A quiz-like exercise to aromatic compounds (Hückel's rule, planarity, conjugation):

Here's the corresponding TikZ code (using chemfig):

\documentclass[12pt,a4paper,landscape]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{shadows}

\usepackage{chemfig}


\begin{document}
\pagestyle{empty}
\begin{tikzpicture}
%% ----------------- A little Quiz ----------------------------
%% Which of the following Compounds are aromatic?
%% 
\node at (3,-1) {\bf Which of the following compounds are aromatic? \rm};

%% Cyclopenta = cyclopentadienylanion!
\coordinate (Cyclopenta) at (0,-3);
\node at (Cyclopenta) {\chemfig{*5(@{C1}-=-=-)}};
\chemmove{
\node[anchor = north] at (C1) {\chemfig{\ominus}};
}

%% Naphtalene
\coordinate (Naphta) at (4,-3);
\node at (Naphta) {\chemfig{[:-60]*6(=-=-=-)-[::-150]=^[::60]-[::60]=^[::60]-[::60]}};

%% Antracene
\coordinate (Antra) at (10,-3);
\node at (Antra) {\chemfig{[:-60]((*6(=-=-(-[::-60]=_[::-60]-[::-60]=_[::-60]-[::-60])%%
=-))-[::-150]=^[::60]-[::60]=^[::60]-[::60])}};

%% Next Line
%% More Molecules

%% Cyclopropanylanion and Cyclopropanylcathion
\coordinate (Propanion) at (0,-6);
\node at (Propanion) {\chemfig{-[::60]@{C2}-[::-120]=[::-120]}};
\chemmove{
\node[ anchor = south ] at (C2) {$\ominus$};
} 

\coordinate (Propcathion) at (2,-6);
\node at (Propcathion) {\chemfig{-[::60]@{C3}-[::-120]=[::-120]}};
\chemmove{
\node[ anchor = south ] at (C3) {$\oplus$};
} 

%% Cyclobutane
\coordinate (Cyclobut) at (4,-6);
\node at (Cyclobut) {\chemfig{*4(-=-=-)}};

%% Cyclobutane-cathion
\coordinate (Cyclobutcathion) at (6,-6);
\node at (Cyclobutcathion) {\chemfig{*4(-=-=)}};
\coordinate (Q) at (5.5,-6.5);
\node[anchor = north east] at (Q) {$\oplus$};

%% Cyclobutane-anion
\coordinate (Cyclobutanion) at (8,-6);
\node at (Cyclobutanion) {\chemfig{*4(-=-=)}};
\coordinate (Q') at (7.5,-6.5);
\node[anchor = north east] at (Q') {$\ominus$};

%% Some more fancy molecules
\coordinate (TwoProp) at (10,-6);
\node at (TwoProp) {\chemfig{@{N}-[::60]=[::-120](-[::120]@{N'}-[::+120])-[::-120]}};
\chemmove{
\node[ anchor=north ] at (N) {$\oplus$};\node[anchor=south] at (N') {$\oplus$};
}

%% Solution:
\node at (-0.6,-8) {\bf Solution:\rm};
\begin{scope}[]
\node[ white,rotate = 180,draw=blue,fill = blue!50!white!50!gray,rounded corners ] at( 3, -11.5 ) {%%
\begin{minipage}{.75\textwidth}
{
\scriptsize
\begin{itemize}
\item 6-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 10-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 14-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 2-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 6-$\pi-\mathrm{e^-}$, planar, not conjugated$=>$anti-aromatic
\item 2-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic,but not stable!
\end{itemize}
}
\end{minipage}
}
;
\end{scope}
\end{tikzpicture}
\end{document}