Give reason :
(a) H3PO4 is triprotic acid but H3PO3 is diprotic acid. Why ?
(b) NCl3 and PCl3 are hydrolysed differently. Why ?

Question

Accepted Answer

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> [!TLDR]
> $H_3PO_3$'s basicity is fixed by how many H atoms are actually bonded to oxygen (ionisable) versus directly to phosphorus (non-ionisable); and $N$ vs $P$ having (or lacking) usable d orbitals decides whether hydrolysis attacks the central atom or the halogen.
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> **(a) $H_3PO_4$ triprotic, $H_3PO_3$ diprotic:**
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> Basicity (the number of ionisable/replaceable $H^+$) depends on the number of $-OH$ groups bonded to the central P atom, **not** simply on the total count of H atoms in the molecular formula. Only an H atom attached to an O (a P–O–H linkage) can ionise as $H^+$; an H atom bonded directly to P (a P–H bond) is non-ionisable, because P–H is a much less polar, non-acidic bond.
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> - **Orthophosphoric acid, $H_3PO_4$:** structure is $PO(OH)_3$ — all three hydrogens are on oxygen (three P–OH bonds), so all three are ionisable → **triprotic**.
> - **Phosphorous acid, $H_3PO_3$:** structure is $HPO(OH)_2$ — only two hydrogens are on oxygen (two P–OH bonds); the third hydrogen is bonded directly to phosphorus (one P–H bond) and does not ionise → only **two** ionisable protons, so it is **diprotic** even though its formula shows three H atoms.
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> **(b) Different hydrolysis of $NCl_3$ and $PCl_3$:**
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> $$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$$
> $$NCl_3 + 3H_2O \rightarrow NH_3 + 3HOCl$$
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> Phosphorus (period 3) has energetically accessible vacant 3d orbitals, so it can expand its coordination number and accept the incoming water molecule directly at the P atom (forming a pentacoordinate transition state), after which HCl is eliminated — giving $H_3PO_3 + 3HCl$, i.e. hydrolysis occurs **at the central atom**.
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Q.Give reason :

Concept understanding — p-Block Elements

Q.Give reason :

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