bis(pentamethylcyclopentadienyl)nickel(ii)

bis(pentamethylcyclopentadienyl)nickel(ii)

molar mass | 329.1 g/mol formula | C_20H_30Ni empirical formula | C_20Ni_H_30 SMILES identifier | CC1=C(C)C(C)(C(=C1C)C)[Ni]C2(C)C(=C(C)C(=C2C)C)C InChI identifier | InChI=1/2C10H15.Ni/c2*1-6-7(2)9(4)10(5)8(6)3;/h2*1-5H3; InChI key | IBWGRWRPKIAFEV-UHFFFAOYSA-N

vertex count | 21 edge count | 22 Schultz index | 3254 Wiener index | 804 Hosoya index | 11480 Balaban index | 2.321

longest chain length | 9 atoms longest straight chain length | 0 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 bis(pentamethylcyclopentadienyl)nickel(ii) 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_20H_30Ni 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 C (carbon) | 20 Ni (nickel) | 1 H (hydrogen) | 30 N_atoms = 20 + 1 + 30 = 51 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 C (carbon) | 20 | 20/51 Ni (nickel) | 1 | 1/51 H (hydrogen) | 30 | 30/51 Check: 20/51 + 1/51 + 30/51 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 20 | 20/51 × 100% = 39.2% Ni (nickel) | 1 | 1/51 × 100% = 1.96% H (hydrogen) | 30 | 30/51 × 100% = 58.8% 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 C (carbon) | 20 | 39.2% | 12.011 Ni (nickel) | 1 | 1.96% | 58.6934 H (hydrogen) | 30 | 58.8% | 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 C (carbon) | 20 | 39.2% | 12.011 | 20 × 12.011 = 240.220 Ni (nickel) | 1 | 1.96% | 58.6934 | 1 × 58.6934 = 58.6934 H (hydrogen) | 30 | 58.8% | 1.008 | 30 × 1.008 = 30.240 m = 240.220 u + 58.6934 u + 30.240 u = 329.1534 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 C (carbon) | 20 | 39.2% | 240.220/329.1534 Ni (nickel) | 1 | 1.96% | 58.6934/329.1534 H (hydrogen) | 30 | 58.8% | 30.240/329.1534 Check: 240.220/329.1534 + 58.6934/329.1534 + 30.240/329.1534 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 20 | 39.2% | 240.220/329.1534 × 100% = 72.98% Ni (nickel) | 1 | 1.96% | 58.6934/329.1534 × 100% = 17.83% H (hydrogen) | 30 | 58.8% | 30.240/329.1534 × 100% = 9.187%

The first step in finding the oxidation states (or oxidation numbers) in bis(pentamethylcyclopentadienyl)nickel(ii) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In bis(pentamethylcyclopentadienyl)nickel(ii) 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 2 carbon-nickel bonds, and 20 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-nickel bonds: element | electronegativity (Pauling scale) | C | 2.55 | Ni | 1.91 | | | Since carbon is more electronegative than nickel, the electrons in these bonds will go to carbon. Decrease the oxidation number for carbon in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for nickel accordingly: 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) | 10 -1 | C (carbon) | 2 0 | C (carbon) | 8 +1 | H (hydrogen) | 30 +2 | Ni (nickel) | 1

vertex count | 51 edge count | 52 Schultz index | 26720 Wiener index | 6990 Hosoya index | 7.063×10^8 Balaban index | 3.641