AL chemistry Question

1.
In P4 molecule, each P atom is sp3 hybridized. The angle between two sp3 orbitals is 109.5o.

The P4 atoms are arranged as a regular tetrahedron. The bond angle is 60o. Therefore, the P-P covalent bond is formed by non-head-on overlapping between two sp3 orbitals. This is called bond angle strain. The P-P covalent bond is thus unstable because of bond angle strain.


2.
Reason 1:
NH3 accepts a proton to become NH4+, while PH3 to become PH3. The positive charge of NH4+ is more dispersed than PH4+, and thus NH4+ is more stable. Therefore, NH3 accepts a proton more readily, and thus more basic.

Reason 2:
Comparing with PH3, the lone pair electrons on the N atom in NH3 are more tightly held by the more electronegative N atom. Therefore, the lone pair of electrons on the N atom is less diffused, and thus more available to accept a proton.


3.
On heating, both NaF and NaBr can react with conc. H2SO4 to form HF and HBr respectively.
NaF + H2SO4 → NaHSO4 + HF
NaBr + H2SO4 → NaHSO4 + HBr

HBr is a reducing agent, while hot concentrated H2SO4 is an oxidizing agent. Redox reaction between HBr and conc. H2SO4 thus occurs to give Br2 vapour and SO2 gas.
2HBr + H2SO4 → 2H2­O + Br2 + SO2
HF is not a reducing agent, which is evolved and collected.


4.
Al3+ itself cannot release H+, and is thus not a Bronsted acid.

Al3+ has a strong polarizing power, because of small size and great charge. After [Al(H2O)6]3+ releasing H+ to form [Al(OH)(H2O)5]2+, the OH- ligand is stabilized.
(An old explanation is that H2O ligand is polarized, thus becomes unstable, and release H+ readily. This explanation is not correct.)


5.
Cu+ ([Ar] 3d10) has a greater size than Cu2+ ([Ar] 3d9). This is because Cu+ has a greater number of electrons, and thus the repulsion between electrons is stronger.


6.
Ca2+ ([Ar]) has a significant smaller size than Ca ([Ar] 4s2), because Ca2+ has one less filled electronic shell.
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