Element atomic number position
Ba 56 group 2, period 6
Ca 12 group 2, period 3
S 16 group 16, period 3
Si `14 group 14, period 3
Now, you need to know the properties of the different type of elements and the tendencies on the periodic table.
The metallic elements are, those placed on the left side of the periodic table, are the ones that release an electron more easily, so they will requiere less energy to give it up when forming chemical bonds.
The higher the metallic character the less the energy need to give up an electron.
The metallic character grows as the group number decreases (goes to the left) period increases (goes downward), so among the elements considered, Barium will require the least amount of energy to give un an electron when forming chemical bonds.
This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength. Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.
Answer:
They have electrons in their 3d- and 4s-orbital for bond formation.
Explanation:
d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.
The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.
If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.
If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2
Phương trình hóa học: 2Mg+O2->2Mgo
