Answer:
Element D would have the lowest ionization.
(D) is correct option
Explanation:
Given that,
Number of electron in element A = 17
Number of electron in element B = 3
Number of electron in element C = 18
Number of electron in element D = 11
We know that,
The name of element A is chlorine which have 17 electron.
The name of element B is lithium which have 3 electron.
The name of element C is argon which have 18 electron.
The name of element D is sodium which have 11 electron.
We know that,
The ionization energy of the element decreases from up to down in group and increases from left to right in a period.
We need to find the element would have the lowest ionization
Using periodic table
According to periodic table the lowest Ionization energy of sodium.
Hence, Element D would have the lowest ionization.
(D) is correct option
