Chlorination of methane

I got a question that asks what is formed in the initiation step in the mechanism of the chlorination of methane.

Before I looked at the possible answers I knew it was chlorine radicals.

Then when I went to circle that it wasn't there..

The possible answers are:

•

hydrogen atoms

•

hydrogen chloride

•

methyl radicals

•

chlorine atoms

Am I correct in thinking this was a typo or am I wrong entirely?

Assistance greatly appreciated!

Reply 1

EierVonSatan

The right answer is amongst the choices :yep:

What is a chlorine radical's electronic configuration?

Reply 2

Jargonium

Original post by EierVonSatan

The right answer is amongst the choices :yep:

What is a chlorine radical's electronic configuration?

A radical is a particle with a single unpaired electron.

Therefore, whenever I write the initiation step I show that under UV light, Cl₂ turns to 2Cl radicals as one electron jumps from the covalent bond to each chlorine (which is homolytic fisson).

That means that the electronic configuration would be:

[Ne] 3s²3p⁴

I still think the answer is chlorine radicals!

Sorry, please explain :colondollar:

Reply 3

EierVonSatan

Original post by Jargonium

A radical is a particle with a single unpaired electron.

[Ne] 3s²3p⁴

No..

Cl₂ is a neutral molecule (34 protons, 34 electrons) and is split into two by UV light:

Cl₂ ---> 2Cl

Leaving two neutral Cl's (17 protons, 17 electrons) with configuration [Ne] 3s²3p⁵

I still think the answer is chlorine radicals!

You're correct. Does this help??

Reply 4

Borek

Original post by Jargonium

[Ne] 3s²3p⁴

Try to assign charge to this configuration.

And note it doesn't have any unpaired electrons.

Reply 5

Jargonium

Original post by EierVonSatan

No..

Cl₂ is a neutral molecule (34 protons, 34 electrons) and is split into two by UV light:

Cl₂ ---> 2Cl

Leaving two neutral Cl's (17 protons, 17 electrons) with configuration [Ne] 3s²3p⁵

You're correct. Does this help??