Question bank — Phosphorus allotropes (white, red, black); P₄O₆, P₄O₁₀; oxoacids of P (H₃PO₃ vs H₃PO₄ basicity)

False. One H is bonded directly to P (P–H, non-ionisable); only the two O–H hydrogens leave as $H^{+}$, so basicity $=2$. Count P–OH groups, not total H.

False. Both are allotropes — the same element (P) in different bonding arrangements; white is strained $P_4$ tetrahedra, red is a polymer chain.

False. Six are bridging (P–O–P edges) and four are terminal ($P=O$ caps); the four terminal ones are exactly why P is $+5$ and not $+3$.

False. $H_{3}P{O}_4$ has no P–H bond, and P is already at its maximum $+5$ state, so it cannot be oxidised further → it is not a reducing agent. Reducing power needs a P–H bond.

False. Colour has nothing to do with it. Black P is a layered, graphite-like network with normal, strain-free bond angles → it is the most stable / least reactive allotrope.

False. It is a rearrangement done in an inert atmosphere at ~$573K$; if air were present the white P would combust to $P_4{O}_{10}$ instead of relaxing into chains.

True. $P_4{O}_6$ already has the 6 bridging O; adding 4 terminal O atoms (capping each corner P) gives $P_4{O}_{10}$ and raises P from $+3$ to $+5$.

True. $H_{3}P{O}_2$ has two P–H bonds versus one in $H_{3}P{O}_3$; more P–H bonds → stronger reducing power (it even reduces $A{g}^{+}$ to metallic Ag).

Error: water does not react with white P. It is stored under water to exclude air, since white P ignites in air at ~$35°C$.

Error: two of those H atoms are P–H (non-acidic). Only one O–H exists, so $H_{3}P{O}_2$ is monobasic (basicity = 1).

Error: P orbitals actually prefer ~$90$–$109°$. The $60°$ is forced by the tetrahedral $P_4$ geometry, creating angular strain — the opposite of a preference.

Error: $P_4{O}_6$ has P in the $+3$ state, so it gives the $+3$ acid, $H_{3}P{O}_3$ (phosphorous acid). $P_4{O}_{10}$ ($+5$) is the one that gives $H_{3}P{O}_4$.

Error: it is a chemical dehydrating agent — it chemically grabs water (even pulling it from $HN{O}_3$ to make $N_2{O}_5$), not merely adsorbing it physically.

Error: dibasic just means two ionisable H atoms; the second ionisation is always weaker (harder to pull H⁺ off an already-negative anion), not stronger.

In O–H the very electronegative O pulls electron density off H, making it easy to release as $H^{+}$; in P–H the electronegativities are nearly equal ($E_P\approx 2.19$, $E_H\approx 2.20$) so the bond is non-polar and won't ionise.

Limited $O_2$ can only bridge the 6 P–P edges (P stays $+3$); abundant $O_2$ additionally caps the 4 corner P atoms with terminal O, raising them to $+5$.

Oxygen doesn't break the P skeleton — it merely inserts into the P–P edges as bridges and adds terminal caps, so the underlying tetrahedral arrangement of the four P atoms is preserved.

We want the strained $P_4$ units to rearrange into chains; in air the reactive white P would just burn to $P_4{O}_{10}$ instead.

More bonds to oxygen (higher ox. state) means more of those oxygens end up as O–H groups after hydrolysis; $+1,+3,+5$ line up with 1, 2, 3 O–H groups respectively.

$H_{3}P{O}_3$ still has one P–H bond and P is only $+3$ (can rise to $+5$); $H_{3}P{O}_4$ has no P–H and P is already at its top state $+5$, so there is nothing left to oxidise.

Yes. $3-2=1$, matching its monobasic behaviour — the formula generalises to all three oxoacids ($H_{3}P{O}_2/H_{3}P{O}_3/H_{3}P{O}_4$ → basicity $1/2/3$).

P–H count goes $2\to 1\to 0$; fewer P–H bonds means weaker reducing power, so reducing ability decreases and basicity increases in that same order.

No. With P at the maximum $+5$ and zero P–H bonds, there is no oxidisable centre — this is a hard boundary, not a "weak reducer" case.

To white P only — its strained, high-energy $P_4$ molecules cause both the glow (phosphorescence) and the toxicity; red and black P are non-toxic and do not glow.

It must contain at least one (likely two) P–H bonds, pointing to a lower oxidation state acid like $H_{3}P{O}_2$ — since only P–H bonds confer strong reducing power.