HomeChemistryQ
Chemistryp-Block Elements
Identify the incorrect statement from the following :
Options
1
P(C₂H₅)₃ and As(C₆H₅)₃ form dπ-dπ bond with transition metals.
2
Nitrogen can form dπ-pπ bond with oxygen.
3
Nitrogen can form pπ-pπ multiple bonds with itself.
4
Phosphorus, arsenic and antimony show catenation property.
Incorrect Statement (Answer)
Option 2 — N cannot form dπ-pπ bond with O
Why Option 2 is INCORRECT
1

dπ-pπ bond requires a d orbital. Nitrogen (Period 2, Z=7) has configuration 1s²2s²2p³ — it has NO d orbitals in its valence shell.

2

Oxygen (Period 2, Z=8) also has no d orbitals.

For dπ-pπ bonding: one atom needs a d orbital (donor) and the other a p orbital. Since BOTH N and O are Period 2 — neither has d orbitals → dπ-pπ bonding is impossible between them.

❌ WRONG: "N can form dπ-pπ bond with O"

✅ CORRECT: N forms only pπ-pπ bonds (N=N in N₂, N=O in NO₂). Both N and O are Period 2 — no d orbitals available.

✅ Option 1 CORRECT: P and As are Period 3+ → have d orbitals → form dπ-dπ bonds with transition metals (back-bonding in metal complexes)

✅ Option 3 CORRECT: N₂ has N≡N (pπ-pπ triple bond), N=N in hydrazine derivatives

✅ Option 4 CORRECT: P (P₄), As, Sb show catenation — form M-M bonds

Theory: Bonding in Group 15 Elements
1. pπ-pπ vs dπ-pπ Bonding

pπ-pπ bonding: both atoms use p orbitals to form a pi bond. Only possible for small atoms where p-p orbital overlap is good — Period 2 elements (C, N, O). Examples: N≡N (in N₂), C=C (alkenes), N=O (in NO, NO₂). dπ-pπ bonding: one atom uses a d orbital and the other uses a p orbital. Requires at least one Period 3+ atom (which has d orbitals). Examples: S=O in SO₂ (S uses 3d, O uses 2p), P=O in phosphoryl compounds. dπ-dπ bonding: both atoms use d orbitals — seen in metal complexes where ligands like P(C₂H₅)₃ donate back electrons via d orbitals.

2. Why N is Different from P in Bonding

📌 Nitrogen (Period 2): 2s²2p³ — only s and p orbitals → max bond order through pπ-pπ only → N₂ has triple bond (N≡N, bond energy = 946 kJ/mol — strongest diatomic)

📌 Phosphorus (Period 3): 3s²3p³3d⁰ — has empty 3d orbitals → can form dπ-pπ bonds → P=O bonds in H₃PO₄ → also expands octet (PCl₅, PF₅)

📌 Key Rule: Period 2 elements (C,N,O,F) — NO dπ bonding. Period 3+ (P,S,Cl,Si...) — CAN form dπ bonding

📌 This explains why N₂O₅ is N-O-N (no N=O with d orbital), but P₂O₅ has P=O bonds

3. Why N₂ is Exceptionally Stable

N₂ has a triple bond (one σ + two π, all pπ-pπ). Bond energy = 946 kJ/mol — highest of any diatomic molecule. This extreme stability makes N₂ chemically inert at room temperature — despite nitrogen being 78% of atmosphere, plants cannot use it directly (nitrogen fixation is needed). The two π bonds in N₂ are both pπ-pπ type (N is Period 2, uses 2p orbitals). For comparison: P₄ has only single P-P bonds (P uses 3p orbitals; pπ-pπ overlap with 3p is poor due to larger size and diffuse orbitals).

4. Catenation in Group 15

Catenation = forming bonds with like atoms. In Group 15: N shows very limited catenation (N-N bond in hydrazine N₂H₄ and a few others) because N-N single bond is relatively weak (163 kJ/mol) due to lone pair-lone pair repulsion between adjacent N atoms. P, As, Sb show much better catenation: P₄ (tetrahedral P₄ molecules), As₄, Sb₄, polyphosphides (P₄²⁻, P₅⁻ chains). The larger atoms have orbitals that overlap better for homonuclear bonding with less lone pair repulsion. Bi shows least catenation in Group 15.

5. Oxoacids of Nitrogen vs Phosphorus

📌 N oxoacids: HNO₂ (nitrous), HNO₃ (nitric) — N uses only pπ-pπ bonding with O

📌 P oxoacids: H₃PO₂, H₃PO₃, H₃PO₄, H₄P₂O₇ — P uses dπ-pπ bonding (P=O)

📌 Basicity of P acids: number of P-OH groups = number of replaceable H (basicity). P=O and P-H are NOT acidic.

📌 H₃PO₂ (hypophosphorous): 1 P-OH → monobasic. H₃PO₃ (phosphorous): 2 P-OH → dibasic. H₃PO₄ (phosphoric): 3 P-OH → tribasic

📌 Both H₃PO₂ and H₃PO₃ are good reducing agents (they have P-H bonds)

6. Allotropy of Phosphorus

White phosphorus (P₄): tetrahedral structure, 60° P-P-P angle (strained), highly reactive and toxic, glows in dark (chemiluminescence), stored under water. Red phosphorus: polymeric, less reactive, non-toxic, used in matchboxes. Black phosphorus: most stable, graphite-like layered structure, semiconductor. Violet phosphorus: monoclinic, most thermodynamically stable. Conversion: white → red (heating, 250°C); white → black (high pressure). White phosphorus dissolves in CS₂ (carbon disulphide); red phosphorus does not — used to distinguish them.

7. Oxides of Nitrogen

📌 N₂O (nitrous oxide, +1): laughing gas, anaesthetic, linear

📌 NO (nitric oxide, +2): colourless, paramagnetic (odd e⁻), turns brown in air

📌 N₂O₃ (dinitrogen trioxide, +3): blue solid, anhydride of HNO₂

📌 NO₂ (nitrogen dioxide, +4): brown gas, paramagnetic, acidic, smog component

📌 N₂O₄ (dinitrogen tetroxide, +4): colourless, NO₂ dimer

📌 N₂O₅ (dinitrogen pentoxide, +5): white solid, anhydride of HNO₃

8. Fixation of Nitrogen

Atmospheric N₂ must be "fixed" (converted to useful compounds) before plants can use it. Natural fixation: lightning converts N₂ + O₂ → NO → HNO₃ (about 5-8%); Biological fixation by Rhizobium bacteria (in legume root nodules) using nitrogenase enzyme — converts N₂ → NH₃. Industrial fixation: Haber process — N₂ + 3H₂ ⇌ 2NH₃ (Fe catalyst, 400-500°C, 200 atm, high pressure). Ostwald process converts NH₃ → HNO₃. About 50% of world food production depends on synthetic nitrogen fertilisers made this way.

Frequently Asked Questions
1. Why can't N form dπ-pπ bond but P can?
N is Period 2 (1s²2s²2p³) — its valence shell (n=2) has only s and p orbitals; no d orbitals exist at the n=2 level. dπ-pπ bonding requires one partner to have a d orbital to accept (or donate) electron density. P is Period 3 (3s²3p³3d⁰) — has empty 3d orbitals. P can use these 3d orbitals to form dπ-pπ bonds with oxygen (in P=O), or to accept back-donated electrons from transition metals (dπ-dπ back-bonding). This is the fundamental reason for the different chemistry of N and P despite being in the same group.
2. What is back-bonding in metal-ligand complexes?
Back-bonding (π back-donation): metal d electrons are donated INTO empty antibonding orbitals of ligand (π-acceptor ligands). Example: P(C₂H₅)₃ has empty 3d orbitals on P. Metal d electrons → P empty 3d → this is dπ-dπ back-bonding (metal d → P d). CO has empty π* antibonding orbital → metal d → CO π* (dπ-pπ back-bonding — most important). Back-bonding strengthens metal-ligand bond, weakens C≡O bond (CO stretching frequency decreases). Strong π-acceptors: CO > NO > CN⁻ > PPh₃ > NH₃ (no back-bonding).
3. Why does N₂ have such a high bond energy?
N₂ has a triple bond: one σ bond (head-on p-p overlap) + two π bonds (sideways p-p overlap) = N≡N. Bond energy = 946 kJ/mol (highest diatomic). Small atomic radius of N → very good p orbital overlap → strong π bonds. Compare to P₂: P-P single bond = 201 kJ/mol (much weaker). P cannot form stable P≡P like N≡N because 3p orbitals are too diffuse for good sideways overlap. That's why P exists as P₄ (single bonds) while N exists as N₂ (triple bond).
4. What makes white phosphorus so reactive?
White phosphorus (P₄) has a tetrahedral structure with P-P-P bond angles of only 60°. Ideal bond angle for sp³ = 109·5°. The actual 60° angle causes enormous ring strain (like cyclopropane but much worse). This strain energy makes the P-P bonds weak and easy to break. High reactivity: P₄ ignites spontaneously in air at ~30°C (pyrophoric), glows in dark (slow oxidation = chemiluminescence). It's kept under water to prevent contact with air. Extremely toxic — 50 mg can be fatal to a human.
5. What is the Haber process for ammonia synthesis?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol (exothermic). Conditions: Fe catalyst (with K₂O and Al₂O₃ as promoters), 400-500°C, 200-300 atm pressure. Equilibrium analysis: exothermic → lower temperature favours NH₃ (Le Chatelier) but rate too slow. High pressure favours NH₃ (4 mol → 2 mol gas). Compromise: 400-500°C gives reasonable rate + acceptable yield (~15-25%). NH₃ is continuously removed → equilibrium shifts right. Fritz Haber (Nobel 1918) + Carl Bosch developed this — considered one of the most important chemical processes ever (feeds ~half of humanity).
6. What are the oxoacids of phosphorus and their basicity?
Rule: basicity = number of P-OH groups (only OH groups are acidic; P=O and P-H are NOT). H₃PO₂ (hypophosphorous acid): structure H₂P(=O)(OH) → 1 P-OH → monobasic. Has 2 P-H bonds → strong reducing agent. H₃PO₃ (phosphorous acid): structure HP(=O)(OH)₂ → 2 P-OH → dibasic. Has 1 P-H bond → reducing agent. H₃PO₄ (phosphoric acid): structure (HO)₃P=O → 3 P-OH → tribasic. No P-H bonds → NOT a reducing agent. H₄P₂O₇ (pyrophosphoric acid): 4 P-OH → tetrabasic. HPO₃ (metaphosphoric acid): 1 P-OH → monobasic.
7. Why does nitrogen show anomalous behaviour compared to other Group 15 elements?
Nitrogen is anomalous because: (1) Small size → high electronegativity (3·0) → forms strong hydrogen bonds. (2) No d orbitals → maximum covalency = 4 (in NH₄⁺); P,As,Sb,Bi can show higher covalency using d orbitals. (3) Forms strong pπ-pπ bonds (N₂, N=O) — most stable diatomic. (4) N-N single bond is weak due to lone pair-lone pair repulsion (adjacent small atoms with LP). (5) Doesn't form N₅ (unlike P forms PCl₅ with d orbitals). (6) N₂ is inert (unlike P₄ which is reactive). This "anomalous" behaviour of first member of a group due to small size and no d orbitals is a general periodic trend.
8. What is the Ostwald process for nitric acid?
Step 1: 4NH₃ + 5O₂ →(Pt/Rh catalyst, 900°C) 4NO + 6H₂O. Step 2: 2NO + O₂ → 2NO₂ (cooled). Step 3: 3NO₂ + H₂O → 2HNO₃ + NO (NO recycled). Product: 68% HNO₃ (azeotrope with water). Concentrated HNO₃ (>90%) by dehydration with H₂SO₄. Uses of HNO₃: fertilisers (NH₄NO₃), explosives (TNT, RDX, nitroglycerin), dyes, pharmaceuticals. HNO₃ is a strong oxidising acid — dissolves most metals except Au, Pt. Aqua regia (3HCl + 1HNO₃) dissolves Au and Pt.
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