Question #b4ab9

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Answer 1 · 2017-10-23T01:04:51

Here's what I got.

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.

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