A molecule with the formula $$AX_4Y$$ has all its elements from p-block. Element $$A$$ is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among$$ A, X$$ and $$Y$$. Elements $$X$$ and $$Y$$ have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:

We need to identify elements A, X, and Y, all from the p-block.

Element X has the first highest electronegativity among all known elements. The most electronegative element is Fluorine (F). So, $$X = F$$.

Element Y has the second highest electronegativity among all known elements. The second most electronegative element is Oxygen (O). So, $$Y = O$$.

Now, element A is from the p-block, is the rarest monoatomic non-radioactive element from its group, and has the lowest ionization enthalpy among A, X, and Y.

Since A has lower ionization enthalpy than both F and O (which have very high ionization enthalpies), A must be a heavier p-block element. The term "monoatomic" tells us A belongs to the noble gas group (Group 18), since noble gases exist as monoatomic species. Among the non-radioactive noble gases (He, Ne, Ar, Kr, Xe), the rarest one is Xenon (Xe). Xenon also has the lowest ionization enthalpy in this context, as it is a large atom with loosely held outer electrons. So, $$A = Xe$$.

The molecule is $$AX_4Y = XeF_4O$$, commonly written as $$XeOF_4$$.

Now let us determine the shape of $$XeOF_4$$.

Xenon has 8 valence electrons. It forms 4 bonds with F atoms and 1 double bond with the O atom (we can treat it as 1 bond domain). So the number of bond pairs around Xe = 5.

Total valence electrons in $$XeOF_4$$:
Xe contributes 8, each F contributes 7 (4 × 7 = 28), and O contributes 6.
Total = 8 + 28 + 6 = 42 electrons.

Around the central Xe atom, there are 5 bonding domains (4 Xe-F bonds + 1 Xe=O bond) and 1 lone pair. This gives a total of 6 electron domains around Xe.

With 6 electron domains, the electron geometry is octahedral. With 5 bonding pairs and 1 lone pair, the molecular shape is square pyramidal. The lone pair occupies one of the positions, and the oxygen and four fluorine atoms form a square pyramid around xenon.

Hence, the correct answer is Option A.