Questions: Write the symbol of a period 3 element that forms a +3 cation. Write the symbol of an ion from group 17 (7A) with 36 electrons.

Transcript text: Write the symbol of a period 3 element that forms a +3 cation. $\square$ Part 2 of 3 Write the symbol of an ion from group 17 (7A) with 36 electrons. $\square$

Solution

Solution Steps

Step 1: Identify Period 3 Elements

Period 3 elements are those in the third row of the periodic table: Na, Mg, Al, Si, P, S, Cl, and Ar.

Step 2: Determine Which Element Forms a +3 Cation

Among the period 3 elements, aluminum (Al) is known to commonly form a +3 cation ($ \text{Al}^{3+} $).

Step 3: Identify Group 17 Elements

Group 17 elements, also known as halogens, include F, Cl, Br, I, and At.

Step 4: Determine the Ion with 36 Electrons

A neutral bromine (Br) atom has 35 electrons. To have 36 electrons, it must gain one electron, forming the bromide ion ($ \text{Br}^- $).

Final Answer

For Part 1: $ \boxed{\text{Al}} $
For Part 2: $ \boxed{\text{Br}^-} $

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