Question
Question: What will be the correct order: A. \[Ne > F > I > Li\, \to IP\] B. \[Cl > S > F > O\, \to EA\] ...
What will be the correct order:
A. Ne>F>I>Li→IP
B. Cl>S>F>O→EA
C. Se>Ti>Ca→Size
D. Na2O>SiO2>Ca2O7→(Acidiccharacter)
Solution
Refer to the periodic table to observe the trend in different periodic properties. The periodic properties ionization potential, electron affinity, size, electronegativity, acidic characters, basic characters, etc depends on the electronic configuration of the element, the nuclear force of attraction, the position of the element in the periodic table.
Complete solution:
The amount of energy required to remove the most loosely bound electron from the isolated gaseous atom or molecule is known as ionization potential or ionization enthalpy. Ionization potential is increasing as we move from left to right of the periodic table and decreases as we move from the top to bottom along the group in the periodic table.
Here, the first order is given as Ne>F>I>Li→IPthe ionization potential of neon is greater than the F which is correct because neon is the noble gas that possesses a stable electronic configuration hence, required the highest amount of energy to remove an electron from it. In F and I, the size of F is small and possesses the greater nuclear force of attraction, therefore, the amount of energy required is greater than the iodine.
Now, in the order, it is given that the ionization potential of the iodine is greater than the lithium which is incorrect. This is because the size of the lithium is smaller but possesses only one valence electron so it can be removed easily and attains a stable electronic configuration. But in the case of iodine more amount of energy is required to remove the valence electron.
Thus, the given order for ionization potential for elements is Ne>F>I>Li→IP hence, the option order is the correct answer for the given question.
When an electron is added into the neutral gaseous atom of an element amount of energy is released known as electron affinity.
Electron affinity increases along the period and decreases down the group.
Here, the order of the electron affinity given is Cl>S>F>O→EAas down the group atomic size increases but electron affinity decreases. As we know that chlorine and fluorine
belong to the same group and oxygen and sulphur in the same group. As the size of the chlorine is greater than the fluorine electron affinity of the chlorine is less than the fluorine but, in the order, it is given that the electron affinity of the chlorine is greater than the fluorine. Hence, the given order is the incorrect order for the electron affinity.
In the periodic table as we move from the left to right size decreases because of the increases in nuclear force of attraction, while along the group from the top to bottom new shells are added hence, the size increases.
Here, the order of size is givenSe>Ti>Ca→Size. All these elements belong to the 4th period in the periodic table and the period from the left to right size decreases. Now, calcium is the 2nd group element, titanium is the 4th group element and the 16th group element hence, the correct order of the size is Ca>Ti>Se, therefore, the given order is incorrect.
The order of the acidic characters is given asNa2O>SiO2>Ca2O7→(Acidiccharacter). The acidic character increases from the left to right in the periodic table. The oxide on the left side elements are basic, and oxides on the right side of the periodic table are acidic while oxides from the middle are amphoteric. As silicon is the p-block element and present on the right side of the periodic table the corresponding oxide is the most acidic among the given oxides. Hence, the given order is the incorrect answer for the given question.
So ,the correct option is (A).
Note: While observing the trend in the periodic properties of the element, atoms it is important to assign the position of the element in the periodic table. The particular periodic property either increases or decreases along with the period or group with some exceptions.