2,3,4,5,6-pentafluorobenzylzinc chloride

2, 3, 4, 5, 6-pentafluorobenzylzinc chloride

molar mass | 281.9 g/mol formula | C_7H_2ClF_5Zn empirical formula | Cl_F_5C_7Zn_H_2 SMILES identifier | C(C1=C(C(=C(C(=C1F)F)F)F)F)[Zn+].[Cl-] InChI identifier | InChI=1/C7H2F5.ClH.Zn/c1-2-3(8)5(10)7(12)6(11)4(2)9;;/h1H2;1H;/q;;+1/p-1/fC7H2F5.Cl.Zn/h;1h;/q;-1;m InChI key | GGNRNMYEDWCPQV-UHFFFAOYSA-M

vertex count | 14 edge count | 13 Schultz index | 869 Wiener index | 220 Hosoya index | 326 Balaban index | 2.787

longest chain length | 7 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 2, 3, 4, 5, 6-pentafluorobenzylzinc chloride 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_7H_2ClF_5Zn 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) | 1 F (fluorine) | 5 C (carbon) | 7 Zn (zinc) | 1 H (hydrogen) | 2 N_atoms = 1 + 5 + 7 + 1 + 2 = 16 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) | 1 | 1/16 F (fluorine) | 5 | 5/16 C (carbon) | 7 | 7/16 Zn (zinc) | 1 | 1/16 H (hydrogen) | 2 | 2/16 Check: 1/16 + 5/16 + 7/16 + 1/16 + 2/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/16 × 100% = 6.25% F (fluorine) | 5 | 5/16 × 100% = 31.3% C (carbon) | 7 | 7/16 × 100% = 43.8% Zn (zinc) | 1 | 1/16 × 100% = 6.25% H (hydrogen) | 2 | 2/16 × 100% = 12.5% 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) | 1 | 6.25% | 35.45 F (fluorine) | 5 | 31.3% | 18.998403163 C (carbon) | 7 | 43.8% | 12.011 Zn (zinc) | 1 | 6.25% | 65.38 H (hydrogen) | 2 | 12.5% | 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) | 1 | 6.25% | 35.45 | 1 × 35.45 = 35.45 F (fluorine) | 5 | 31.3% | 18.998403163 | 5 × 18.998403163 = 94.992015815 C (carbon) | 7 | 43.8% | 12.011 | 7 × 12.011 = 84.077 Zn (zinc) | 1 | 6.25% | 65.38 | 1 × 65.38 = 65.38 H (hydrogen) | 2 | 12.5% | 1.008 | 2 × 1.008 = 2.016 m = 35.45 u + 94.992015815 u + 84.077 u + 65.38 u + 2.016 u = 281.915015815 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) | 1 | 6.25% | 35.45/281.915015815 F (fluorine) | 5 | 31.3% | 94.992015815/281.915015815 C (carbon) | 7 | 43.8% | 84.077/281.915015815 Zn (zinc) | 1 | 6.25% | 65.38/281.915015815 H (hydrogen) | 2 | 12.5% | 2.016/281.915015815 Check: 35.45/281.915015815 + 94.992015815/281.915015815 + 84.077/281.915015815 + 65.38/281.915015815 + 2.016/281.915015815 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 6.25% | 35.45/281.915015815 × 100% = 12.57% F (fluorine) | 5 | 31.3% | 94.992015815/281.915015815 × 100% = 33.70% C (carbon) | 7 | 43.8% | 84.077/281.915015815 × 100% = 29.82% Zn (zinc) | 1 | 6.25% | 65.38/281.915015815 × 100% = 23.19% H (hydrogen) | 2 | 12.5% | 2.016/281.915015815 × 100% = 0.7151%

The first step in finding the oxidation states (or oxidation numbers) in 2, 3, 4, 5, 6-pentafluorobenzylzinc chloride is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2, 3, 4, 5, 6-pentafluorobenzylzinc chloride 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 5 carbon-fluorine bonds, 1 carbon-zinc bond, and 7 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-fluorine bonds: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, 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 carbon accordingly: Next look at the carbon-zinc bond: element | electronegativity (Pauling scale) | C | 2.55 | Zn | 1.65 | | | Since carbon is more electronegative than zinc, the electrons in this bond will go to carbon: 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 -1 | Cl (chlorine) | 1 | F (fluorine) | 5 0 | C (carbon) | 1 +1 | C (carbon) | 5 | H (hydrogen) | 2 +2 | Zn (zinc) | 1

vertex count | 16 edge count | 15 Schultz index | 1239 Wiener index | 318 Hosoya index | 582 Balaban index | 2.969