Ionic bonds result from the electrostatic attraction between oppositely charged ions, which form when valence electrons are transferred from one atom to another.
Well to solve this problem, we must remember that Oxygen
in ionic form is Oxide O2- (an electric charge of negative 2). Hence the
correct formulas and names taking into account the charge of Oxygen are:
c) Co2O3, cobalt(III) oxide
f) CoO, cobalt(II) oxide
The periodic table of the elements are describe the electronic configuration of the elements on which the properties of the elements depends. Among the given groups only metal, non-metal and semi-metal group are the part of periodic table. The metallic property depends upon the binding energy of the electrons with the nucleus. Thus the elements which have the valence electrons more near to the nucleus that is s-block elements are more metallic in nature. On the other hand the elements which have the valence electrons far from the nucleus are more non-metallic in nature like p-block elements. However the binding energy or the attraction of the outermost electrons to the nucleus depends not only its valence electrons position but also some other factors like shielding effect, effective nuclear charge etc.
The elements which are in between the metals and non-metals can be classified as semi-metals.
Although the conductivity of a material is an inherent property of the metals but sometime the nonmetals or semi-metals are also behave like a conductor due to presence of the other elements, thus it cannot be a p[property of the periodic table. Similarly acidity, flammable gases are not part of the periodic table.
Answer:
The fluoride which precipitates first is CaF₂
Explanation:
When F⁻ is added, CaF₂ and BaF₂ are produced following the ksp equation:
For CaF₂:
Ksp = 3.2x10⁻¹¹ = [Ca²⁺] [F⁻]²
<em>Where [Ca²⁺] = 0.075M * {35mL / (25mL + 35mL)} = 0.04375M</em>
3.2x10⁻¹¹ = [0.04375M] [F⁻]²
[F⁻]² = 7.31x10⁻¹⁰
[F⁻] = 2.7x10⁻⁵M
<h3>CaF₂ begins precipitation when [F⁻] = 2.7x10⁻⁵M.</h3>
For BaF₂:
Ksp = 1.5x10⁻⁶ = [Ba²⁺] [F⁻]²
<em>Where [Ba²⁺] = 0.090M * {25mL / (25mL + 35mL)} = 0.0375M</em>
1.5x10⁻⁶ = [0.0375M] [F⁻]²
[F⁻]² = 4x10⁻⁵
[F⁻] = 6.3x10⁻³M
BaF₂ begins precipitation when [F⁻] = 6.3x10⁻³M
Thus, the fluoride which precipitates first is CaF₂
