bromic acid

bromic acid

molar mass | 128.9 g/mol formula | BrHO_3 empirical formula | Br_O_3H_ SMILES identifier | Br(=O)(=O)O InChI identifier | InChI=1/BrHO3/c2-1(3)4/h(H, 2, 3, 4)/f/h2H InChI key | SXDBWCPKPHAZSM-UHFFFAOYSA-N

Draw the Lewis structure of SMILES:Br(=O)(=O)O. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: n_Br, val + n_H, val + 3 n_O, val = 26 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): n_Br, full + n_H, full + 3 n_O, full = 34 Subtracting these two numbers shows that 34 - 26 = 8 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.20 (hydrogen), 2.96 (bromine), and 3.44 (oxygen). Because the electronegativity of bromine is smaller than the electronegativity of oxygen, expand the valence shell of bromine to 5 bonds. Therefore we add a total of 2 bonds to the diagram: Answer: | |

melting point | -58.61 °C boiling point | 63.59 °C critical temperature | 493.4 K critical pressure | 11.84 MPa critical volume | 188.5 cm^3/mol molar heat of vaporization | 50.5 kJ/mol molar heat of fusion | 12.37 kJ/mol molar enthalpy | -664.4 kJ/mol molar free energy | -675 kJ/mol (computed using the Joback method)

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 | 1 atom

Find the elemental composition for SMILES:Br(=O)(=O)O in terms of the atom and mass percents: atom percent = N_i/N_total × 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_total and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: BrHO_3 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_total: | number of atoms Br (bromine) | 1 O (oxygen) | 3 H (hydrogen) | 1 N_total = 1 + 3 + 1 = 5 Divide each N_i by N_total to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction Br (bromine) | 1 | 1/5 O (oxygen) | 3 | 3/5 H (hydrogen) | 1 | 1/5 Check: 1/5 + 3/5 + 1/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/5 × 100% = 20.0% O (oxygen) | 3 | 3/5 × 100% = 60.0% H (hydrogen) | 1 | 1/5 × 100% = 20.0% 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 Br (bromine) | 1 | 20.0% | 79.904 O (oxygen) | 3 | 60.0% | 15.999 H (hydrogen) | 1 | 20.0% | 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 Br (bromine) | 1 | 20.0% | 79.904 | 1 × 79.904 = 79.904 O (oxygen) | 3 | 60.0% | 15.999 | 3 × 15.999 = 47.997 H (hydrogen) | 1 | 20.0% | 1.008 | 1 × 1.008 = 1.008 m = 79.904 u + 47.997 u + 1.008 u = 128.909 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 Br (bromine) | 1 | 20.0% | 79.904/128.909 O (oxygen) | 3 | 60.0% | 47.997/128.909 H (hydrogen) | 1 | 20.0% | 1.008/128.909 Check: 79.904/128.909 + 47.997/128.909 + 1.008/128.909 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 20.0% | 79.904/128.909 × 100% = 61.98% O (oxygen) | 3 | 60.0% | 47.997/128.909 × 100% = 37.23% H (hydrogen) | 1 | 20.0% | 1.008/128.909 × 100% = 0.7819%

The first step in finding the oxidation states (or oxidation numbers) in SMILES: Br(=O)(=O)O is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In SMILES: Br(=O)(=O)O 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 3 bromine-oxygen bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-oxygen bonds: element | electronegativity (Pauling scale) | Br | 2.96 | O | 3.44 | | | Since oxygen is more electronegative than bromine, 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 bromine accordingly: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 3 +1 | H (hydrogen) | 1 +5 | Br (bromine) | 1

First draw the structure diagram for SMILES:Br(=O)(=O)O, and for every non-hydrogen atom, count the σ-bonds. Note that double and triple bonds consist of one σ-bond together with one or two π-bonds: Identify those atoms with lone pairs: Find the steric number by adding the lone pair count to the number of σ-bonds: Consult the following chart to determine the hybridization from the steric number: steric number | hybridization 2 | sp 3 | sp^2 4 | sp^3 5 | dsp^3 6 | d^2sp^3 7 | d^3sp^3 Now assign the hybridization for each atom: Answer: | |

vertex count | 5 edge count | 4 Schultz index | 68 Wiener index | 18 Hosoya index | 7 Balaban index | 2.54