370 likes | 831 Views
The Nomenclature of Binary Compounds. Formula of Binary Compounds (Compounds composed of two elements). (1). Shows the number of atoms of each type that are found in the compound. These numbers are indicated by subscripts Al 2 O 3 NaCl S 4 Cl 2 SF 6 Subscripts
E N D
Formula of Binary Compounds(Compounds composed of two elements) (1). Shows the number of atoms of each type that are found in the compound. These numbers are indicated by subscripts Al2O3 NaCl S4Cl2 SF6 Subscripts (2). In the formulas the less electronegative atom is written first
Compounds Covalent (Molecular) Ionic I-Cl NaCl Cl- Cl- Cl- Na+ Na+ Cl- Cl1- Na+ Na+ Na+
NO NO2 N2O3 N2O5 Na2O N2O
NaCl CO CO2 Be able to summerize the nomenclature of binary ionic and binary covalent ( i.e. molecular) compounds contrasting differences in the way that the two classes of compounds are named.
Nomenclature of Binary Compounds ionic compounds (typically compounds of metallic and non-metallic elements) Name the Name the less electronegative element more electronegative element Name of the Element Root + ide Prefixes like di, or tri (indicating the number of atoms) are never used when naming ionic compounds
Nomenclature of Binary Compounds Covalent (molecular) compounds (typically compounds of non-metallic elements) Name the Name the less electronegative element more electronegative element Name of the Element Prefix+ Root + ide Prefix+ prefix: used to indicate the number of less electronegative atoms ( if more than one) and is always used to indicate the number of more electronegative atoms
Brom Arsen RootRoot O oxygen ox Br Bromine Cl chlorine chlor As Arsenic N nitrogen nitr Te Tellurium H hydrogen hydr I iodine Se selenium P phosphorus F fluorine S sulfur C carbon tellur Iod Selen Phosph fluor sulf carb
Numerical Prefixes 1 mono 8 octa 2 di 9 nona 3 tri 10 deca 4 tetra 11 undeca 5 penta 12 dodeca 6 hexa 13 trideca 7 hepta 14 tetradeca
Name the following binary compounds diarsenic trisulfide Sodium sulfide As2S3 Na2S Mo2Cl8 (name as a covalent compound) Al4C3 Cl2O7 B3N3 ICl RbCl MgI2 SF6 KF dimolydenum octachloride Aluminum carbide dichlorine heptoxide Triboron trinitride Iodine monochloride rubidium chloride magnesium iodide Sulfur hexafluoride potassium fluoride
Name the following binary compounds Sulfur trioxide Sodium chloride Iodine monochloride diphosphorus pentasulfide 1. NaCl 13. SO3 2. ICl 14. P2S5 3. Al2O3 4. C2O3 15. SF6 5. Al4C3 16. XeF4 6. CS2 7. CO 8. AlP 17. S4Cl2 9. SO3 10. NI3 11. Ba3N2 12. OF2 Aluminum oxide Dicarbon trioxide Sulfur hexafluoride Xenon tetrafluoride Aluminum carbide Carbon disulfide Carbon monoxide tetrasulfur dichloride Aluminum phosphide Sulfur trioxide Nitrogen triiodide Barium nitride Oxygen difluoride
Name the following binary compounds (name the following metal - containing compounds as covalent compounds) 18. CrO3 19. TiCl4 20. W2Cl8 21. PbO2 Chromium trioxide Titanium tetrachloride ditungsten octachloride lead dioxide
Representative Elements that may form only 1 cation 1A 2A 3A 1+ Named as: Element name+ ion H 2+ 1+ Li Be 3+ Al 2+ 1+ Na Mg (1). lose all electrons from the valence shell 2+ 1+ K Ca 1+ (2). form a cation, Mn+; where n is the number of electrons lost( designated by the Group Nb.) 2+ Rb Sr 1+ 2+ Cs Ba
Non metals and metalloids Forming Monatomic Anions 1A 4A 5A 6A 7A H C N O F Si P S Cl As Se Br Te I • These elements gain one or more electrons into the outer shell forming an anion, An- • The charge, n-, of the anion is the number of electrons gained • The nb of electrons gained is that nb. needed to achieve an octet of electrons in the valance electron shell, i.e. 8-Group Nb.
Non metals and metalloids Forming Monatomic Anions 1A 4A 5A 6A 7A H1- C4- N3- O2- F1- Si4- P3- S2- Cl1- As3- Se2- Br1- Te2- I1- • These elements gain one or more electrons into the outer shell forming an anion, An- • The charge, n-, of the anion is the number of electrons gained • The nb of electrons gained is that nb. needed to achieve an octet of electrons in the valance electron shell, i.e. 8-Group Nb.
Nomenclature of Monatomic Anions Root + ide identifies the element identifies the ion as a monatomic anion
Representative Elements that may form only 1 cation 1A 2A 3A 1+ Named as: Element name+ ion H 2+ 1+ Li Be 3+ Al 2+ 1+ Na Mg (1). lose all electrons from the valence shell 2+ 1+ K Ca 1+ (2). form a cation, Mn+; where n is the number of electrons lost( designated by the Group Nb.) 2+ Rb Sr 1+ 2+ Cs Ba
Non metals and metalloids Forming Monatomic Anions 1A 4A 5A 6A 7A H C N O F Si P S Cl As Se Br Te I H1- C4- N3- O2- F1- Si4- P3- S2- Cl1- As3- Se2- Br1- Te2- I1- Names: Root + ide • These elements gain one or more electrons into the outer shell forming an anion, An- • The charge, n-, of the anion is the number of electrons gained • The nb of electrons gained is that nb. needed to achieve an octet of electrons in the valance electron shell, i.e. 8-Group Nb.
The reaction of the elements Li and F Li F 9+ 3+ Li1+ F1-
Predicting the Formula of Binary Ionic Compounds (Formed Through the Reaction of the Elements) • Determine the identity of the cation and the anion that the metal and non-metal would form. • Determine the stoichiometry of the • formula unit for the compound.
Following the outline given on the previous slide determine the formula of the following combinations of elements Al and O Ba and Se
Predicting the Formula of Binary Ionic Compounds (Formed Through the Reaction of the Elements) Al and O • Determine the identity of the cation and the anion that the metal and non-metal would form. Al3+ O2- • Determine the stoichiometry of the formula unit for the compound. Al3+O2- Al2O3 2 3
Predicting the Formula of Binary Ionic Compounds (Formed Through the Reaction of the Elements) Ba and Se • Determine the identity of the cation and the anion that the metal and non-metal would form. Ba2+ Se2- • Determine the stoichiometry of the formula unit for the compound. Ba2Se2 Ba2+O2- 2 2 BaSe
Give formula for, and name, the binary ionic compounds that would be formed through the reaction of the following pairs of elements: Formula Ions Formed CaCl2 1. Ca and Cl Ca2+ Cl1- Calcium chloride 2. C and Al Al3+ C4- Al4C3 C4- Al3+ Aluminum carbide 3. Na and F Na1+ F1- NaF Sodium fluoride 4. Ba and O Ba2- O2- BaO Barium oxide
Give formula for, and name, the binary ionic compounds that would be formed through the reaction of the following pairs of elements: Formula Ions Formed Mg3N2 Mg2+ N3- N3- Mg2- 5. N and Mg 6. Li and Se 7. Ga and As Magnesium nitride Li2Se Li1+ Se2- Lithium selenide GaAs Ga3+ As3- Gallium arsenide
Metals that form more than 1 cation: Transition and Post-Transition Metals Transition MetalsPost Transition Metals 3B 4B 5B 6B 7B 8B 1B 2B 3A 4B 5A Sc Tii V Cr Mn Fe Co Ni Cu Zn Ga Ag Cd In Sn Au Hg Tl Pb Bi
Metals that form more than 1 cation:Transiton Metals 3B 4B 5B 6B 7B 8B 1B 2B Sc2+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu1+ Zn2+ Cation of Lower Charge: Formed by losing Ag1+ Cd2+ all electrons from the outermost electron shell ( 2 e- for all, except for Group 1B Au1+ Hg1+ where 1 e- is lost)
Metals that form more than 1 cation:Transiton Metals 3B 4B 5B 6B 7B 8B 1B 2B Sc3+ Ti3+ V3+ Cr3+ Mn3+ Fe3+ Co3+ Ni3+ Cu2+ Zn3+ Ag2+Cd3+ Cation of Higher Charge: Possesses a charge Au2+ Hg2+ 1 higher than the cation of lower charge (ions in red are uncommon)
Nomenclature of Transition and Post-Transition Metal Ions I. Stock System Name of Metal (Ionic Charge:Roman Numerals) Fe2+ iron(II) ion Sn4+ tin(IV) ion Cu1+ copper(I) ion Cr3+ chromium(III) ion
II. Older Nomenclature -ic (ion of higher charge) Root + endings identifies the metal-ous (ion of lower charge) Fe2+ferrous ion Sn4+stannic ion Cu1+cuprous ion Cr3+chromic ion
Fe iron ferrum • Cu copper cuprum • Sn tin stannum • Pb lead plumbum • Ag silver argentenum • Au gold aurum
Name the following compounds: Fe2O3 Iron(III) or ferric oxide CuCl Copper (1) or cuprous chloride PbCl4 Lead (IV) or plumbic chloride SnO Tin (II) or stannous oxide Iron (II) or ferrous sulfide FeS AgO Silver (II) or argentenic oxide