Question

Question #b4ab9

Answer 1

Here's what I got.

Explanation:

For starters, calculate the charge of the ion by using the equation

`color(blue)(ul(color(black)("net charge" = "no. of protons " - " no. of electrons"`

In your case, you have

`"net charge" = 52 - 54 = -2`

This tells you that the ion carries a `2-` charge, which implies that you are dealing with a negatively charged ion, or anion.

Now, the isotope notation of this ion requires

• the atomic number of the element, `Z`
• the mass number of the isotope, `A`

![)

In your case, you know that the element has `52` protons, which means that

`Z = 52`

In order to find the mass number of the isotope, simply add the number of protons and the number of neutrons present in the nucleus.

`A = "52 protons + 76 neutrons"`

`A = "128 nucleuons"`

This means that the symbol of the neutral isotope will look like this

`""_(color(white)(1)52)^128"X"`

A quick look in the Periodic Table will reveal that you're dealing with tellurium-128, an isotope of tellurium, `"Te"`.

`""_(color(white)(1)52)^128"Te"`

Finally, to show that this is an anion and not a neutral atom, add the net charge.

`""_(color(white)(1)52)^128"Te"^(2-)`