PBr 3. EXAMPLES FOR DRAWING LEWIS STRUCTURES (see section 4.1 in the textbook for more details and examples)
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1 EXAMPLES FOR DRAWING LEWIS STRUCTURES (see section 4.1 in the textbook for more details and examples) Example 1: Draw the Lewis structure for a phosphorus tribromide PBr 3 The two remaining electrons are paired up on the phosphorus atom. The phosphorus atoms has 5 valence electrons, and each bromine atom has 7 valence electrons (7) = 26 valence electrons The phosphorus atom is surrounded by 8 electrons (3 bonding pairs and 1 lone pair). 6 electrons used (20 electrons remaining) The phosphorus atom "owns" 5 electrons (one in each of the three bonds and the lone pair). Since a neutral phosphorus atom has 5 valence electrons, there is no formal charge on the phosphorus atom. Each bromine atom "owns" 7 electrons (three lone pairs plus one in the bond). Since neutral bromine atoms have 7 valence electrons, there are no formal charges on the bromine atoms. 24 electrons used (2 electrons remaining)
2 Example 2: Draw the Lewis structure for a formaldehyde CH 2 O There are no electrons remaining to put on central atom. The carbon atom has 4 valence electrons, each hydrogen atom has 1 valence electron, and the oxygen has 6 valence electrons (1) + 6 = 12 valence electrons 6 electrons used (6 electrons remaining) The carbon atom is surrounded by only 6 electrons (3 bonding pairs). One of the lone pairs on the oxygen atom must be reassigned as a second bond between oxygen and carbon. The carbon atom "owns" 4 electrons (one from each bond). Since a neutral carbon atom has 4 valence electrons, there is no formal charge on the carbon atom. The oxygen atom "owns" 6 electrons (one from each bond plus the two lone pairs). Since a neutral oxygen atom has 6 valence electrons, there is no formal charge on the oxygen atom. Hydrogen atoms never have a formal charge. 12 electrons used (0 electrons remaining)
3 Example 3: Draw the Lewis structure for a nitrite ion. NO 2 The two remaining electrons are paired up on the nitrogen atom. The nitrogen atom has 5 valence electrons, and each oxygen atom has 6 valence electrons. One more electron must be added to account for the 1 charge (6) + 1 = 18 valence electrons The nitrogen atom is surrounded by only 6 electrons (2 bonding pairs and one lone pair). One of the lone pairs on an oxygen atom must be reassigned as a second bond between an oxygen and the nitrogen. 4 electrons used (14 electrons remaining) The oxygen on the right "owns" 7 electrons (one from the bond and 3 lone pairs). Since a neutral oxygen atom has only 6 valence electrons, there is a formal charge of 1 on the oxygen atom. The formal charge cannot be reduced. The final 16 electrons used (2 electrons remaining)
4 Example 4: Draw the Lewis structure for a krypton tetrafluoride KrF 4 The 4 remaining electrons are paired up on the krypton atom. The krypton atom has 8 valence electrons, and each fluorine atom has 7 valence electrons (7) = 36 valence electrons The krypton atom is surrounded by 12 electrons (4 bonding pairs and 2 lone pairs). The krypton atom has an expanded octet. 8 electrons used (28 electrons remaining) The krypton atom "owns" 8 electrons (one from each bond and two lone pairs). Since a neutral krypton atom has 8 valence electrons, there is no formal charge on the carbon atom. Each fluorine atom "owns" 7 electrons (one from the bond and three lone pairs). Since a neutral fluorine atom has 7 valence electrons, there are no formal charges on the fluorine atoms. 32 electrons used (4 electrons remaining)
5 Example 5: Draw the Lewis structure for a chlorous acid HClO 2 The 4 remaining electrons are paired up on the chlorine atom. The hydrogen atom has 1 valence electron, the chlorine atom has 7 valence electrons, and each oxygen atom has 6 valence electrons. The chlorine atom is surrounded by 8 electrons (2 bonding pairs and 2 lone pairs) (6) = 20 valence electrons Note that in oxoacids (HNO 3, H 2 SO 4, etc) the hydrogen atoms are bonded to oxygen atoms rather than the central atom. 6 electrons used (14 electrons remaining) The oxygen atom on the left "owns" 7 electrons, and therefore has a formal charge of 1. The chlorine atom owns 6 electrons and therefore has a formal charge of electrons used (4 electrons remaining) If one of the lone pairs on the oxygen atom is reassigned as a second bond between the oxygen and the chlorine then the formal charges are eliminated. The chlorine atom has an expanded octet.
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