SSC Chemistry

Solution: The given atomic number of element A is 16. Therefore, element A is Sulfur S. That means the two isotopes of Sulfur, denoted by ³²A and ³⁴A, are ³²S and ³⁴S. Let the abundance of the ³²S isotope be x.

Therefore, the abundance of the 34S isotope = 1 - x.

We know,

(Abundance of ³²S * Mass of ³²S) + (Abundance of 34S * Mass of 34S)  = Relative Atomic Mass

x * 32 + (1-x) * 34 = 32.07

x = 0.965

Therefore, percentage abundance of ³²S is 0.965 * 100% = 96.5% and percentage abundance of ³⁴S is (1-0.965) * 100% = 3.5%

Problem: Analyze whether the electron affinity and ionization energy of the elements 11X, 12Y, 15Z, and 16Q are same or not.

The four elements are ₁₁Na, ₁₂Mg, ₁₅P, and ₁₆S respectively. The order of their electron affinity and ionization energy will not be the same.

Order of Ionization Energy: Ionization energy is a periodic property. The more one moves to the right across any period the more the atomic number increases, the more the ionization energy increases. This is because as the atomic number increases, the attraction of the nucleus for the outermost electrons increases. Consequently, more energy is required to remove an electron from the outermost shell. That is, the value of the ionization energy is higher.

The four elements belong to the same period in the periodic table. Na is the leftmost and S is the rightmost. Therefore, their expected order of ionization energy is: 

Na < Mg < P < S.

However, let's look at the electron configurations of P and S:

P(15) = 1s² 2s² 2p⁶ 3s² 3p³​ (Stable)

S(16) = 1s2 2s2 2p6 3s2 3p4  ^​​_​​(Not stable)

It is seen that the electron configuration of P(15) is more stable because its valence subshell 3p is half-filled. Therefore, it requires more energy to remove an electron e^- from its valence shell than S. This is why the ionization energy of P is greater than S.

So, the correct order of ionization energy is : Na < Mg < S < P.