tetrafluoro copper

tetrafluoro copper

molar mass | 139.5 g/mol formula | CuF_4 empirical formula | F_4Cu_ SMILES identifier | [Cu](F)(F)(F)F InChI identifier | InChI=1/Cu.4FH/h;4*1H/q+4;;;;/p-4/fCu.4F/h;4*1h/qm;4*-1 InChI key | LELTVLRYLALSFG-UHFFFAOYSA-J

vertex count | 5 edge count | 4 Schultz index | 64 Wiener index | 16 Hosoya index | 5 Balaban index | 3.024

longest chain length | 3 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for tetrafluoro copper in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: CuF_4 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms F (fluorine) | 4 Cu (copper) | 1 N_atoms = 4 + 1 = 5 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction F (fluorine) | 4 | 4/5 Cu (copper) | 1 | 1/5 Check: 4/5 + 1/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 4 | 4/5 × 100% = 80.0% Cu (copper) | 1 | 1/5 × 100% = 20.0% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u F (fluorine) | 4 | 80.0% | 18.998403163 Cu (copper) | 1 | 20.0% | 63.546 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u F (fluorine) | 4 | 80.0% | 18.998403163 | 4 × 18.998403163 = 75.993612652 Cu (copper) | 1 | 20.0% | 63.546 | 1 × 63.546 = 63.546 m = 75.993612652 u + 63.546 u = 139.539612652 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction F (fluorine) | 4 | 80.0% | 75.993612652/139.539612652 Cu (copper) | 1 | 20.0% | 63.546/139.539612652 Check: 75.993612652/139.539612652 + 63.546/139.539612652 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 4 | 80.0% | 75.993612652/139.539612652 × 100% = 54.46% Cu (copper) | 1 | 20.0% | 63.546/139.539612652 × 100% = 45.54%

The first step in finding the oxidation states (or oxidation numbers) in tetrafluoro copper is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 4 copper-fluorine bonds in tetrafluoro copper. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the copper-fluorine bonds: element | electronegativity (Pauling scale) | Cu | 1.90 | F | 3.98 | | | Since fluorine is more electronegative than copper, the electrons in these bonds will go to fluorine. Decrease the oxidation number for fluorine in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for copper accordingly: Now summarize the results: Answer: | | oxidation state | element | count -1 | F (fluorine) | 4 +4 | Cu (copper) | 1

vertex count | 5 edge count | 4 Schultz index | 64 Wiener index | 16 Hosoya index | 5 Balaban index | 3.024