How to determine capacitor and inductor values for parallel tank circuit?
Here's the first formula to start with: -



This allows you to understand the ratio of L to C given that you already know the Q-factor yhat you need. R is the series resistance of the inductor.

If you don't know what Q you want then here's a crash course: -



Q is centre frequency (f0) divided by 3 dB band width. In other words if you want a really tight filter you want a really high Q: -



Pictures taken from here (an answer I previously answered that may provide useful information).

So, given Q and some amount of series resistance that you will want to set to help determine Q you can get to the ratio of L to C. It's just math now - C becomes a value related to L and your first formula is used to get the resonant frequency.

Given the operating frequency you mention a first pass guesstimate is around 4,7 nH for L. The formula then tells you that C is going to be about 1 pF - this gives you about 2.32 GHz resonance. Bear in mind that at this frequency the coil capacitance and PCB parasitic capacitances are going to make this problematic to control precisely and it might prove more useful making the inductor from two parallel tracks a few mm long with the capacitor soldered between them - keep the heat on and power on and move the cap with the soldering iron until things look spot on frequency wise. 

Don't underestimate how much trickier this will be at 2.4 GHz compared to running at (say) 500 MHz (where it is fairly easy to do but I'd still probably use a stripline inductor as I've just described at 500 MHz +.



EDIT to give some math: -

If you mess around with the formulas for resonance and Q you can get C in terms of L, Q and R: -

\$C = \dfrac{L}{Q^2R^2}\$

If you plug that into the resonance formula and do some math you find that...

\$L = \dfrac{Q\cdot R}{2\pi f}\$ or \$Q = \dfrac{\omega_0 L}{R}\$ as per the 2nd part of my original formula.

So, let's say you want a Q of 100 and you assume the AC resistance (not DC resistance) is 1 ohm. Plug the numbers in and L comes out at 6.63 nH. But of course you have the problem of what R actually is and you will have to read the data sheets carefully but many will provide you the Q value of the inductor. 

However you have to find a part that has Q specified near or at the operating frequency and, what you might find is that the Q is quite low such as 10. Then you have to calculate twhat the series resistance is. A 4.7 nH inductor has an impedance of 71 ohms at 2.4 GHz and, if it only has a Q of 10, the effective series resistance is 7.1 ohms.

So, plug those numbers back into the formula and you get: -

\$L = \dfrac{10\times 7.1}{2\pi f}\$ = 4.7 nH.

I guess from a new starting point it is a bit daunting - I kind of naturally knew the value would be about 4.7nH and calculated capacitance based on that knowledge but, hopefully there is a helpful (if slightly tortuous) path in the words above.