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Ions are very reactive for obvious reasons (i.e. Coulomb force); $\ce{Cl-}$ ions will be very quick to form an ionic bond with positive ions (unless inhibited, e.g. through being in a solution; let's consider both $\ce{Cl}$ and $\ce{Cl^-}$ to be in gaseous phase).

Chlorine atoms without reduction ($\ce{Cl}$), aren't charged, yet tend to combine into $\ce{Cl2}$, utilising a covalent bond.

Thus, both $\ce{Cl}$ and $\ce{Cl-}$ tend to form bonds. My confusion lies in the following:

  • Noble gases are chemically inert, $\ce{Cl-}$ has the electron configuration of $\ce{Ar}$, yet it's very chemically reactive.
  • Neutral atoms aren't as attracted to each other as ions, and could be considered more "chemically stable" than ions with higher charge. However, even though $\ce{Cl}$ has no charge, it still seems eager to react and get to the electron configuration of a noble gas.

In both cases, information seems conflicting, and perhaps a more clear explanation will make me understand the situation. Is one more prone to react than the other?

(N.B.: My knowledge on chemistry, maths, physics etc … stretches to (but doesn't include) quantum physics, so feel free to go all-out.)

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  • $\begingroup$ In addition to the answers you got it is evident fron your text that you are considering ions to be reactive as the same of to form ionic bond. Altough they reactivity is dictated by being charged, they can react in different ways and end up in a covalent bond. $\endgroup$ – Alchimista Dec 15 '18 at 17:40
  • $\begingroup$ The apparent example of chloride ions rapidly forming ionic bonds isn't really a chemical reaction but a change of state. The ionic solid still contains the (stable) chloride but in a solid crystal with counterbalancing ions. The stability of that solid drives the physical change of state not any reaction of the chloride. $\endgroup$ – matt_black Dec 16 '18 at 20:37
  • $\begingroup$ @matt_black I disagree with that statement. The combination of positive and negative ions to form an ionic solid can certainly be considered as a chemical reaction simply because there is a chemical change occurring. Ionic bonds (i.e. electrostatic forces of attraction) are being formed, with the release of heat energy. The potential energy of the ions experience an overall decrease as the ions experience the mutual attraction and repulsion forces, with net attraction dominating. $\endgroup$ – Tan Yong Boon Dec 17 '18 at 13:21
  • $\begingroup$ @TanYongBoon But that makes the freezing of water a chemical reaction too. Few would agree with that. $\endgroup$ – matt_black Dec 17 '18 at 13:25
  • $\begingroup$ @matt_black In that case, what do you take to be the distinguishing features of a chemical reaction? $\endgroup$ – Tan Yong Boon Dec 17 '18 at 14:36
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I think we can also approach this from an energetics perspective. Reactivity can also be seen to be related to thermodynamic stability in the following manner: The more thermodynamically stable a chemical species is, the less reactive it is. With this in mind, we can look at the enthalpies of formation of $\ce {Cl.}$ and $\ce {Cl^-}$.

Your question is slightly vague as you did not state the specified state of the substance. This can critically affect the answer. The reactivity of a heavily solvated $\ce {Cl^-}$ ion is certainly different from that of a free $\ce {Cl^-}$ ion in the gaseous phase.

By definition, the enthalpy change of formation, $\Delta H_f$, of an element in its standard state (e.g. $\ce {Cl2 (g)}$) is $\ce {0}$. $\ce {2 Cl.}$ are produced from the homolytic cleavage of the $\ce {Cl-Cl}$ bond in $\ce {Cl2}$. Knowing the bond dissociation energy of this bond to be $\ce {+242 kJ/mol}$, we can approximate $\Delta H_f$ of $\ce {Cl.}$ to be $\ce {+121 kJ/mol}$.

Similarly, the gaseous chloride ion can be produced from the gain of an electron by $\ce {Cl.}$. The first electron affinity, $\ce {1st E_a}$, of chlorine is the energy change when one mole of electrons is acquired by one mole of these gaseous chlorine atoms to give one mole of singly, negatively-charged gaseous $\ce {Cl-}$ ions. $\Delta H_f$ of $\ce {Cl- (g)}$ can thus be approximated as $\Delta H_f$ of $\ce {Cl.}$ + $\ce {1st E_a}$ of chlorine . Since this $\ce {1st E_a}$ is $\ce {-349 kJ/mol}$, this enthalpy change of formation for $\ce {Cl^- (g)}$ can be approximated as $\ce {-228 kJ/mol}$.

As can be seen from energetics data, $\ce {Cl^-}$ is more stable than $\ce {Cl.}$ in the gaseous phase. Thus, we may say that the $\ce {Cl.}$ is more reactive in the gaseous phase. Furthermore, being charged, $\ce {Cl^-}$ is solvated by water molecules in aqueous solution, further increasing its stability as it has ion-dipole interactions with the neighbouring water molecules. Electrically-neutral chlorine radicals would not experience this solvation effect, thus they would also be more reactive in the aqueous phase. In non-polar solvents, $\ce {Cl^-}$ and $\ce {Cl.}$ would have similar reactivity as in the gaseous phase since they do not experience starkly different solvation effects. Overall, based on energetics, we may say that $\ce {Cl.}$ is generally more reactive than $\ce {Cl.}$.

Additionally, it is important to point out the difference in the type of reactivity of the two different chemical species. Being negatively-charged, chloride ions participate in ionic reactions, involving charged species. These reactions include nucleophilic substitutions, as well as nucleophilic additions. On the other hand, chlorine radicals mostly participate in non-polar reactions, involving mostly radical species, such as free radical substitutions.

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  • $\begingroup$ A quantitative approach indeed seems to be appropriate for an answer, and based on the reasoning through electron affinity, I'm assuming that qualitatively, one may conclude that the noble gas configuration is the least energetic state an element could reach, thus the most stable? N.B.: As you pointed out, I was in fact referring to gaseous states, so I've edited the question to make that more clear. $\endgroup$ – Mew Dec 16 '18 at 12:07
  • $\begingroup$ @Mew That is not true at all. The idea of the noble gas configuration as the state of the lowest energy is very wrong, if you are referring to the chemical species in the gaseous phase. For example, to make Mg2+ (g) from Mg (s) in standard state, the enthalpy change is very large and endothermic. What makes this ionisation of Mg an energetically favourable process is the bonding interactions that the Mg2+ ion can undergo (e.g. ion-dipole interactions with water molecules, ionic bonds with Cl- ions etc.)... I hope I have made you aware of how important it is to specify the physical state... $\endgroup$ – Tan Yong Boon Dec 16 '18 at 12:35
  • $\begingroup$ Then why is the noble gas configuration so commonly observed in compounds if it is not the most stable configuration energetically? The reason is simply this coincidentally allows the chemical maximise the energy loss overall from the exothermic bonding interactions and the endothermic electron affinity/ionisation/atomisation processes. $\endgroup$ – Tan Yong Boon Dec 16 '18 at 12:41
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Ions actually aren’t particularly reactive (chemically speaking). Furthermore, most of the reactions they participate in are either full-blown redox reactions (i.e. the product will be a completely different ion or nothing at all) or simple ligand exchange reactions. There are some examples in organic chemistry, especially concerning nucleophilic substitutions, where ions will react in a non-redox fashion to form a new stable compound—but these don’t often lead to high reaction enthalpies.

Radicals, on the other hand, are particularly reactive. The most obvious indication of this is a chlorine radical being able to abstract a hydrogen atom from a $\ce{C-H}$ bond in radical chain reactions:

$$\ce{R-H + Cl. -> R. + H-Cl}\tag{1}$$

The bond dissociation enthalpy of a typical $\ce{C-H}$ bond is in the area of $\pu{420 kJ/mol}$ which is quite a lot of energy. You will be hard-pressed to find an example of an anion abstracting a hydrogen from a hydrocarbon unless you are using a very, very strong base or have a particularly acidic carbon atom. Most notably, the chloride anion is such a weak base that it will never deprotonate any neutral, unactivated hydrocarbon.

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You've more or less walked all around the answer. Think about this partial statement of yours:

Noble gases are chemically inert, [and] $\ce{Cl^-}$ has the electron configuration of $\ce{Ar}$.

The free chlorine atom is a radical with one unpaired electron. This arrangement is highly unstable and the chlorine radical either wants to either extract an electron from some atom to become a $\ce{Cl^-}$ anion, or to form a covalent bond somehow.

So, in general, the free chlorine atom is more reactive than the chloride ion. However which is more reactive in a particular reaction, depends on the reaction. For instance in interstellar space the $\ce{Cl^-}$ anion would react more with $\ce{H^+}$ cations because of the coulombic attraction.

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Using chlorine as an example, $\ce{Cl-}$ is reactive for the reasons you would think - it is negative so it is drawn to other species with positive charges or electron deficient areas. $\ce{Cl-}$ does have a stable electron configuration, but that doesn't mean it not reactive. $\ce{Ar}$ is only not reactive like $\ce{Cl-}$ as $\ce{Ar}$ has no charge meaning it is not drawn to species other than by van de Waal's forces.

$\ce{Cl}$ atoms are a little bit of a different case. chlorine atoms will reactive together to form $\ce{Cl2}$ very rapidly giving both atoms a stable electron configuration. But one individual chlorine atom is very very reactive as it has an unpaired electron - a so called chlorine free radical. Any free radical is very reactive as the unpaired electron results in a less stable atomic orbital - so it reacts very quickly to increase the atom's stability. when chlorine atoms bond together to form $\ce{Cl2}$ they are still reactive but no where near as reactive as a chlorine free radical - such so you cannot just find chlorine free radicals they are only formed during reactions and act as very reactive intermediates.

So to answer to your original question individual $\ce{Cl}$ atoms are much more reactive that $\ce{Cl-}$ but as we don;'t get chlorine free radicals outside of reactions a better question would be is $\ce{Cl2}$ or $\ce{Cl-}$ more reactive? and the answer to this depends on what environment they are in as the two reactive in different ways.

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  • $\begingroup$ So then, if I understand correctly, an atom's orbital instability (the absence of filled or semi-filled subshells) is a greater motivator for a reaction to happen than an instability - if we may call it that - in charge? As a follow-up: What kind of force is at the base of radicals being so violent? I can't see how, for example, the Coulomb force has any direct role in this. $\endgroup$ – Mew Dec 15 '18 at 18:16
  • $\begingroup$ a semi-filled sub-shell ISN'T unstable to have that clear. But each sub-shell is made of orbitals, of which each can hold two electrons and it is unstable when an orbital only has one electron in it. the reason why free radicals react so violently is that they want to fill the orbital as they are thermodynamically unstable with it only having one electron in it. There isn't really a force that explains their reactivity as they are neutrally charged. It is only due to stability of the atom $\endgroup$ – H.Linkhorn Dec 15 '18 at 18:56
  • $\begingroup$ I feel like that is explaining what instability is, but not which physical phenomenon underlies the effect. Based off of Tan Yong Boon's response, it seems that this is part of the second law of thermodynamics, but I might be overlooking a specific actor (i.e., it'd seem strange that $Cl$ bonds to $Cl$ because of its radical state, but without any force making the two approach each other, based on their neutral state). $\endgroup$ – Mew Dec 16 '18 at 11:55

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