Cycloheptatrienyl anion antiaromatic or non-aromatic?
At first, it seems like it is antiaromatic due to the presence of $8 \pi$ electrons, but some sources say it is non-aromatic as $sp^3$ carbon destroys its planarity and the compound become non-planar and thus non-aromatic.
What is it actually?
It's a little bit of both, really. From Salikov et al.[1]:
The ground state of cycloheptatrienyl anion in the gas phase is triplet, planar and Baird-aromatic. In DMSO, it assumes a singlet distorted allylic form with a paratropic ring current. The other derivatives in both phases assume either allylic or diallylic conformations depending on the substituent pattern.
Decoding things:
The term "Baird-aromatic" for the gas-phase structure refers to the triplet state often called "antiaromatic". This concept is explained here in reference to the cyclopropenyl system. The triplet ring state is actually more resonance-stabilized than a corresponding acyclic structure with the same triplet spin state, but significantly less stablized than a Hückel-aromatic ring following the usual $4n+2$ rule would be.
The "allylic/diallylic" structure of the substituted rings, and the unsubstituted ring in DMSO solvent, actually break the ring conjugation along certain bonds, producing isolated double bonds and allylic groups. These structures would then properly be called nonaromatic.
The calculations cited above are theoretical, but the existence of cycloheptatrienyl anion has some experimental evidence[2]. A solution of cycloheptatriene in basic liquid ammonia ("basic" here meaning they added $\ce{KNH2}$) will dimerize the hydrocarbon to a mixture of tricyclic products; the above reference reports evidence for the cycloheptatrienyl ion as an intermediate in this reaction.
References
Salikov RF, Belyy AY, Ilyushchenko MK Platonov DN, Sokolova AD, Tomilov YV. "Antiaromaticity of Cycloheptatrienyl Anions: Structure, Acidity, and Magnetic Properties". Chemistry. 2024 Jul 19; 30(41):e202401041. doi: 10.1002/chem.202401041. Epub 2024 Jul 5. PMID: 38785416.
Stuart W. Staley, Arnold W. Orvedal. "Mechanism of the bass[sic]-promoted cyclodimerization of cycloheptatriene". J. Am. Chem. Soc. 1974, 96, 5, 1618–1620. doi: 10.1021/ja00812a070.
I want to preface this by saying that the geometry determines the hybridization, not the other way around.
That said, the molecule is not anti-aromatic. For that, the molecule would need to be cyclic, have a conjugated π-electron system, be planar, and have 4n π-electrons within the system. The cycloheptadienyl anion does not meet the 4n π-electron requirement (it has 6 π-electrons, four in the double bonds and two more from the lone pair), and neither does it meet the conjugation or planarity requirements.
In this case, the lone pair does not contribute to making the molecule planar like it would with something like the cyclopentadienyl anion. The carbon with the negative charge on it will be somewhere between $\text{sp}^3$ and $\text{sp}^2$ hybridized. What carbon the anion is on is inconsequential here; a pure $\text{sp}^2$ hybrid is not possible because of the significant ring strain that would develop with this geometry. Regardless, there are always two other tetrahedral carbons (in their $\text{sp}^3$ hybridized geometry) which make the molecule non-planar. You would classify this molecule as non-aromatic.
Cycloheptatrienyl is the more common example, and perhaps what you meant in your question considering you mention 8 π-electrons. Consider (in the photo below) the carbanion; if this molecule was planar, it would be anti-aromatic (4n π-electrons, conjugated, and cyclic) and consequently incredibly unstable. The 'why' behind the instability of this anti-aromaticity (for pretty much all molecules) has to do with the molecular orbitals (which I encourage you to investigate more about). For a shortcut, you can try drawing the Frost circle for the cycloheptatrienyl anion and find the presence two unpaired electrons both in the antibonding MOs! Not very stable.
As pointed out by Mithoron, practically the molecule will bend out of planarity to create a tetrahedral carbon (with the anion on it) to avoid the anti-aromatic destabilization, facilitated by the large enough ring system, making it non-aromatic too.
