Question

Hybridization of p in h3 po5

• A. sp
• B. sp2
• C. sp3
• D. dsp2

Answer 1

Answer: (C)

sp3

Answer 2

The molecule is peroxomonophosphoric acid, $\text{H}_3\text{PO}_5$.

The central atom is Phosphorus (P). To determine the hybridization of the central atom, we need to find its steric number. The steric number is the sum of the number of sigma bonds and the number of lone pairs around the central atom.

The structure of $\text{H}_3\text{PO}_5$ is: Phosphorus is bonded to one oxygen atom via a double bond (P=O), two oxygen atoms via single bonds (P-OH), and one oxygen atom via a single bond within a peroxide linkage (P-O-O-H).

Let's count the sigma bonds around the P atom:

1. P=O double bond contains 1 sigma bond and 1 pi bond.
2. P-OH single bond (first one) contains 1 sigma bond.
3. P-OH single bond (second one) contains 1 sigma bond.
4. P-O single bond (in P-O-O-H linkage) contains 1 sigma bond.

Total number of sigma bonds around P = 1 + 1 + 1 + 1 = 4 sigma bonds.

Now, let's count the lone pairs on the P atom. Phosphorus is in Group 15 and has 5 valence electrons. In this structure, Phosphorus forms 4 sigma bonds and 1 pi bond. Assuming P utilizes its valence electrons to form these bonds (octet rule can be expanded for P), there are typically no lone pairs on the P atom in such oxoacids where it forms 5 bonds (one double, three single). We can verify this by considering formal charges, but for hybridization calculation using the steric number method, we primarily count sigma bonds and lone pairs. Based on the bonding arrangement (4 sigma bonds and 0 lone pairs), the steric number for P is 4 + 0 = 4.

The hybridization corresponding to a steric number of 4 is $\text{sp}^3$.

Since the steric number of P in $\text{H}_3\text{PO}_5$ is 4, its hybridization is $\text{sp}^3$.