Ahydrogen-like ion is an ion containing only one electron. the energy of the electron in a hydrogen-like ion is given by en = ? (2.18 × 10? 18j) z2 ( 1 n2 ) where n is the principal quantum number and z is the atomic number of the element. plasma is a state of matter consisting of positive gaseous ions and electrons. in the plasma state, a mercury atom could be stripped of its 80 electrons and therefore could exist as hg80+. use the equation above to calculate the energy required for the last ionization step: hg79+(g) ? hg80+(g)+ e?

Energy required for the ionization of mercury is

Explanation: Energy of the electron in a hydrogen-like ion is given by the equation:

where, Z = atomic number

n = principle quantum number

Ionization energy is the energy required to release the outermost electron from an isolated gaseous atom.

For the ionization of mercury, the equation follows:

Mercury has an atomic number = 80

As, in this element 79 protons are already released, which means that the electronic configuration for is

and the principle quantum number for the last ionization step = 1

Putting the values in energy equation, we get

(Conversion Factor: 1kJ = 1000J)

c. orbital shape i hope this is correct