Question

Why is the answer C for this problem?

Answer 1

Answer:

$\boxed{\text{(C) X} _{3} P _{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X} _{3} P _{2}} }$.