bismuth silicate tetrasodium

bismuth silicate tetrasodium

molar mass | 393 g/mol formula | (BiNa_4O_4Si)^3+ empirical formula | Na_4Si_O_4Bi_ SMILES identifier | [Bi+3].[Na+].[Na+].[Na+].[Na+].[O-][Si]([O-])([O-])[O-] InChI identifier | InChI=1/Bi.4Na.O4Si/c;;;;;1-5(2, 3)4/q+3;4*+1;-4 InChI key | VEYDVEWDEPKHKQ-UHFFFAOYSA-N

vertex count | 10 edge count | 4 Schultz index | 64 Wiener index | 16 Hosoya index | 5 Balaban index | 3.024

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 | 4 atoms H-bond donor count | 0 atoms

Find the elemental composition for bismuth silicate tetrasodium 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: (BiNa_4O_4Si)^3+ 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 Na (sodium) | 4 Si (silicon) | 1 O (oxygen) | 4 Bi (bismuth) | 1 N_atoms = 4 + 1 + 4 + 1 = 10 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 Na (sodium) | 4 | 4/10 Si (silicon) | 1 | 1/10 O (oxygen) | 4 | 4/10 Bi (bismuth) | 1 | 1/10 Check: 4/10 + 1/10 + 4/10 + 1/10 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Na (sodium) | 4 | 4/10 × 100% = 40.0% Si (silicon) | 1 | 1/10 × 100% = 10.00% O (oxygen) | 4 | 4/10 × 100% = 40.0% Bi (bismuth) | 1 | 1/10 × 100% = 10.00% 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 Na (sodium) | 4 | 40.0% | 22.98976928 Si (silicon) | 1 | 10.00% | 28.085 O (oxygen) | 4 | 40.0% | 15.999 Bi (bismuth) | 1 | 10.00% | 208.98040 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 Na (sodium) | 4 | 40.0% | 22.98976928 | 4 × 22.98976928 = 91.95907712 Si (silicon) | 1 | 10.00% | 28.085 | 1 × 28.085 = 28.085 O (oxygen) | 4 | 40.0% | 15.999 | 4 × 15.999 = 63.996 Bi (bismuth) | 1 | 10.00% | 208.98040 | 1 × 208.98040 = 208.98040 m = 91.95907712 u + 28.085 u + 63.996 u + 208.98040 u = 393.02047712 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 Na (sodium) | 4 | 40.0% | 91.95907712/393.02047712 Si (silicon) | 1 | 10.00% | 28.085/393.02047712 O (oxygen) | 4 | 40.0% | 63.996/393.02047712 Bi (bismuth) | 1 | 10.00% | 208.98040/393.02047712 Check: 91.95907712/393.02047712 + 28.085/393.02047712 + 63.996/393.02047712 + 208.98040/393.02047712 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Na (sodium) | 4 | 40.0% | 91.95907712/393.02047712 × 100% = 23.40% Si (silicon) | 1 | 10.00% | 28.085/393.02047712 × 100% = 7.146% O (oxygen) | 4 | 40.0% | 63.996/393.02047712 × 100% = 16.28% Bi (bismuth) | 1 | 10.00% | 208.98040/393.02047712 × 100% = 53.17%

The first step in finding the oxidation states (or oxidation numbers) in bismuth silicate tetrasodium is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 4 oxygen-silicon bonds in bismuth silicate tetrasodium. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the oxygen-silicon bonds: element | electronegativity (Pauling scale) | O | 3.44 | Si | 1.90 | | | Since oxygen is more electronegative than silicon, the electrons in these bonds will go to oxygen. Decrease the oxidation number for oxygen in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for silicon accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 4 +1 | Na (sodium) | 4 +3 | Bi (bismuth) | 1 +4 | Si (silicon) | 1

hybridization | element | count sp^3 | O (oxygen) | 4 | Si (silicon) | 1

Orbital hybridization Structure diagram

vertex count | 10 edge count | 4 Schultz index | 64 Wiener index | 16 Hosoya index | 5 Balaban index | 3.024