Can fluorescent & phosphorescent molecules still be fluorescent or phosphorescent as ions and not as salts?
Fluorescence derives from an electronically excited state after the molecule has been excited by absorbing a photon, usually in the visible or uv parts of the spectrum. Almost all organic molecules fluoresce to a greater of lesser extent, most common are aromatics for example, benzene and its derivatives as well as most dye molecules such rhodamines or coumarins. They can be ions or neutral. 

When a molecule is electronically excited an electron is transferred from the highest occupied molecular orbital (HOMO) into an anti-bonding orbital usually the lowest one (LUMO). From here the molecule, which now has excess energy, can emit a photon and return to the ground state. The lifetime of the singlet excited state  varies between different types of molecules but is generally speaking around a few nanoseconds and not longer than microseconds. 

By definition the excited state produced has the same spin as the ground state, which is usually means spin paired electrons $\ce{^ (v)}$. They remain paired even in the excited state (called a singlet excited state). Fluorescence is emission from a molecule with spin paired electrons, the transition is called 'allowed' and happens rapidly.  

Occasionally the molecule can suffer a perturbation which causes one of the spins to be flipped to produce 'parallel' spins $\ce{^ ^}$ (heavy atoms or paramagnetic species can cause this). In this case a triplet state is produced and now the transition back to the ground state is far more difficult as the spin-flip has to be undone. The emission from such a state is called phosphorescence, it is long lived vs fluorescence, microseconds and longer, and as the triplet state is lower in energy than the singlet phosphorescence occurs at a longer wavelength in the same molecule, as fluorescence. 

The figure  shows the energy levels. Different process compete with one another, thus the singlet can loose energy by fluorescence (with rate constant $k_f$) or by non-radiatively crossing to a triplet state (intersystem crossing) or non-radiatively to the ground state (internal conversion). 


continued after question below.

The breaking of formally ‘ forbidden’ transitions is very common indeed, possibly the best example is in the observation of phosphorescence from almost any aromatic or dye molecule. Absorption in these molecules is generally a spin allowed transition, (single to singlet), however, triplet states, from which phosphorescence is observed,  can only be produced by a spin forbidden process, singlet to triplet, and just as in phosphorescence, is a spin forbidden process and becomes allowed due to spin-orbit coupling. In the case of the manganese ions and in other transition metal the forbidden transition is in absorption.  

The general treatment starts with allowing the two states to interact via some  operator. In the case of photons (absorption / emission) it is an electric dipole operator, i.e. dipole moment operator. If the operator is $H_0$, the rate of the transition from one state to another is  

$$W_{if}^0 \approx \int \psi_f H_0 \psi_i d\tau = \int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau $$

where $\sigma$ represents the spin wavefunction and $\psi$ the spatial one. As the operator does not depend on spin then
$$\int \psi_f\sigma_f H_0 \psi_i \sigma_i d\tau = \int \psi_f H_0 \psi _i d\tau\int\sigma _f \sigma_i dtau = 0$$
and the  result is zero because the spin wavefunctions belonging to i and f are orthogonal (as singlet &amp; triplet) and so the last integral is zero. 

If we suppose that it is possible to mix a little of the initial state wavefunction into the final one and generate a new state with wavefunction $\psi_m$ then $$\psi_m=\psi_f + b\psi_i$$ where b is some mixing constant we can determine later on. Now using this wavefunction as the final state and repeating the calculation above and separating out terms then as orthogonal spin terms are zero the only term left is

$$W_{if} \approx \int \psi_m H_0 \psi_i d\tau = b\int \psi_i \sigma_i H_0 \psi_i \sigma _i d\tau$$
in which case the integral is no longer zero and evaluates to a number.(Note that all the indices are the same unlike the equation for $W_{if}^0$)  This equation shows that there is a finite rate for the process with operator $H_0$. In the case in point, spin-orbit coupling, the operator $\Omega$ depends on the mass of the atoms involved and the product of orbital and spin angular momentum. This is why it is also called a ‘heavy atom effect’ when applied to emission processes.

The constant b is deribed using perturbation theory and turns out to have the form 
$$b=\frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i}$$
where $\Omega$ is the spin-orbit operator. Thus the rate for the process finally becomes

$$ W_{if} \approx \frac{\int\psi _f \Omega \psi _i d\tau}{E_f-E_i} $$

and we can use symmetry to decide if this integral is zero or is finite. The dependence on energy gap between states is clear, rapid rate constants for small energy gap. (The degenerate case when the energy gap is zero needs special and a more complex treatment)