5,10,15,20-tetraphenyl-21 h,23 h-porphine nickel(ii)

5, 10, 15, 20-tetraphenyl-21 h, 23 h-porphine nickel(ii)

molar mass | 673.4 g/mol formula | (C_44H_30N_4Ni)^2+ empirical formula | Ni_C_44N_4H_30 SMILES identifier | C1=CC=C(C=C1)C2=C3C=CC(=C(C4=CC=CC=C4)C5=NC(=C(C6=CC=CC=C6)C7=CC=C(C(=C8C=CC2=N8)C9=CC=CC=C9)N7)C=C5)N3.[Ni+2] InChI identifier | InChI=1/C44H30N4.Ni/c1-5-13-29(14-6-1)41-33-21-23-35(45-33)42(30-15-7-2-8-16-30)37-25-27-39(47-37)44(32-19-11-4-12-20-32)40-28-26-38(48-40)43(31-17-9-3-10-18-31)36-24-22-34(41)46-36;/h1-28, 45, 48H;/q;+2 InChI key | QYYOZCCOYWFLGO-UHFFFAOYSA-N

vertex count | 49 edge count | 58 Schultz index | 34224 Wiener index | 7428 Hosoya index | (data not available) Balaban index | 1.062

longest chain length | 17 atoms longest straight chain length | 0 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 44 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for 5, 10, 15, 20-tetraphenyl-21 h, 23 h-porphine 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_44H_30N_4Ni)^2+ Use the chemical formula, (C_44H_30N_4Ni)^2+, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Ni (nickel) | 1 C (carbon) | 44 N (nitrogen) | 4 H (hydrogen) | 30 N_atoms = 1 + 44 + 4 + 30 = 79 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 Ni (nickel) | 1 | 1/79 C (carbon) | 44 | 44/79 N (nitrogen) | 4 | 4/79 H (hydrogen) | 30 | 30/79 Check: 1/79 + 44/79 + 4/79 + 30/79 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Ni (nickel) | 1 | 1/79 × 100% = 1.27% C (carbon) | 44 | 44/79 × 100% = 55.7% N (nitrogen) | 4 | 4/79 × 100% = 5.06% H (hydrogen) | 30 | 30/79 × 100% = 38.0% 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 Ni (nickel) | 1 | 1.27% | 58.6934 C (carbon) | 44 | 55.7% | 12.011 N (nitrogen) | 4 | 5.06% | 14.007 H (hydrogen) | 30 | 38.0% | 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 Ni (nickel) | 1 | 1.27% | 58.6934 | 1 × 58.6934 = 58.6934 C (carbon) | 44 | 55.7% | 12.011 | 44 × 12.011 = 528.484 N (nitrogen) | 4 | 5.06% | 14.007 | 4 × 14.007 = 56.028 H (hydrogen) | 30 | 38.0% | 1.008 | 30 × 1.008 = 30.240 m = 58.6934 u + 528.484 u + 56.028 u + 30.240 u = 673.4454 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 Ni (nickel) | 1 | 1.27% | 58.6934/673.4454 C (carbon) | 44 | 55.7% | 528.484/673.4454 N (nitrogen) | 4 | 5.06% | 56.028/673.4454 H (hydrogen) | 30 | 38.0% | 30.240/673.4454 Check: 58.6934/673.4454 + 528.484/673.4454 + 56.028/673.4454 + 30.240/673.4454 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Ni (nickel) | 1 | 1.27% | 58.6934/673.4454 × 100% = 8.715% C (carbon) | 44 | 55.7% | 528.484/673.4454 × 100% = 78.47% N (nitrogen) | 4 | 5.06% | 56.028/673.4454 × 100% = 8.320% H (hydrogen) | 30 | 38.0% | 30.240/673.4454 × 100% = 4.490%

The first step in finding the oxidation states (or oxidation numbers) in 5, 10, 15, 20-tetraphenyl-21 h, 23 h-porphine nickel(ii) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 5, 10, 15, 20-tetraphenyl-21 h, 23 h-porphine 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 8 carbon-nitrogen bonds, and 48 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-nitrogen bonds: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in these bonds will go to nitrogen. Decrease the oxidation number for nitrogen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon 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 | N (nitrogen) | 4 -1 | C (carbon) | 28 0 | C (carbon) | 8 +1 | C (carbon) | 6 | H (hydrogen) | 30 +2 | C (carbon) | 2 | Ni (nickel) | 1