Using the lewis dot structure, what does if6+ look like?
i need this as soon as possible tytyy

This ion is in an octahedral geometry.

See the diagram attached for the Lewis Dot Structure of the iodide hexafluoride cation . (Created with Google Drawings.)

Note that in this diagram,

A pair of double dots on an atom represent a lone pair of electrons.A single dash represents a single chemical bond.The square bracket and the superscript indicates that this structure is charged.

Explanation:

The iodine in forms an expanded octet. There are twelve valence electrons in total around this atom.

Each of the six fluorine atom needs 8 - 7 = 1 electron to achieve an octet of eight electrons.The iodine atom needs 12 - 7 = 5 electrons to achieve an expanded octet of twelve electrons.The ion carries a positive charge of +1. Atoms in this ion lacks one extra electron.

Overall, there needs to be more electrons for seven atoms to achieve an octet. They will form half that number of chemical bonds. That's bonds.

Now consider: what will be the geometry of this ion? There are six chemical bonds but no lone pair around the central iodine atom. The six bonds repel each other equally. They will stay as far apart from each other as possible. As a result, the shape of the ion will be octahedral. Each of the fluorine atoms occupies a vertex.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.

B. nuclear reactions could first occur.