Question

The first eight ionization energies of a third row element are 999.6, 2252, 3357, 4556, 7004, 8496, 27107, and 31719 kJ/mol. Identify the element.

Choose one:A. AlB. SC. MgD. PE. NaF. ArG. ClH. Si

Answer 1

Answer -

Given,

First ionization energies of third row element

999.6, 2252, 3357, 4556, 7004, 8496, 27107, and 31719 kJ/mol

Element = ?

We know that,

Ionization Energy is the amount of energy required to remove electron from the valance shell.

First Ionization Energy = energy for removing first electron.

Second Ionization Energy = energy for removing Second electron.

third Ionization Energy = energy for removing third electron.

Same for rest.

NOTE - small difference in the IE means electron is in outer shell. Large difference means electron is in core.

Now,

in the given, difference of given IE

IE 2- IE1 = 2252 -999.6 = 1252.4 kJ/mol

IE 3- IE2 = 3357 - 2252 = 1105 kJ/mol

IE 4 - IE3 = 4556 - 3357 = 1199 kJ/mol

IE 5 - IE4 = 7004 - 4556 = 2448 kJ/mol

IE 6- IE5 = 8496 -7004 = 1492 kJ/mol

IE 7 - IE6 = 27107 - 8496 = 18611 kJ/mol

IE 8 - IE7 = 31719 -27107 = 4612 kJ/mol

We can analyse that in IE 7-IE 6, there is sudden increase in IE which means this electron is in the core. i.e 7 th electron is in core and 6 are in valence shell

Also, the element in the third row which have 6 valence electron is Sulfur. So, This element is Sulfur.

So, OPTION B (SULFUR) is correct