– Empirical Lewis Diagram Central Atom(s) Central shape(s) Bond dipoles Polar or – formula / Show valence e- lone pairs Atoms(s) around each Non-polar Molecular only formula bonding e- central formula atom – –N – 13 N2H3F(g) N 5 valence e- H N F N has 1 lone pair N has 3 Pyramidal N H polar – │ H 1 valence e- bonding e- and Nitrogen is more F – 7 valence e- H H pyramidal electronegative N N no dipole N F Fluorine is more electronegative 14 C2H5OH(l) C – 4 valence e- H H C has 0 lone pairs C – 4 bonding Tetrahedral C O polar – │ │ O 6 valence e- O has 2 lone pairs e- Tetrahedral Oxygen is more – – – – H 1 valence e- H -- C C O O 2 bonding Angular or electronegative │ e- bent or O H H H H Oxygen is more electronegative H C C is more electronegative – – 15 NH4+ (aq) N 5 valence e- H+ N has 1 lone pair N 3 bonding Tetrahedral N H So polar H – 1 valence e- │ e Nitrogen is more that it H – N--- H One electronegative forms an ion coordinate H covalent bond with the H+ 16 CO32-(aq) C has 4 valence e- O- - C = O C – 0 lone pairs C – 4 bonding Trigonal C O So polar │ – O has 6 valence e- O 2 lone pairs e- pyramidal O is more that it O- has 7 valence e O- O- has 2 lone O – 2 bonding electronegative forms an ion pairs e- C O- O- has 1 O- is more bonding e- electronegative
Empirical formula / Molecular formula Lewis Diagram Show valence e- only formula Central Atom(s) – lone pairs Central Atoms(s) – bonding e- shape(s) around each central atom Bond dipoles Polar or Non-polar 13 N2H3F(g) N – 5 valence e- H – 1 valence e- F – 7 valence e- H –N – N F │ H H N has 1 lone pair N has 3 bonding e- Pyramidal and pyramidal N H Nitrogen is more electronegative N N no dipole N F Fluorine is more electronegative polar 14 C2H5OH(l) C – 4 valence e- O – 6 valence e- H – 1 valence e- H H │ │ H -- C – C – O │ H H H C has 0 lone pairs O has 2 lone pairs C – 4 bonding e- O – 2 bonding e- Tetrahedral Tetrahedral Angular or bent or C O Oxygen is more electronegative O H Oxygen is more electronegative H C C is more electronegative polar 15 NH4+ (aq) N – 5 valence e- H – 1 valence e- H+ │ H – N--- H H N has 1 lone pair N – 3 bonding e One coordinate covalent bond with the H+ Tetrahedral N H Nitrogen is more electronegative So polar that it forms an ion 16 CO32-(aq) C has 4 valence e- O has 6 valence e- O- has 7 valence e O- - C = O │ O- C – 0 lone pairs O – 2 lone pairs O- has 2 lone pairs C – 4 bonding e- O – 2 bonding e- O- has 1 bonding e- Trigonal pyramidal C O O is more electronegative C O- O- is more electronegative So polar that it forms an ion
Teacher_notes_Stereochemistry_Lab.docx