How can I write the formula for magnesium phosphide?
Answer
581.4k+ views
Hint As the name, magnesium phosphide suggests that there are two atoms in the compound that is magnesium and phosphorus. Magnesium is the element of group 2 and its valence electrons are 2, and phosphorus is the element of group 15 and its valence electrons are 3.
Complete step by step answer:
As the name suggests, magnesium phosphide suggests that there are two atoms in the compound that are magnesium and phosphorus.
Magnesium is the element of group 2, and its atomic number is 12 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$. Therefore, the number of valence electrons is 2 which it can lose to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ . It is written as $M{{g}^{2+}}$ and two electrons take place in the bond formation.
Phosphorus is the element of group 15, and its atomic number is 15 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$. Therefore, the number of valence electrons is 3 and it can gain three electrons to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$. It is written as ${{P}^{3-}}$ .
The atoms in the compound in such a way that the electrons donated by one atom must be equal to the number of electrons gained by the other atom. So, when three magnesium atoms donate 2 + 2 + 2 = 6 electrons then two phosphorus atoms will gain 3 + 3 = 6 electrons.
Therefore, the formula will be $M{{g}_{3}}{{P}_{2}}$.
Note: The bond between the magnesium and phosphorus in the magnesium phosphide is ionic because the bond is formed between a metal and a nonmetal in which the magnesium is the metal and phosphorus is the non-metal.
Complete step by step answer:
As the name suggests, magnesium phosphide suggests that there are two atoms in the compound that are magnesium and phosphorus.
Magnesium is the element of group 2, and its atomic number is 12 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$. Therefore, the number of valence electrons is 2 which it can lose to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ . It is written as $M{{g}^{2+}}$ and two electrons take place in the bond formation.
Phosphorus is the element of group 15, and its atomic number is 15 so, its electronic configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{3}}$. Therefore, the number of valence electrons is 3 and it can gain three electrons to achieve the noble gas configuration of $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$. It is written as ${{P}^{3-}}$ .
The atoms in the compound in such a way that the electrons donated by one atom must be equal to the number of electrons gained by the other atom. So, when three magnesium atoms donate 2 + 2 + 2 = 6 electrons then two phosphorus atoms will gain 3 + 3 = 6 electrons.
Therefore, the formula will be $M{{g}_{3}}{{P}_{2}}$.
Note: The bond between the magnesium and phosphorus in the magnesium phosphide is ionic because the bond is formed between a metal and a nonmetal in which the magnesium is the metal and phosphorus is the non-metal.
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