Relationship Between Keto Enol Forms and Geometric Isomers

To determine the correct statements concerning the structures $\mathbf{E}$, $\mathbf{F}$, and $\mathbf{G}$:

1. Relation between $\mathbf{E}$, $\mathbf{F}$, and $\mathbf{G}$ (Tautomerism vs Resonance):

   • Structure $\mathbf{E}$ is a ketone (keto form).
   • Structures $\mathbf{F}$ and $\mathbf{G}$ are enols (enol forms) differing in the connectivity of hydrogen and the position of the double bond:
     $\text{Keto form } (\mathbf{E}) \rightleftharpoons \text{Enol forms } (\mathbf{F} \text{ and } \mathbf{G})$
   • Since the interconversion involves the migration of a proton accompanied by a shift of a $\pi$-bond (change in atomic connectivity), $\mathbf{E}$ and $\mathbf{F}$, as well as $\mathbf{E}$ and $\mathbf{G}$, are tautomers (structural isomers in dynamic equilibrium), not resonance structures (which differ only in electron distribution).
   • Therefore, statement (A) is incorrect and statement (B) is correct.

2. Relation between $\mathbf{F}$ and $\mathbf{G}$ (Geometrical Isomerism and Diastereomerism):

   • Both $\mathbf{F}$ and $\mathbf{G}$ are stereoisomers with restricted rotation about the carbon-carbon double bond ($\text{C}=\text{C}$).
   • In structure $\mathbf{F}$, the $-\text{OH}$ group and the top substituent are oriented on the same side (or relative spatial arrangement), whereas in $\mathbf{G}$, their spatial orientation across the double bond is reversed (one is $(E)$-isomer and the other is $(Z)$-isomer).
   • Because they differ in the spatial arrangement of atoms around the rigid double bond and are non-superimposable non-mirror images, $\mathbf{F}$ and $\mathbf{G}$ are geometrical isomers.
   • All geometrical isomers that are non-mirror images are, by definition, diastereomers.
   • Therefore, statements (C) and (D) are correct.

Thus, the correct statements are B, C, and D.