Question

3 attempts left Check my work Re Enter r your answer in the provided box in scientific notation. A hydrogen-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 x 10-18)2+(5 Guf 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: Hg?''(8) Hg80(8)+e 1x 10 kJ/mol H < Prev 4 of 16 Next >

Answer 1

79+ Hght (8) Hg 807 (9) te The electron con figuration [x] 6s? " 10 5d song = 79+ 1 Hg 15 79+. Hg In the last ionization step, the election nemoves from the 1s onbital of The principal quantum number of is onbitalo (n) = 1 so the enengy mequired for the lastjonization step, we can use the enengy of election in a hydrogen - like ion, is given by En = -(2.18 x 10-48 ) 22 (na) 2= Atomic Number= For mencury 5 it is so

1.39 52 X 1074 79+ For High -18 13 ) J (2.18 x 10 (80)? (12 Eu - 1.395 2x 10-14 The electron nemoves from is onbital (n=1) of Highet goes to the n=00 (2:18 X10-18 3 (80) (0)2 E00 79+ is - Ex - The total energy nequined for the removal of the last election of Hg 794 - El 0 ( - 1:29 52 x 10-1) J/atom 1.3952x10-14 g/atom X 6.022 X10 8.402 x 109 I moll 8.402x 100 kJ mol-1 23 g/mol
The answer is 8.402 × 106 kJ/mol