SMILES:BrF

SMILES: BrF

molar mass | 98.9 g/mol formula | BrF empirical formula | Br_F_ SMILES identifier | BrF InChI identifier | InChI=1/BrF/c1-2 InChI key | MZJUGRUTVANEDW-UHFFFAOYSA-N

Draw the Lewis structure of SMILES:BrF. Start by drawing the overall structure of the molecule: Count the total valence electrons of the bromine (n_Br, val = 7) and fluorine (n_F, val = 7) atoms: n_Br, val + n_F, val = 14 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8) and fluorine (n_F, full = 8): n_Br, full + n_F, full = 16 Subtracting these two numbers shows that 16 - 14 = 2 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There is 1 bond and hence 2 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 14 - 2 = 12 electrons left to draw: Answer: | |

melting point | -123 °C boiling point | -8.32 °C critical temperature | 436.3 K critical pressure | 7.014 MPa critical volume | 115.5 cm^3/mol molar heat of vaporization | 21.2 kJ/mol molar heat of fusion | 4.12 kJ/mol molar enthalpy | -213.1 kJ/mol molar free energy | -231.4 kJ/mol (computed using the Joback method)

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

Find the elemental composition for SMILES:BrF 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: BrF 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 F (fluorine) | 1 N_total = 1 + 1 = 2 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 | 1/2 F (fluorine) | 1 | 1/2 Check: 1/2 + 1/2 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/2 × 100% = 50.0% F (fluorine) | 1 | 1/2 × 100% = 50.0% 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 | 50.0% | 79.904 F (fluorine) | 1 | 50.0% | 18.998403163 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 | 50.0% | 79.904 | 1 × 79.904 = 79.904 F (fluorine) | 1 | 50.0% | 18.998403163 | 1 × 18.998403163 = 18.998403163 m = 79.904 u + 18.998403163 u = 98.902403163 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 | 50.0% | 79.904/98.902403163 F (fluorine) | 1 | 50.0% | 18.998403163/98.902403163 Check: 79.904/98.902403163 + 18.998403163/98.902403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 50.0% | 79.904/98.902403163 × 100% = 80.79% F (fluorine) | 1 | 50.0% | 18.998403163/98.902403163 × 100% = 19.21%

The first step in finding the oxidation states (or oxidation numbers) in SMILES: BrF is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There is 1 bromine-fluorine bond in SMILES: BrF. For this bond, assign the bonding electrons to the most electronegative element. First examine the bromine-fluorine bond: element | electronegativity (Pauling scale) | Br | 2.96 | F | 3.98 | | | Since fluorine is more electronegative than bromine, the electrons in this bond will go to fluorine. Decrease the oxidation number for fluorine (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 -1 | F (fluorine) | 1 +1 | Br (bromine) | 1

First draw the structure diagram for SMILES:BrF, 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 | 2 edge count | 1 Schultz index | 4 Wiener index | 1 Hosoya index | 2 Balaban index | 1