trichloro(pentamethylcyclopentadienyl)titanium(iv)

trichloro(pentamethylcyclopentadienyl)titanium(iv)

molar mass | 289.5 g/mol formula | C_10H_15Cl_3Ti empirical formula | Cl_3Ti_C_10H_15 SMILES identifier | CC1=C(C)C(C)(C(=C1C)C)[Ti](Cl)(Cl)Cl InChI identifier | InChI=1/C10H15.3ClH.Ti/c1-6-7(2)9(4)10(5)8(6)3;;;;/h1-5H3;3*1H;/q;;;;+3/p-3/fC10H15.3Cl.Ti/h;3*1h;/q;3*-1;m InChI key | QCEOZLISXJGWSW-UHFFFAOYSA-K

vertex count | 14 edge count | 14 Schultz index | 982 Wiener index | 252 Hosoya index | 357 Balaban index | 3.109

longest chain length | 6 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 trichloro(pentamethylcyclopentadienyl)titanium(iv) 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: C_10H_15Cl_3Ti 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 Cl (chlorine) | 3 Ti (titanium) | 1 C (carbon) | 10 H (hydrogen) | 15 N_atoms = 3 + 1 + 10 + 15 = 29 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 Cl (chlorine) | 3 | 3/29 Ti (titanium) | 1 | 1/29 C (carbon) | 10 | 10/29 H (hydrogen) | 15 | 15/29 Check: 3/29 + 1/29 + 10/29 + 15/29 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 3 | 3/29 × 100% = 10.3% Ti (titanium) | 1 | 1/29 × 100% = 3.45% C (carbon) | 10 | 10/29 × 100% = 34.5% H (hydrogen) | 15 | 15/29 × 100% = 51.7% 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 Cl (chlorine) | 3 | 10.3% | 35.45 Ti (titanium) | 1 | 3.45% | 47.867 C (carbon) | 10 | 34.5% | 12.011 H (hydrogen) | 15 | 51.7% | 1.008 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 Cl (chlorine) | 3 | 10.3% | 35.45 | 3 × 35.45 = 106.35 Ti (titanium) | 1 | 3.45% | 47.867 | 1 × 47.867 = 47.867 C (carbon) | 10 | 34.5% | 12.011 | 10 × 12.011 = 120.110 H (hydrogen) | 15 | 51.7% | 1.008 | 15 × 1.008 = 15.120 m = 106.35 u + 47.867 u + 120.110 u + 15.120 u = 289.447 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 Cl (chlorine) | 3 | 10.3% | 106.35/289.447 Ti (titanium) | 1 | 3.45% | 47.867/289.447 C (carbon) | 10 | 34.5% | 120.110/289.447 H (hydrogen) | 15 | 51.7% | 15.120/289.447 Check: 106.35/289.447 + 47.867/289.447 + 120.110/289.447 + 15.120/289.447 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 3 | 10.3% | 106.35/289.447 × 100% = 36.74% Ti (titanium) | 1 | 3.45% | 47.867/289.447 × 100% = 16.54% C (carbon) | 10 | 34.5% | 120.110/289.447 × 100% = 41.50% H (hydrogen) | 15 | 51.7% | 15.120/289.447 × 100% = 5.224%

The first step in finding the oxidation states (or oxidation numbers) in trichloro(pentamethylcyclopentadienyl)titanium(iv) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In trichloro(pentamethylcyclopentadienyl)titanium(iv) hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond: With hydrogen out of the way, look at the remaining bonds. There are 1 carbon-titanium bond, 3 chlorine-titanium bonds, and 10 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-titanium bond: element | electronegativity (Pauling scale) | C | 2.55 | Ti | 1.54 | | | Since carbon is more electronegative than titanium, the electrons in this bond will go to carbon. Decrease the oxidation number for carbon (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for titanium accordingly: Next look at the chlorine-titanium bonds: element | electronegativity (Pauling scale) | Cl | 3.16 | Ti | 1.54 | | | Since chlorine is more electronegative than titanium, the electrons in these bonds will go to chlorine: Next look at the carbon-carbon bonds: element | electronegativity (Pauling scale) | C | 2.55 | C | 2.55 | | | Since these elements are the same the bonding electrons are shared equally, and there is no change to the oxidation states: Now summarize the results: Answer: | | oxidation state | element | count -3 | C (carbon) | 5 -1 | C (carbon) | 1 | Cl (chlorine) | 3 0 | C (carbon) | 4 +1 | H (hydrogen) | 15 +4 | Ti (titanium) | 1

vertex count | 29 edge count | 29 Schultz index | 5950 Wiener index | 1599 Hosoya index | 86424 Balaban index | 4.342