 0.0) < C–H ( Δ EN = 0.4) < C–O ( Δ EN = 1.0) < O–H ( Δ EN = 1.4) This order makes sense because the further apart the atoms are on the periodic table, the more polar the bond. (d) Structural formula of the glycinate ion is produced by removing the H atom on O–H and changing the net charge to –1. H N H C H H C O O _ (e) Lewis structures are molecular structures that include the non-bonded electrons as pairs of dots. The atoms that need dots are N and O, since the H atoms already have the two they need in the single bond, and the C atoms already have the eight electrons they need in the bonds. The N atom has six electrons in bonds, so they need two more electrons in the form of a pair of dots. The double-bonded O atom has four electrons in bonds, so it needs four more in the form of two pairs of dots. The single-bonded O atom has two electrons in a bond, so it

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 8: Covalent Bonding 337 needs six more in the form of three pairs of dots. The resonance structure will alternate the position of the double bond. The following two Lewis structures represent the resonance hybrid for glycinate ion: H N H C H H C O O _ H N H C H H C O O _ . .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 35

chapter08_part1 - 336 Chapter 8 Covalent Bonding...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online