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Question: The amount of energy released when \({{10}^{0}}\)atoms of iodine in vapor state are converted to \({...

The amount of energy released when 100{{10}^{0}}atoms of iodine in vapor state are converted to I−{{I}^{-}}ions is 4.9×10−13J4.9\times {{10}^{-13}}J. What is the electron affinity of iodine in eV per atom?
(A)- 2.0
(B)- 2.5
(C)- 2.75
(D)- 3.0

Explanation

Solution

The amount of energy released when an electron is added to a neutral atom to form an anion is known as the electron affinity.

Complete Step by step answer:
-In other words, the potential energy change of the atom when an electron is added to a neutral atom in its gaseous state forming a negative ion is known as the electron affinity of the particular atom.
-In respect to the question, electron affinity can be described as the amount of energy released when one mole of iodine or 6.022×10236.022\times {{10}^{23}} atoms of iodine in their gaseous state are converted into I−{{I}^{-}}ion. Let us now write the chemical equation for the question-
I+e−→I−I+{{e}^{-}}\to {{I}^{-}}
-According to the question, electron affinity (E.A)=4.9×10−18J=4.9\times {{10}^{-18}}J
Therefore, the amount of energy released when 6.022×10236.022\times {{10}^{23}}atoms of iodine in their gaseous state are converted into I−{{I}^{-}}ions =4.9×10−13106×6.022×1023=29.5×104Jmol−1=295KJmol−1=\dfrac{4.9\times {{10}^{-13}}}{{{10}^{6}}}\times 6.022\times {{10}^{23}}=29.5\times {{10}^{4}}Jmo{{l}^{-1}}=295KJmo{{l}^{-1}}
-Since 96.3KJmol−1=1eV atom−196.3KJmo{{l}^{-1}}=1eV\text{ ato}{{\text{m}}^{-1}}
Therefore, 295KJmol−1=1×29596.3eV atom−1=3.06eV atom−1295KJmo{{l}^{-1}}=\dfrac{1\times 295}{96.3}eV\text{ ato}{{\text{m}}^{-1}}=3.06eV\text{ ato}{{\text{m}}^{-1}}

So, the correct answer is option D.

Note: -The electron affinity has a general trend along the groups and periods in the periodic table. Electron affinity on going from left to right across a period increases as the nuclear attraction increases. Electron affinity on going down in the group decreases as the electrons are less tightly bonded and therefore closer in energy to a free electron.
-Electron affinity depends on the following factors-
(i) Atomic size- Smaller the size of the atom, the greater will the electron gain enthalpy because greater will be the nuclear forces.
(ii) Nuclear charge- Greater the nuclear charger, the larger the value for electron gain enthalpy as an increase in nuclear charge will increase the effective nuclear force on valence electrons.
(iii) Screening effect- Electron affinity is inversely proportional to the screening effect, that is larger the screening effect lesser the electron gain enthalpy because lesser will be the nuclear force.
-In general electron affinity follows the following trend-
Halogens Oxygen family Carbon family Nitrogen family Metals of group 1 and 13 Metals of group 2\text{Halogens Oxygen family Carbon family Nitrogen family Metals of group 1 and 13 Metals of group 2}