Calculate the molarity of 0.300 mol of Na2S in 1.75 L of solution.
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The picture of Au₃N is attached below.
The first part of the picture shows the formation of Au and N ions.
Formation of Au⁺¹ :
As Gold has one valence electron in 6s¹ therefore, it will loose it to form Au⁺¹. In case of Au₃N three atoms of Au looses three electrons to form three Au⁺¹ ions.
Formation of N⁻³ :
As Nitrogen has 5 valence elctrions therefore, it will gain three electrons that lost by Au to form Nitrite (i.e. N⁻³)
Formation of Au₃N:
Three cations of Au⁺ combines with one anion of N⁻³ to form a neutral ionic compound i.e. Au₃N as shown in second part of the picture.
