potassium trans-1-decenyltrifluoroborate

potassium trans-1-decenyltrifluoroborate

molar mass | 246.2 g/mol formula | C_10H_19BF_3K empirical formula | C_10B_F_3K_H_19 SMILES identifier | CCCCCCCC/C=C/[B-](F)(F)F.[K+] InChI identifier | InChI=1/C10H19BF3.K/c1-2-3-4-5-6-7-8-9-10-11(12, 13)14;/h9-10H, 2-8H2, 1H3;/q-1;+1/b10-9+; InChI key | SJPMLIRYZHTWRI-RRABGKBLSA-N

vertex count | 15 edge count | 13 Schultz index | 1570 Wiener index | 424 Hosoya index | 411 Balaban index | 2.994

longest chain length | 12 atoms longest straight chain length | 12 atoms longest aliphatic chain length | 10 atoms aromatic atom count | 0 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for potassium trans-1-decenyltrifluoroborate 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_19BF_3K Use the chemical formula, C_10H_19BF_3K, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms C (carbon) | 10 B (boron) | 1 F (fluorine) | 3 K (potassium) | 1 H (hydrogen) | 19 N_atoms = 10 + 1 + 3 + 1 + 19 = 34 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) | 10 | 10/34 B (boron) | 1 | 1/34 F (fluorine) | 3 | 3/34 K (potassium) | 1 | 1/34 H (hydrogen) | 19 | 19/34 Check: 10/34 + 1/34 + 3/34 + 1/34 + 19/34 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 10 | 10/34 × 100% = 29.4% B (boron) | 1 | 1/34 × 100% = 2.94% F (fluorine) | 3 | 3/34 × 100% = 8.82% K (potassium) | 1 | 1/34 × 100% = 2.94% H (hydrogen) | 19 | 19/34 × 100% = 55.9% 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) | 10 | 29.4% | 12.011 B (boron) | 1 | 2.94% | 10.81 F (fluorine) | 3 | 8.82% | 18.998403163 K (potassium) | 1 | 2.94% | 39.0983 H (hydrogen) | 19 | 55.9% | 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) | 10 | 29.4% | 12.011 | 10 × 12.011 = 120.110 B (boron) | 1 | 2.94% | 10.81 | 1 × 10.81 = 10.81 F (fluorine) | 3 | 8.82% | 18.998403163 | 3 × 18.998403163 = 56.995209489 K (potassium) | 1 | 2.94% | 39.0983 | 1 × 39.0983 = 39.0983 H (hydrogen) | 19 | 55.9% | 1.008 | 19 × 1.008 = 19.152 m = 120.110 u + 10.81 u + 56.995209489 u + 39.0983 u + 19.152 u = 246.165509489 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) | 10 | 29.4% | 120.110/246.165509489 B (boron) | 1 | 2.94% | 10.81/246.165509489 F (fluorine) | 3 | 8.82% | 56.995209489/246.165509489 K (potassium) | 1 | 2.94% | 39.0983/246.165509489 H (hydrogen) | 19 | 55.9% | 19.152/246.165509489 Check: 120.110/246.165509489 + 10.81/246.165509489 + 56.995209489/246.165509489 + 39.0983/246.165509489 + 19.152/246.165509489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 10 | 29.4% | 120.110/246.165509489 × 100% = 48.79% B (boron) | 1 | 2.94% | 10.81/246.165509489 × 100% = 4.391% F (fluorine) | 3 | 8.82% | 56.995209489/246.165509489 × 100% = 23.15% K (potassium) | 1 | 2.94% | 39.0983/246.165509489 × 100% = 15.88% H (hydrogen) | 19 | 55.9% | 19.152/246.165509489 × 100% = 7.780%

The first step in finding the oxidation states (or oxidation numbers) in potassium trans-1-decenyltrifluoroborate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In potassium trans-1-decenyltrifluoroborate 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 boron-carbon bond, 3 boron-fluorine bonds, and 9 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the boron-carbon bond: element | electronegativity (Pauling scale) | B | 2.04 | C | 2.55 | | | Since carbon is more electronegative than boron, 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 boron accordingly: Next look at the boron-fluorine bonds: element | electronegativity (Pauling scale) | B | 2.04 | F | 3.98 | | | Since fluorine is more electronegative than boron, the electrons in these bonds will go to fluorine: 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) | 1 -2 | C (carbon) | 8 -1 | C (carbon) | 1 | F (fluorine) | 3 +1 | H (hydrogen) | 19 | K (potassium) | 1 +3 | B (boron) | 1

hybridization | element | count sp^2 | C (carbon) | 2 sp^3 | B (boron) | 1 | C (carbon) | 8 | F (fluorine) | 3

Orbital hybridization Structure diagram

vertex count | 34 edge count | 32 Schultz index | 10128 Wiener index | 2750 Hosoya index | 417778 Balaban index | 6.829