The EN (Electronegativity) of element A is E_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the Mulliken scale,the electronegativity $(EN)$ of an element is the average of its ionization potential $(IP)$ and electron affinity

Vedclass · Accepted Answer

$2E_1 - E_2$

Vedclass · Answer

(A) According to the Mulliken scale,the electronegativity $(EN)$ of an element is the average of its ionization potential $(IP)$ and electron affinity $(EA)$.The formula is given by: $EN = \frac{IP + EA}{2}$.Given that $EN = E_1$ and $IP = E_2$,we substitute these values into the formula:$E_1 = \frac{E_2 + EA}{2}$.Multiplying both sides by $2$ gives:$2E_1 = E_2 + EA$.Rearranging to solve for $EA$:$EA = 2E_1 - E_2$.

The $EN$ (Electronegativity) of element $A$ is $E_1$ and its $IP$ (Ionization Potential) is $E_2$. Hence,the $EA$ (Electron Affinity) will be:

Similar Questions

If coordinate $X$ is atomic number,then select the best option for the $Y$ coordinate based on the given graph for $Li$,$Na$,and $K$.

The property of attracting electrons by the halogen atom in a molecule is called

The electronegativity of the following elements increases in the order:

With respect to chlorine,hydrogen will be:

Aqueous solutions of two compounds $M_1OH$ and $M_2OH$ are prepared in two different beakers. If the $EN$ of $M_1 = 3.4$,$EN$ of $M_2 = 1.2$,$EN$ of $O = 3.5$,and $EN$ of $H = 2.1$,then the nature of the two solutions will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module