Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁴
(b) 1s² 2s² 2p⁶ 3s² 3p⁵
(c) sp³
(d) No valence orbital remains unhybridized.
Explanation:
<em>Consider the SCl₂ molecule. </em>
<em>(a) What is the electron configuration of an isolated S atom? </em>
S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
<em>(b) What is the electron configuration of an isolated Cl atom? </em>
Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>
SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.
<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>
No valence orbital remains unhybridized.
Answer:
.
Explanation:
Lithium is in the first column of the periodic table, so it will have 1 valence electron.
Bromine is in the seventh column of the periodic table, so it will have seven valence electrons.
They must combine in a way to reach 8.
When combining elements to form compounds, the "crisscross method" is used. Above Li would be a charge of +1, and above Br would be a charge of -1.
Cross the 1 from the top of Li to the bottom of Br, and so there is 1 Br.
Cross the 1 from the top of Br to the bottom of Li, and so there is 1 Li.
It is not written BrLi because chemists decided to order them the other way. Technically speaking, it isn't wrong, but the positive charge is normally put on the left and the negative charge is normally put on the right.
Answer:
Explanation:
a. Oxidation : 2O + 4e^- ------> 2O^2-
b. Reduction: 2Sr - 4e- -------> Sr^2+
c. Balanced redox reaction
2Sr + O2 ------------> 2Sr O
Oxidation and reduction can be defined by various means, addition of oxygen, removal of hydrogen, removal of electrons. For this reaction, this definition is used, oxidation is the loss of electrons while reduction is the gaining of electrons.
In (a) oxidation half reaction, the valency of oxygen is zero and then moves into lossing two electrons resulting into -2 valency.
In (b) reduction half reaction, the valency of Sr is zero and gains electrons resulting into valency of 2.
In the overall redox reaction, Sr and O2 with valency of 0 each reacts together and form SrO with valency of 2 and -2 respectively, which gives 0 and then balances the equation.
Remember....
mass number= atomic number + number of neutrons
If the mass number is 19 and the atomic number is 9, then the number of neutrons is 19-9 which is 10.
