For acidic solutions, which element is added to balance half-reactions
2 answers:
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Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2
Or,
Bond Order = 0.5
Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2
Bond Order = (4 - 3) ÷ 2
Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2
Or,
Bond Order = 0.5
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
1) Calculate Number of Valence Electrons:
# of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.
3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,