What are the formulas for the following ionic compounds: sodium sulphide, caesium bromide and magnesium nitride?

Answer 1

Look at the positions of the parent atoms in the Periodic Table...

#"Sodium sulfide"#...sodium is a Group 1 metal, and thus has one valence electron to lose to give the #Na^+# ion. On the other hand, sulfur is a Group 16 atom, that is likely to pick up two electrons to form the #S^(2-)#...and thus we got #Na_2S#..

#"Caesium bromide"#...caesium is a Group 1 metal, and thus has one valence electron to lose to give the #Cs^+# ion. On the other hand, bromine is a Group 17 atom, that is likely to pick up ONE electrons to form the #Br^(-)#...and thus we got #CsBr#..

#"Magnesium nitride"#...magnesium is a Group 2 metal, and thus has two valence electron to lose to give the #Mg^(2+)# ion. Nitride anion is #N^(3-)#...i.e. it picks up THREE electrons to fill its valence shell...and so again the salt is electrically neutral, i.e. #Mg_3N_2#.

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Answer 2

Ellie Adams

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.