What is the correct formula for the compound made by magnesium and nitrogen?

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What is the correct formula for the compound made by magnesium and nitrogen?

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Answer 1 · 2015-10-30T22:45:09

The correct formula for the Magnesium Nitrogen compound is $M g_{3} N_{2}$ $Mg_3N_2$ .

Explanation:

Magnesium has two valence electrons, and Nitrogen has five valence electrons.

Answer 2 · 2015-10-30T22:51:25

$M g_{3} N_{2}$ $Mg_3N_2$ , magnesium nitride.

Explanation:

Magnesium is in Group 2, nitrogen is in Group 15; the formula is as you would predict from magnesium ion, $M g^{2 +}$ $Mg^(2+)$ , and nitride anion, $N^{3 -}$ $N^(3-)$ . This is an ionic solid that hydrolyzes to give ammonia, and magnesium hydroxide, I.e.:

$M g_{3} N_{2}$ $Mg_3N_2$ $+$ $+$ $6 H_{2} O \to$ $6H_2O rarr$ $2 N H_{3}$ $2NH_3$ $+$ $+$ $3 M {g (O H)}_{2}$ $3Mg(OH)_2$

Answer 3 · 2015-10-30T22:56:52

Magnesium and nitrogen combine to form magnesium nitride, ${Mg}_{3} N_{2}$ $"Mg"_3"N"_2"$ .

Explanation:

Magnesium nitride is an ionic compound. Magnesium forms ions with a $2+$ $"2+"$ charge, and nitrogen forms nitride ions with a $3-$ $"3-"$ charge. The overall charge of the compound must be zero. Therefore, the numerical value of the positive and negative charge must be $6$ $"6"$ .

$3 {[Mg]}^{2 +} + 2 {[N]}^{3 -}$ $3["Mg"]^(2+) + 2 ["N"]^(3-)$ $\to$ $rarr$ ${Mg}_{3} N_{2}$ $"Mg"_3"N"_2"$

$3 \times (2 +) = 6 +$ $3xx(2+)=6+$ and $2 \times (3 -) = 6 -$ $2xx(3-)=6-$

$(6 +) + (6 -) = 0$ $(6+)+(6-)=0$

An easy way to determine the formula unit of an ionic compound is called the crisscross method, as demonstrated in the video below.

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What is the correct formula for the compound made by magnesium and nitrogen?

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