Lai Yung asked in 科學及數學化學 · 1 decade ago

AL chemistry Question

1.Why is the P4 molecule unstable?

2.Explain why NH3 is more basic than PH3.

3.Hrdorgen flouride can be prepared by the action of sulfuric acid on sodium flouride . Explain why hydrogen bromide cannot bre prepared by the action of the same acid on sodium bromide.

4.Al³+ is not a Brønsted acid but Al(H2O)6³+ is .Explain

5.Explain which of the following cations is larger , and why : Cu + or Cu²+

6. Explain why the atomic radius of calcium is singificantly greater than the ionic radius of calcium.

1 Answer

  • ?
    Lv 7
    1 decade ago
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    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.


    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.


    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.


    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.)


    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.


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


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