Animated Solution for Chemistry - s and p-Block Elements: The total number of compounds having at least one bridging oxo group among the molecules given below is______. ${\text{N}}_2{\text{O}}_3$, ${\text{N}}_2{\text{O}}_5$, ${\text{P}}_4{\text{O}}_6$, ${\text{P}}_4{\text{O}}_7$, ${\text{H}}_4{\text{P}}_2{\text{O}}_5$, ${\text{H}}_5{\text{P}}_3{\text{O}}_{10}$, ${\text{H}}_2{\text{S}}_2{\text{O}}_3$, ${\text{H}}_2{\text{S}}_2{\text{O}}_5$

When we look at a chemical formula like ${\text{H}}_2{\text{S}}_2{\text{O}}_5$, it is incredibly tempting to just imagine a symmetrical line of atoms. However, chemistry is not just about counting atoms; it is about understanding the 3D architecture that holds them together. In this problem, we are tasked with finding the number of molecules that possess a bridging oxo group.

But what exactly is a bridging oxo group? Imagine a bridge connecting two distinct cities. In the molecular world, a bridging oxo group is an oxygen atom that acts as a bridge, covalently bonded between two central atoms. This creates an $M-O-M$ linkage. Let's embark on a journey through these eight molecules to see which ones have built these bridges.

A Tale of Two Forms

We start our investigation with the oxides of nitrogen. ${\text{N}}_2{\text{O}}_3$ is a fascinating molecule because it suffers from a bit of an identity crisis. It exists in an equilibrium between two forms. The more common, unsymmetrical form features a direct $N-N$ bond (${\text{ON-NO}}_2$). However, it also has a symmetrical form that features a clear $N-O-N$ bridge. Because of this dual nature, it can technically be counted as having a bridging oxo group.

${\text{N}}_2{\text{O}}_5$, on the other hand, is much more straightforward. Its structure is unequivocally ${\text{O}}_2{\text{N-O-NO}}_2$. The central oxygen atom proudly bridges the two nitrogen atoms, making it a definite 'yes' for our count.

Next, we encounter the magnificent oxides of phosphorus: ${\text{P}}_4{\text{O}}_6$ and ${\text{P}}_4{\text{O}}_7$. These are not simple linear chains; they are intricate, cage-like structures.

Imagine a tetrahedron where the four corners are occupied by phosphorus atoms. In ${\text{P}}_4{\text{O}}_6$, an oxygen atom sits on every single edge of this tetrahedron, connecting the phosphorus atoms. This means there are a staggering six bridging oxo groups in a single molecule! ${\text{P}}_4{\text{O}}_7$ shares this exact same cage framework, but adds a terminal double-bonded oxygen to one of the phosphorus atoms. Both of these molecules are absolute champions of the bridging oxo group.

Moving to the oxyacids of phosphorus, we look at ${\text{H}}_4{\text{P}}_2{\text{O}}_5$ (pyrophosphorous acid) and ${\text{H}}_5{\text{P}}_3{\text{O}}_{10}$ (triphosphoric acid).

Whenever you see the prefix 'pyro-' or 'tri-' in phosphorus oxyacids, it is a strong hint that condensation has occurred, linking phosphorus atoms together via oxygen bridges. ${\text{H}}_4{\text{P}}_2{\text{O}}_5$ features a single $P-O-P$ linkage. ${\text{H}}_5{\text{P}}_3{\text{O}}_{10}$ extends this chain even further, utilizing two $P-O-P$ linkages to connect its three central phosphorus atoms. Both of these molecules confidently join our tally.

Finally, we arrive at the trap of the problem: the sulfur oxyacids ${\text{H}}_2{\text{S}}_2{\text{O}}_3$ and ${\text{H}}_2{\text{S}}_2{\text{O}}_5$.

It is incredibly easy to look at ${\text{H}}_2{\text{S}}_2{\text{O}}_5$ (disulfurous acid) and assume that the oxygen atoms must be bridging the sulfurs. However, sulfur has a strong affinity for catenation—bonding directly to itself. In ${\text{H}}_2{\text{S}}_2{\text{O}}_3$ (thiosulfuric acid), one sulfur is double-bonded to the central sulfur ($S=S$). In ${\text{H}}_2{\text{S}}_2{\text{O}}_5$, the two sulfur atoms are connected by a direct single bond ($S-S$). Neither of these molecules contains a bridging oxo group.

Let's review our findings. The molecules that definitively contain at least one bridging oxo group are ${\text{N}}_2{\text{O}}_5$, ${\text{P}}_4{\text{O}}_6$, ${\text{P}}_4{\text{O}}_7$, ${\text{H}}_4{\text{P}}_2{\text{O}}_5$, and ${\text{H}}_5{\text{P}}_3{\text{O}}_{10}$. This gives us a solid count of 5.

If we choose to include the symmetrical form of ${\text{N}}_2{\text{O}}_3$, the count rises to 6. In the actual JEE Advanced examination, both 5 and 6 were awarded full marks, acknowledging the nuanced reality of chemical equilibria. This problem serves as a beautiful reminder: in chemistry, you must always look beyond the formula and visualize the structure!