When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc
ture of this compound has a formal charge of zero on all elements in the compound. Draw the Lewis structure for the vile-smelling orange liquid.
1 answer:
Answer:
The structure is shown below.
Explanation:
The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:
FC = V - (L + S/2)
Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.
The possible formulas, from the empiric one, are:
SCl, S₂Cl₂, and S₃Cl₃.
To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:
FC = 7 - (6 + 2/2) = 0
So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)
FC = 6 - (4+ 2/2) = +1
For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so
FC = 6 - (4 + 4/2) = 0
So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.
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Answer:
a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b) the Q-value of this reaction is 5.789 MeV
c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
Explanation:
a)
The decay equation for the alpha decay is expressed as;
²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q
b)
Calculate the Q-value (in MeV) of this reaction.
Q = Mparent - Mdaughter -Mg
Q = MRa - MRn -Mg
= 224.020202 - 220.011384 - 4.00260305
= 0.00621495 amu
= 5.789 MeV
therefore the Q-value of this reaction is 5.789 MeV
c)
Energy of alpha particle is expressed as;
E∝ = MQ / ( m + M)
now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;
The energy of the alpha particle gives;
E∝ = 220(5.789) / ( 4 + 220) = 5.69 MeV
as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.
Therefore the energy of this alpha is
E∝ = 5.69 - 0.241 = 5.449 Mev
Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev
d)
Sketch of the nuclear decay scheme have been uploaded along side this answer.
