Text/character alignment in table tabular environment

Question

I'm having some difficulty with the text alignment in some table columns and was wondering if someone could help me work out what's going on? I've provided a MWE and image of the table output.

Before using the \multicolumn command for the last 2 columns of the top row, all the text was centered as a result of the \begin{tabular}{*{7}{c}} command. However, when using \multicolumn, this appears to override the centering for everything in the last two columns. How can I change this so that the text is centered again?

Also, I'd like to make the text in columns 1, 3 and 5 vertically aligned to the left, so that the first letter/number starts in the same position for each entry, but have it so the text is centered to the width of the column i.e. so the text is centered for the longest character entry for a column and everything else is left aligned to it. Is this possible?

I know that I could specify \begin{tabular}{l c l c l c c }, to left align the columns, but this doesn't quite achieve what I want.

\documentclass[12pt,a4paper]{report}
\usepackage{amsmath}
\usepackage{caption}
\usepackage{array,booktabs,threeparttable}
\usepackage{chemformula}
\usepackage[utf8]{inputenc}

\begin{document}

\begin{table}
\centering
\caption[Alpha emitting radioactive decay series.]{Alpha emitting radioactive decay series.}
\label{tab:alpha decay series}
\begin{threeparttable}
\resizebox{\columnwidth}{!}{%
\begin{tabular}{*{7}{c}}
\toprule
            &             &                &                            &                & \multicolumn{2}{c}{Number of decays} \\
Series type & Series name & Parent nucleus & Half-life ($10^{9}$ years) & Stable nucleus & $\alpha$ & $\beta^{-}$ \\
\midrule
$A = 4n$    & Thorium           & \ch{^{232}Th} & 14.05     & \ch{^{208}Pb}     & 6 & 4 \\
$A = 4n+1$  & Neptunium         & \ch{^{237}Np} & 0.0021    & \ch{^{205}Tl}     & 8 & 5 \\
$A = 4n+2$  & Radium (Uranium)  & \ch{^{238}U}  & 4.4683    & \ch{^{206}Pb}     & 8 & 6 \\
$A = 4n+3$  & Actinium          & \ch{^{235}U}  & 0.7038    & \ch{^{207}Pb}     & 7 & 4 \\
\bottomrule
\end{tabular}%
}
\end{threeparttable}
\end{table}
\end{document}

Answer 1

If \ spanning cell is wider than the natural width of the cells that it spans then all the extra width is placed in the last spanned cell, which is almost never what you want.

One way is to force the alpha and beta to be wider, so the width is shared between the last two columns.

\documentclass[12pt,a4paper]{report}
\usepackage{amsmath}
\usepackage{caption}
\usepackage{array,booktabs,threeparttable}
\usepackage{chemformula}
\usepackage[utf8]{inputenc}

\begin{document}

\begin{table}
\centering
\caption[Alpha emitting radioactive decay series.]{Alpha emitting radioactive decay series.}
\label{tab:alpha decay series}
\begin{threeparttable}
\resizebox{\columnwidth}{!}{%
\begin{tabular}{*{7}{c}}
\toprule
            &             &                &                            &                & \multicolumn{2}{c}{Number of decays} \\
Series type & Series name & Parent nucleus & Half-life ($10^{9}$ years) & Stable nucleus & 
\makebox[1.5cm]{$\alpha$} &
\makebox[1.5cm]{$\beta^{-}$} \\
\midrule
$A = 4n$    & Thorium           & \ch{^{232}Th} & 14.05     & \ch{^{208}Pb}     & 6 & 4 \\
$A = 4n+1$  & Neptunium         & \ch{^{237}Np} & 0.0021    & \ch{^{205}Tl}     & 8 & 5 \\
$A = 4n+2$  & Radium (Uranium)  & \ch{^{238}U}  & 4.4683    & \ch{^{206}Pb}     & 8 & 6 \\
$A = 4n+3$  & Actinium          & \ch{^{235}U}  & 0.7038    & \ch{^{207}Pb}     & 7 & 4 \\
\bottomrule
\end{tabular}%
}
\end{threeparttable}
\end{table}
\end{document}