Redox+ ionic bond strength

2010-05-11 4:37 am
1. What are all the factors affecting ionic bond strength?
2. Why does more reactive metals form more stable compounds? (KCl is more
stable than NaCl as K is higher than Na in the reactivity series. However, it
seems that K has a larger atomic radius, meaning that the Na atom may hold the electron in the ionic bond more tightly.)
3. In other words, is ionic bond strength related to stability of compound?
4. Why isn't iron in the reaction between iron and chlorine gas a reducing agent?

Thank you for your help.

回答 (1)

2010-05-11 6:46 am
✔ 最佳答案
1.

There are 2 factors affecting ionic bond strength:
(1) charges of the ions
(2) size of ions

(1) e.g. ionic strength: MgO > NaCl

Mg2+(g) + 2Cl-(g) à MgCl2(s) △Hlattice = -788 kJmol-1
Na+(g) + Cl-(g) à NaCl(s) △Hlattive = -2602 kJmol-1

△Hlattice is more –ve, the ionic bonding is stronger.
The electrostatic attraction between Mg2+ and Cl- is stronger than that of Na+ and Cl-.
Thus the ionic bond strength: MgCl2 > NaCl

(2) e.g. ionic strength: NaCl > KCl

When the size of ions increases, the electrostatic attraction between cations & anions decreases and thus the strength of ionic bonding decreases.

2.

Na+ is smaller than K+ in size.
Na+ has a stronger polarizing power due to its smaller size.
The electron cloud of Cl- becomes more distorted and thus the stability of NaCl decreases.

3.

Yes.
But, the lattice enthalpy is only one of the factors of the stability of the compound.
However, the stability also depends on other enthalpy and the entropy involved in the reaction.

4.

Fe(s) à Fe2+(aq) + 2e- EΦ = + 0.44V
Cl2(g) + 2e- à 2Cl-(aq) EΦ = + 1..36V

Fe(s) is a reducing agent during the aboved reaction.

2010-05-10 23:30:19 補充:
Mg2+(g) + 2Cl-(g) à MgCl2(s) △Hlattice = -2602 kJmol-1


Na+(g) + Cl-(g) à NaCl(s) △Hlattive = -788kJmol-1
參考: 撼頭埋牆


收錄日期: 2021-04-13 17:14:16
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