The P Block Elements Class 12 NCERT

Key Periodic Trends and the Inert Pair Effect

The p-block elements are those in which the last electron enters the outermost p-orbital. Spanning Groups 13 to 18, this block is incredibly diverse as it contains metals, non-metals, and metalloids. For CBSE Class 12, we focus heavily on Groups 15 (Nitrogen family), 16 (Oxygen family), 17 (Halogens), and 18 (Noble Gases).

A primary concept governing their chemistry is the Inert Pair Effect—the reluctance of the outer s-electrons to participate in bonding due to poor shielding of intervening d and f orbitals. Consequently, down a group, lower oxidation states become progressively more stable (e.g., Bi(III) is far more stable than Bi(V)). Additionally, the first element of each group (N, O, F) exhibits anomalous properties due to its exceptionally small atomic size, high electronegativity, high ionization enthalpy, and non-availability of d-orbitals in the valence shell. This absence of d-orbitals limits their maximum covalency to four.

Step-by-Step Analysis: Chemical Trends in Group 15 Hydrides

1. Step 1: Size of the Central Atom — As we move down the group from Nitrogen (N) to Bismuth (Bi), the atomic size of the central atom increases progressively (N < P < As < Sb < Bi) due to the addition of new electronic shells.
2. Step 2: Bond Dissociation Enthalpy — With the increase in atomic size of the central atom, the M-H bond length increases. Consequently, the orbital overlap becomes less effective, resulting in a steady decrease in M-H bond dissociation enthalpy from NH3 to BiH3.
3. Step 3: Determining Thermal Stability — Because the bond dissociation enthalpy decreases down the group, the thermal stability of these hydrides also decreases in the order: NH3 > PH3 > AsH3 > SbH3 > BiH3.
4. Step 4: Deducing Reducing Character — Lower thermal stability means that the hydride can release hydrogen gas more easily. Therefore, the reducing power increases down the group: NH3 is a mild reducing agent, whereas BiH3 is the strongest reducing agent among them.

Exam Trap: Oxoacids of Phosphorus & Basicity

In your CBSE board exams, questions on the structures and acidity of phosphorus oxoacids are highly common.

• The Basicity Trap: Basicity is determined only by the number of ionizable hydrogen atoms, which are those directly attached to oxygen atoms as P-OH bonds. Do not count hydrogen atoms directly bonded to phosphorus (P-H bonds) towards basicity!
• Reducing Nature: Hydrogen atoms directly bonded to phosphorus (P-H) cannot be lost as H+ ions, but they impart reducing properties to the acid. For example, Phosphinic acid ($H_3PO_2$) has two P-H bonds and is a strong monobasic reducing agent, while Phosphonic acid ($H_3PO_3$) has one P-H bond and is dibasic.

Practice Questions with Solutions

• Q: Why is BiH3 the strongest reducing agent among all the hydrides of Group 15 elements? A: Step 1: Analyze the atomic size of the central atoms down the group. As we move down Group 15, the size of the central atom increases from N to Bi. Step 2: Determine bond strength. Due to the large size of the Bi atom, the orbital overlap between the large Bi atom and the small H atom is poor, resulting in a weak and long Bi-H bond. Step 3: Relate bond strength to reducing nature. The weak Bi-H bond has a low bond dissociation enthalpy, meaning it breaks easily to release hydrogen gas. Final answer: BiH3 is the strongest reducing agent because it has the lowest bond dissociation enthalpy and releases hydrogen most easily.
• Q: Explain why H3PO3 is diprotic (dibasic) whereas H3PO4 is triprotic (tribasic). A: Step 1: Draw or describe the structural formula of H3PO3. Phosphorous acid, H3PO3, features one P=O double bond, one P-H bond, and two P-OH single bonds. Step 2: Identify the ionizable hydrogens in H3PO3. Only the hydrogens attached to oxygen (P-OH) are ionizable. Since there are two P-OH groups, H3PO3 can donate two protons, making it dibasic. Step 3: Draw or describe the structural formula of H3PO4. Orthophosphoric acid, H3PO4, features one P=O double bond and three P-OH single bonds. Step 4: Identify the ionizable hydrogens in H3PO4. All three hydrogens are attached to oxygen as P-OH groups, allowing them to ionize completely in aqueous solution. Final answer: H3PO3 is dibasic because it contains only two ionizable P-OH bonds, while H3PO4 is tribasic because it contains three ionizable P-OH bonds.
• Q: Complete and balance the following chemical equations for the hydrolysis of Xenon fluorides: (i) XeF4 + H2O -> ? (ii) XeF6 + H2O -> ? A: Step 1: Recall the hydrolysis products of XeF4. Xenon tetrafluoride undergoes slow disproportionation with water to yield Xenon gas, Xenon trioxide (XeO3), Hydrofluoric acid (HF), and Oxygen (O2). Step 2: Balance the equation for XeF4 hydrolysis: 6 XeF4 + 12 H2O -> 4 Xe + 2 XeO3 + 24 HF + 3 O2 Step 3: Recall the complete hydrolysis of XeF6. Complete hydrolysis of Xenon hexafluoride yields Xenon trioxide (XeO3) and Hydrofluoric acid (HF). Step 4: Balance the equation for XeF6 hydrolysis: XeF6 + 3 H2O -> XeO3 + 6 HF Final answer: (i) 6 XeF4 + 12 H2O -> 4 Xe + 2 XeO3 + 24 HF + 3 O2 (ii) XeF6 + 3 H2O -> XeO3 + 6 HF
• Q: Why does NO2 dimerize to form N2O4? A: Step 1: Count the total number of valence electrons in a nitrogen dioxide (NO2) molecule. Nitrogen has 5 valence electrons, and each oxygen has 6, giving a total of 17 valence electrons. Step 2: Identify the molecular nature of NO2. Since 17 is an odd number, NO2 behaves as an odd-electron molecule with an unpaired electron on the nitrogen atom. Step 3: Analyze the driving force for dimerization. To pair up this lone electron and achieve a stable octet configuration (paired electronic state), two NO2 molecules readily join together. Final answer: NO2 dimerizes because it is an odd-electron molecule containing an unpaired electron; by pairing with another molecule, it forms stable, diamagnetic N2O4.

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