Question d996c

Dec 8, 2016

Here's what I got.

Explanation:

The thing to remember about net charge is that it depends on the balance that exists between the number of protons present inside the nucleus and the number of electrons that surround the nucleus.

More specifically, you have

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{no. of protons" = "no. of e}}^{-}}}}$

NO net charge, the atom is neutral

• $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{no. of protons" > "no. of e}}^{-}}}}$

an overall positive net charge, the atom is now a positively charged ion, i.e. a cation

• color(blue)(ul(color(black)("no. of protons" < "no. of e"^(-)

an overall negative net charge, the atom is now a negatively charged ion, i.e. an anion

You can sum up the three cases by writing something like this

$\textcolor{red}{\underline{\textcolor{b l a c k}{\text{no. of protons " - " no. of electrons" = "net charge}}}}$

In your case, you are given a net charge of $2 +$. Right from the start, this should tell you that the given cation contains more protons inside its nucleus than electrons surrounding its nucleus.

So how would you determine the number of electrons if told that the number of protons is equal to $7$ ?

Well, you know that $7$ electrons would make for a neutral atom, since you would have

${\text{7 protons " - " 7 e}}^{-} = 0$

But you know that

${\text{7 protons " - color(purple)(?)color(white)(.) "e}}^{-} = + 2$

This means that the number of electrons must be equal to

color(darkgreen)(ul(color(black)(color(purple)(?) = 7 -2 = "5 e"^(-))))

Let's try this for a $2 -$ ion. You know that the negative charge means an excess of electrons. Set up your calculation the same way

${\text{7 protons " - color(purple)(?)color(white)(.)"e}}^{-} = - 2$

This time, you will have

color(darkgreen)(ul(color(black)(color(purple)(?) = 7 + 2 = "9 e"^(-))))#

So, to sum this up, net charge only deals with protons and electrons. More specifically, the difference between the two gives you the net charge.