Find the ionization energy of #"C"^(5+)#?

Look for the sixth ionization energy of carbon.

Why the sixth? The ionization energy measures the energy required to completely remove one mole of electrons from one mole of the atoms at gaseous state.

The first ionization energy of an element, use one mole of neutral atoms as the reactant. Taking carbon as an example, the equation

#"C"(g)->"C"^(+)(g)+e^(-)" "DeltaH=-"1st"color(white)(l)"IE"#

characterizes this process.

Similarly
#"C"^(+)(g)->"C"^(2+)(g)+e^(-)" "DeltaH=-"2st"color(white)(l)"IE"#

#"C"^(5+)(g)->"C"^(6+)(g)+e^(-)" "DeltaH=-"6th"color(white)(l)"IE"#

Note that each carbon nucleus contains #6# protons and each neutral atom shall possess #6# electrons. Therefore, ripping a #"C"^(5+)(g)# ion of its last electron would yield a bare carbon nucleus, and therefore the #"6th" color(white)(l)"IE"# happens to be the last possible ionization energy of carbon.

Reference
[1] Winter, Mark. “Carbon: Properties of Free Atoms.” Carbon»Properties of Free Atoms [WebElements Periodic Table], 27 Apr. 2018, www.webelements.com/carbon/atoms.html.

#"C"^"5+"# is a hydrogen-like atom. It has one electron and a nucleus with six protons.

Thus, you can use the Rydberg formula to calculate the energy.

Rydberg's original formula did not look directly at energies, but we can rewrite his formula to have these units.

The Rydberg formula for the change in energy is

#color(blue)(bar(ul(|color(white)(a/a) DeltaE = -R_HZ^2(1/n_text(f)^2 - 1/n_text(i)^2)color(white)(a/a)|)))" "#

where

#R_H = "the Rydberg constant", 2.178 × 10^"-18"color(white)(l) "J"#
#Z =# the number if charges in the nucleus
#n_text(i)# and #n_text(f)# are the initial and final energy levels.

To calculate the ionization energy from the ground state, we set #n_text(i) = 1# and #n_text(f) = oo#, Then,

#DeltaE_(1->oo) = R_HZ^2 = 2.180 × 10^"-18"color(white)(l) "J" ×6^2 = 7.848 × 10^"-17"color(white)(l)"J"#

Thus, the energy required to ionize one atom is #7.848 × 10^"-17"color(white)(l)"J"#.

On a molar basis, the ionization energy is

#"IE"_1(C^(5+)) = "IE"_6(C)#

#= (7.848 × 10^"-17"color(white)(l)"J")/(1 color(red)(cancel(color(black)("atom")))) × (6.022 × 10^23 color(red)(cancel(color(black)("atoms"))))/("1 mol")#

#= 4.726 × 10^7 color(white)(l)"J/mol "= "47 260 kJ·mol"^"-1"#