barium sulfite dichloro palladium

barium sulfite dichloro palladium

molar mass | 394.7 g/mol formula | BaCl_2O_3PdS empirical formula | Cl_2Pd_S_O_3Ba_ SMILES identifier | [Ba+2].Cl[Pd]Cl.O=S([O-])[O-] InChI identifier | InChI=1/Ba.2ClH.H2O3S.Pd/c;;;1-4(2)3;/h;2*1H;(H2, 1, 2, 3);/q+2;;;;+2/p-4/fBa.2Cl.O3S.Pd/h;2*1h;;/qm;2*-1;-2;m InChI key | QWOXUGIROWXGRH-UHFFFAOYSA-J

Structure diagram

longest chain length | 3 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 0 atoms H-bond acceptor count | 3 atoms H-bond donor count | 0 atoms

Find the elemental composition for barium sulfite dichloro palladium 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: BaCl_2O_3PdS 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) | 2 Pd (palladium) | 1 S (sulfur) | 1 O (oxygen) | 3 Ba (barium) | 1 N_atoms = 2 + 1 + 1 + 3 + 1 = 8 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) | 2 | 2/8 Pd (palladium) | 1 | 1/8 S (sulfur) | 1 | 1/8 O (oxygen) | 3 | 3/8 Ba (barium) | 1 | 1/8 Check: 2/8 + 1/8 + 1/8 + 3/8 + 1/8 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/8 × 100% = 25.0% Pd (palladium) | 1 | 1/8 × 100% = 12.5% S (sulfur) | 1 | 1/8 × 100% = 12.5% O (oxygen) | 3 | 3/8 × 100% = 37.5% Ba (barium) | 1 | 1/8 × 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) | 2 | 25.0% | 35.45 Pd (palladium) | 1 | 12.5% | 106.42 S (sulfur) | 1 | 12.5% | 32.06 O (oxygen) | 3 | 37.5% | 15.999 Ba (barium) | 1 | 12.5% | 137.327 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) | 2 | 25.0% | 35.45 | 2 × 35.45 = 70.90 Pd (palladium) | 1 | 12.5% | 106.42 | 1 × 106.42 = 106.42 S (sulfur) | 1 | 12.5% | 32.06 | 1 × 32.06 = 32.06 O (oxygen) | 3 | 37.5% | 15.999 | 3 × 15.999 = 47.997 Ba (barium) | 1 | 12.5% | 137.327 | 1 × 137.327 = 137.327 m = 70.90 u + 106.42 u + 32.06 u + 47.997 u + 137.327 u = 394.704 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) | 2 | 25.0% | 70.90/394.704 Pd (palladium) | 1 | 12.5% | 106.42/394.704 S (sulfur) | 1 | 12.5% | 32.06/394.704 O (oxygen) | 3 | 37.5% | 47.997/394.704 Ba (barium) | 1 | 12.5% | 137.327/394.704 Check: 70.90/394.704 + 106.42/394.704 + 32.06/394.704 + 47.997/394.704 + 137.327/394.704 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 25.0% | 70.90/394.704 × 100% = 17.96% Pd (palladium) | 1 | 12.5% | 106.42/394.704 × 100% = 26.96% S (sulfur) | 1 | 12.5% | 32.06/394.704 × 100% = 8.123% O (oxygen) | 3 | 37.5% | 47.997/394.704 × 100% = 12.16% Ba (barium) | 1 | 12.5% | 137.327/394.704 × 100% = 34.79%

The first step in finding the oxidation states (or oxidation numbers) in barium sulfite dichloro palladium is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 2 chlorine-palladium bonds, and 3 oxygen-sulfur bonds in barium sulfite dichloro palladium. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the chlorine-palladium bonds: element | electronegativity (Pauling scale) | Cl | 3.16 | Pd | 2.20 | | | Since chlorine is more electronegative than palladium, the electrons in these bonds will go to chlorine. Decrease the oxidation number for chlorine in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for palladium accordingly: Next look at the oxygen-sulfur bonds: element | electronegativity (Pauling scale) | O | 3.44 | S | 2.58 | | | Since oxygen is more electronegative than sulfur, the electrons in these bonds will go to oxygen: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 3 -1 | Cl (chlorine) | 2 +2 | Ba (barium) | 1 | Pd (palladium) | 1 +4 | S (sulfur) | 1

vertex count | 8 edge count | 5 Schultz index | Wiener index | Hosoya index | Balaban index |