A brief note on Ionic and Covalent Bonds . Read it before a test. Try each question at the end before you open its answer.
Lewis structures Octet rule: atoms lose, gain or share electrons to reach eight in the outer shell. Hydrogen needs two. Counting: add the valence electrons, then add one per negative charge or take one per positive charge. Drawing: join the outer atoms to the centre, complete their octets, put the rest on the centre, then make multiple bonds if the centre is short. Formal charge: FC = V − L − 1 2 B \text{FC} = V - L - \tfrac{1}{2}B FC = V − L − 2 1 B . Keep it small, with any negative charge on the more electronegative atom. Exceptions: incomplete octet ( BCl 3 \text{BCl}_3 BCl 3 ), odd electrons (NO), expanded octet ( SF 6 \text{SF}_6 SF 6 ). The ionic bond A metal of low ionisation enthalpy gives electrons to a non-metal of negative electron gain enthalpy. Making the gaseous ions costs energy. The lattice enthalpy pays for it. Lattice enthalpy is larger for higher charges and smaller ions. Born-Haber: formation = = = sublimation + + + half bond + + + ionisation + + + electron gain − - − lattice. Ionic solids melt high, are brittle, and conduct only when molten or dissolved. Bond parameters Higher bond order: shorter and stronger bonds. Average bond enthalpy: the mean of the steps that break the same kind of bond. Resonance: the real molecule is one hybrid of its canonical forms, and more stable than any of them. Dipole moment: μ = q × d \mu = q \times d μ = q × d , in debye. A molecule's dipole is the vector sum of its bond dipoles. Fajans: small, highly charged cations and large anions give more covalent character. A worked example Find the formal charges in the cyanate ion, OCN − \text{OCN}^- OCN − , drawn as O=C=N − \text{O=C=N}^- O=C=N − .
Count. 6 + 4 + 5 + 1 = 16 6 + 4 + 5 + 1 = 16 6 + 4 + 5 + 1 = 16 electrons. Bonds. Two double bonds use 8. The other 8 go as 2 lone pairs on oxygen and 2 on nitrogen. Oxygen. 6 − 4 − 2 = 0 6 - 4 - 2 = 0 6 − 4 − 2 = 0 . Carbon. 4 − 0 − 4 = 0 4 - 0 - 4 = 0 4 − 0 − 4 = 0 . Nitrogen. 5 − 4 − 2 = − 1 5 - 4 - 2 = -1 5 − 4 − 2 = − 1 . The sum, − 1 -1 − 1 , is the charge of the ion. Practice with new numbers
A program to try The program finds the formal charges in the cyanate ion and checks that they add up to its charge.
Run it. Compare each charge with the worked example.
1. Change nitrogen to ("N", 5, 2, 6) and oxygen to ("O", 6, 6, 2), for O-C ≡ N \text{O-C}\equiv\text{N} O-C ≡ N . Which atom now carries the charge?
Mistakes to avoid Ignoring the charge when counting electrons for an ion. Counting lone pairs instead of lone pair electrons in a formal charge. Adding the lattice enthalpy in a Born-Haber cycle. Forming the solid gives it out. Calling a molecule with polar bonds polar. Check whether the bond dipoles cancel. Check yourself How many valence electrons are in the hydroxide ion, OH − \text{OH}^- OH − ? Stuck? Show a hint Show the answerWhich has the larger lattice enthalpy, KF or CaO? Stuck? Show a hint Show the answerWhat is the bond order of each oxygen to oxygen bond in ozone? Stuck? Show a hint Show the answerBoron trifluoride has polar bonds. Is the molecule polar? Stuck? Show a hint Show the answer