Why does stannous chloride occur despite the octet rule?

Question

Shouldn't reaching an octet be any atom's "goal"?

However, I've recently learned about cases that are either expanding octets, or have lesser than "enough" electrons for an octet abiding. e.g.:

S in sulfur hexafluoride (Expanding octet into $\text{3d}$)

Source

B in boron trifluoride (It's "hextet", instead of octet)

Source

However, both $\ce{SnCl2}$ and $\ce{SnCl4}$ are existent. The latter is explainable with octet, but not the former. Surprisingly, $\ce{SnCl2}$ is more stable!

1. How?
2. Why is the phenomenon happening?

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Dave pointed out that the octet nonabidingness is happening in the vapor phase. So I redirect the question to ask about the vapor phase; since that's what I'm looking for.

Answer 1

According to Theoretical and experimental studies of the electronic structure of SnCl2 by photoionization with synchrotron radiation AIP Conf. Proc. 258, 60 (1992)

$\ce{SnCl2}$ has a molecular orbital structure as below (HOMO at the top)

Sigma-antibonding: 36% Sn s, 22% Sn p, 41% Cl s

Nonbonding 99% Cl p

Nonbonding 99% Cl p

Nonbonding 7% Sn p , 92% Cl p

Nonbonding 3% Sn s, 6% Sn p , 90% Cl p

Sigma bonding 14% Sn p, 83% Cl p

Sigma bonding 48% Sn s, 6% Cl s 45% Cl p

Nonbonding 3% Sn p, 96% Cl s

Nonbonding 5% Sn s, 2% Sn p, 92% Cl s

So overall it seems:

There is a total bond order of only 1.

Much of the Sn p and some Sn s electron density has been donated to Cl p orbitals.

There is little mixing of Sn s and p orbitals expect in the antibonding orbital.

additional references:

A Symmetry Rule for Predicting Molecular Structures (This reference explains how, based upon second order Jahn Teller effect, the symmetry of molecular orbitals predicts geometry, for example bent vs linear, and says that the symmetry predicts bent for SnCl2)

and Metal oxidation state effect in photoionization of gas‐phase metal halides (which further discusses the molecular orbitals of SnCl2 discussed above)

Answer 2

$\ce{SF6}$ does not expand it octet by way of using 3d orbitals. It is often taught that way, particularly in introductory classes. A better description involves treating it as a hypercoordinated molecule. See these earlier answers for an explanation of how hypercoordination can be applied. When finished you should be able to apply the concept to $\ce{SF6}$.

• reference 1
• reference 2

$\ce{BF3}$ is electron deficient and reacts rapidly with molecules containing lone pairs of electrons (amines, oxygen containing compounds, etc.) to stabilize the octet configuration. It has also been observed that the $\ce{B-F}$ bond length is somewhat shorter than expected. This has been explained by invoking resonance structures such as

and

To whatever degree these types of resonance structures contribute to the true description of $\ce{BF3}$ they help explain the shortened bond length and also achieve an octet around the central boron.

A similar explanation can be applied to $\ce{SnCl2}$, drawing a resonance structure with a $\ce{Sn=Cl}$ double bond creates an octet around the central tin atom. To whatever degree such resonance structures contribute to the real description of $\ce{SnCl2}$, they will help satisfy the octet electronic configuration around the central tin atom.

Answer 3

There is another rather more simple answer to the question:

$\ce{SnCl2}$ can be thought of as a salt to a much greater extent than $\ce{SnCl4}$. In fact, tin is so often thought of as a metal, that it takes some people time to understand that $\ce{SnCl4}$ is in fact not a salt but much more a molecule.

Then, when looking at $\ce{Sn(II)}$, it becomes clear that the unaccompanied ion would have the electron configuration $\ce{[Kr] 4d^{10} 5s^2}$ — doesn’t that look stable to you? Add two chloride ions that took the electrons originally having been in the $\ce{5p}$-orbitals and voilà.

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I know that this answer simplifies a lot of things that maybe shouldn’t be simplified in all cases. But to the best of my knowledge, it explains the compound and its stability well, and should therefore be considered. Once the question turns into ‘why is $\ce{Sn(II)}$ ionic, while $\ce{Sn(IV)}$ better described as molecular?’, it of course becomes invalid.

Answer 4

in a simple way since tin belong to group 14 the stability of lower oxidation state i.e +2,is comparitively more stable than +4 oxidation state due to inert pair effect (reluctance of s orbital electron to take part in bond formation due to poor shielding ability of d orbital which causes the outer s orbital electron to be attracted towards nucleus.)thus tin loses two p electrons only.the outer s orbital electron does not take part in bond formation might.i hope i am right.