handwritten style SMILES:BrNO

SMILES: BrNO

molar mass | 111.9 g/mol formula | BrH_2NO empirical formula | Br_N_O_H_2 SMILES identifier | BrNO InChI identifier | InChI=1/BrH2NO/c1-2-3/h2-3H InChI key | BIBSEZOIHKWNMQ-UHFFFAOYSA-N

Draw the Lewis structure of SMILES:BrNO. Start by drawing the overall structure of the molecule: Count the total valence electrons of the bromine (n_Br, val = 7), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: n_Br, val + 2 n_H, val + n_N, val + n_O, val = 20 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): n_Br, full + 2 n_H, full + n_N, full + n_O, full = 28 Subtracting these two numbers shows that 28 - 20 = 8 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 4 bonds and hence 8 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 20 - 8 = 12 electrons left to draw: Answer: | |

melting point | -10.11 °C boiling point | 134.8 °C critical temperature | 597.1 K critical pressure | 9.175 MPa critical volume | 151.5 cm^3/mol molar heat of vaporization | 45.1 kJ/mol molar heat of fusion | 10.14 kJ/mol molar enthalpy | -115.8 kJ/mol molar free energy | -83.99 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 | 0 atoms H-bond donor count | 2 atoms

Find the elemental composition for SMILES:BrNO 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: BrH_2NO 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 N (nitrogen) | 1 O (oxygen) | 1 H (hydrogen) | 2 N_total = 1 + 1 + 1 + 2 = 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 N (nitrogen) | 1 | 1/5 O (oxygen) | 1 | 1/5 H (hydrogen) | 2 | 2/5 Check: 1/5 + 1/5 + 1/5 + 2/5 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/5 × 100% = 20.0% N (nitrogen) | 1 | 1/5 × 100% = 20.0% O (oxygen) | 1 | 1/5 × 100% = 20.0% H (hydrogen) | 2 | 2/5 × 100% = 40.0% 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 N (nitrogen) | 1 | 20.0% | 14.007 O (oxygen) | 1 | 20.0% | 15.999 H (hydrogen) | 2 | 40.0% | 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 N (nitrogen) | 1 | 20.0% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 1 | 20.0% | 15.999 | 1 × 15.999 = 15.999 H (hydrogen) | 2 | 40.0% | 1.008 | 2 × 1.008 = 2.016 m = 79.904 u + 14.007 u + 15.999 u + 2.016 u = 111.926 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% | 79.904/111.926 N (nitrogen) | 1 | 20.0% | 14.007/111.926 O (oxygen) | 1 | 20.0% | 15.999/111.926 H (hydrogen) | 2 | 40.0% | 2.016/111.926 Check: 79.904/111.926 + 14.007/111.926 + 15.999/111.926 + 2.016/111.926 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 20.0% | 79.904/111.926 × 100% = 71.39% N (nitrogen) | 1 | 20.0% | 14.007/111.926 × 100% = 12.51% O (oxygen) | 1 | 20.0% | 15.999/111.926 × 100% = 14.29% H (hydrogen) | 2 | 40.0% | 2.016/111.926 × 100% = 1.801%

The first step in finding the oxidation states (or oxidation numbers) in SMILES: BrNO is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In SMILES: BrNO 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 bromine-nitrogen bond, and 1 nitrogen-oxygen bond. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-nitrogen bond: element | electronegativity (Pauling scale) | Br | 2.96 | N | 3.04 | | | Since nitrogen is more electronegative than bromine, the electrons in this bond will go to nitrogen. Decrease the oxidation number for nitrogen (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for bromine accordingly: Next look at the nitrogen-oxygen bond: element | electronegativity (Pauling scale) | N | 3.04 | O | 3.44 | | | Since oxygen is more electronegative than nitrogen, the electrons in this bond will go to oxygen: Now summarize the results: Answer: | | oxidation state | element | count -2 | O (oxygen) | 1 -1 | N (nitrogen) | 1 +1 | Br (bromine) | 1 | H (hydrogen) | 2

First draw the structure diagram for SMILES:BrNO, 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