Which metal will form binary halide, #"MX"_2# ?
ELEMENT I II
IE1 520 900
IE2 7300 1760
ELEMENT I II
IE1 520 900
IE2 7300 1760
1 Answer
Element II.
Explanation:
The problem essentially wants you to figure out which of the two elements given are located in group 2 of the periodic table by examining the first and second ionization energies.
The problem provides you with the following data
#color(white)(aaaaaaaacolor(black)(bb"Element I")aaaaaaaaaacolor(black)(bb"Element II")aaaa)#
#color(purple)("IE"_1)color(white)(aaaaacolor(black)("520 kJ mol"^(-1))aaaaaaaacolor(black)("900 kJ mol"^(-1))aaa)#
#color(purple)("IE"_2)color(white)(aaaacolor(black)("7300 kJ mol"^(-1))aaaaaaacolor(black)("1760 kJ mol"^(-1))aaa)#
As you know, ionization energy is a term used to denote the amount of energy needed in order to remove the outermost electron from one mole of atoms in the gaseous state to form one mole of cations.
The first ionization energy,
#"M"_ ((g)) + "IE"_ 1 -> "M"_ ((g))^(+) + "e"^(-)#
The second ionization energy,
#"M"_ ((g))^(+) + "IE"_ 2 -> "M"_ ((g))^(2+) + "e"^(-)#
Now, in order for an element to form
Notice that there's a significant jump between
You can thus look at these values and say that Element I has
By comparison, Element II has a relatively small difference between
The third ionization energy is not given, but you can presume that it will be significantly higher than
This means that Element II is probably located in group 2 of the periodic table, which of course implies that it forms
As a conclusion, you can say that Element I will require a total of
#E_"total 1" = "IE"_1 + "IE"_2#
#E_"total 1" = "520 kJ mol"^(-1) + "7300 kJ mol"^(-1) = "7820 kJ mol"^(-1)#
By comparison, Element II will require
#E_"total 2" = "900 kJ mol"^(-1) + "1760 kJ mol"^(-1) = "2660 kJ mol"^(-1)#
As you can see, you need significantly less energy to get Element II to form