Bohr proposed his atomic model based on:

List $I$ (Spectral Series for Hyddrogen)	List $II$ (Spectral Region/Higher Energy State)
$A$. Lyman	$I$. Infrared region
$B$. Balmer	$II$. $UV$ region
$C$. Paschen	$III$. Infrared region
$D$. Pfund	$IV$. Visible region

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The correct statements from the following are :

$(A)$ The decreasing order of atomic radii of group $13$ elements is $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$.

$(B)$ Down the group $13$ electronegativity decreases from top to bottom.

$(C)$ $\mathrm{Al}$ dissolves in dil. $\mathrm{HCl}$ and liberate $\mathrm{H}_2$ but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface.

$(D)$ All elements of group 13 exhibits highly stable +1 oxidation state.

$(E)$ Hybridisation of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion is $\mathrm{sp}^3 \mathrm{~d}^2$.

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If the ionization enthalpy and electron gain enthalpy of an element are $275$ and $86\, kcal\, mol^{-1}$ respectively, then the electronegativity of the element on the Pauling scale is

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Taking into account of hybridization and resonance effects, rank the following bonds in order of decreasing bond length.