Frist Ionisation Engery Graph

Nice question!

I think the first thing to understand is what exactly ionisation enthalpy really is. In simple terms, it's really a measure of the energy required to remove (one mole of) electrons from (one mole of) an atom (in its gaseous state).

The first ionisation enthalpy is just the concerned with the energy required to remove the first electron from the atom.The trend across a period (overall one) is that as you go across the period the (first) ionisation energy will increase.

The reasons are:
1. Removing from an atom that has a more highly positively charged nucleus (i.e. more protons).
2. Similar shielding so overall greater effective nuclear charge
3. Distance and repulsion effects are minimal (for the overall trend that is)

Hopefully you can see that element Y must therefore be in group 1, since compared to the element before, it has a significantly lower ionisation energy. We can therefore infer that Y has one valence electron in its outer energy level. To see how many valence electrons Z has, we just need to count backwards from Y. The element before Y must have 8 (since Y is the start of a new period), the next element before is 7 and so element Z has 6 valence electrons (i.e. group 6).

So now we have a situation where:
1. Y has 1 valence electron
2. Z has 6 valence electrons

To match it up nicely, the compound (which, correct me if i'm wrong, sort of implies an ionic interaction), having two Y atoms each donate an electron to the Z atom will lead to a stable configuration for both:

So the answer is most likely Y2Z (sorry I don't know how to subscript)

Please take what I say with a pinch of salt though, since my A level knowledge is probably the same (if not worse!) than yours.

Quick summary is
Reasons for abnormalities in ionisation energies are
Change in shielding
Nuclear charge(more protons means stronger force of attraction)
Change in distance from nucleus(the electron that is)

Massive drops to Y suggests Y is in Group 1
If Y in in Group 1 then the 2 elements before it must be in Group 0 and 7
If you're not sure remember there's a drop due to replusion of electrons in the same orbital, drop between group 5 and 6 elements. This is Z
Group 0 elements have a full shell
Group 7 elements need 1 electron to complete the orbital

The question says compound which suggests ionic bonding
Ionic bonding is transfer of electrons so the group 0 element transfers 2 electrons, 1 to each group 7 element to make an ionic compound. Thus you need 2 group 7 elements for every 1 group 0 element

In conclusion most likely is Y2Z

Nice question!

I think the first thing to understand is what exactly ionisation enthalpy really is. In simple terms, it's really a measure of the energy required to remove (one mole of) electrons from (one mole of) an atom (in its gaseous state).

The first ionisation enthalpy is just the concerned with the energy required to remove the first electron from the atom.The trend across a period (overall one) is that as you go across the period the (first) ionisation energy will increase.

The reasons are:
1. Removing from an atom that has a more highly positively charged nucleus (i.e. more protons).
2. Similar shielding so overall greater effective nuclear charge
3. Distance and repulsion effects are minimal (for the overall trend that is)

Hopefully you can see that element Y must therefore be in group 1, since compared to the element before, it has a significantly lower ionisation energy. We can therefore infer that Y has one valence electron in its outer energy level. To see how many valence electrons Z has, we just need to count backwards from Y. The element before Y must have 8 (since Y is the start of a new period), the next element before is 7 and so element Z has 6 valence electrons (i.e. group 6).

So now we have a situation where:
1. Y has 1 valence electron
2. Z has 6 valence electrons

To match it up nicely, the compound (which, correct me if i'm wrong, sort of implies an ionic interaction), having two Y atoms each donate an electron to the Z atom will lead to a stable configuration for both:

So the answer is most likely Y2Z (sorry I don't know how to subscript)

Please take what I say with a pinch of salt though, since my A level knowledge is probably the same (if not worse!) than yours.

Thank you so much!!