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In $\ce{PF2(CH3)3},$ what is the maximum number of atoms lying in one plane?

$\ce{P}$ is the central atom with the hybridization $\mathrm{sp^3d}.$ Two $\ce{F}$ atoms will form axial bonds with it, while three $\ce{CH3}$-groups form equatorial bonds. Each of these $\ce{CH3}$-groups has tetrahedral geometry, which means that one $\ce{H}$ atom lies in the same plane as the $\ce{C}$ central atom. Thus, two $\ce{F}$ atoms, one $\ce{P}$ atom, one $\ce{C}$ atom and one $\ce{H}$ atom with it, totalling five atoms, lie in the same plane.

Is my attempt correct?

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  • $\begingroup$ What if you try another plane? $\endgroup$ Sep 27 '20 at 6:25
  • $\begingroup$ @Ivan Neretin Which? $\endgroup$ Sep 27 '20 at 6:41
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    $\begingroup$ For one, you state $\ce{F-P-F}$ are aligned in axial manner, without noting H, nor C. Based on this assumption, this adds up to three. But since the axial direction is orthogonal to the equatorial plane, count what's possible there; build a model and don't forget that methyl / $\ce{CH3}$-groups bound by $\sigma$-bonds may rotate like propellers, thus form multiple conformers. $\endgroup$
    – Buttonwood
    Sep 27 '20 at 6:46
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    $\begingroup$ @Buttonwood Ah, I see. There should be 7 atoms in the equatorial plane ($P$, 3 $C$ atoms and 3 $H$ atoms). $\endgroup$ Sep 27 '20 at 6:50
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    $\begingroup$ @IshanJMukherjee At maximum, true; because molecules are flexible and dynamic / vibrate (in IR spectroscopy occasionally you occasionally hear a colloquial «breathing vibration»). $\endgroup$
    – Buttonwood
    Sep 27 '20 at 7:01
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Seven is correct (the OP found this in the comments). The methyl carbons are in the equatorial plane and each one can rotate a hydrogen atom into this plane. In fact the hydrogen atoms fit well with this conformation so it's actually favorable.

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