The electronegativity order of the following | vedclass

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Question

(C) Electronegativity is inversely proportional to the atomic size. Smaller the size,greater is the electronegativity.Nitrogen $(N)$ is in the second

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$N > I > S > Te$

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(C) Electronegativity is inversely proportional to the atomic size. Smaller the size,greater is the electronegativity.Nitrogen $(N)$ is in the second period,so it has the smallest size and the highest electronegativity.Iodine $(I)$ is a halogen,which are generally highly electronegative.For sulfur $(S)$ and tellurium $(Te)$,as we move down the group,the atomic size increases,so the electronegativity decreases $(S > Te)$.Comparing these,the electronegativity values are approximately: $N (3.04) > I (2.66) > S (2.58) > Te (2.10)$.Thus,the correct order is $N > I > S > Te$.

The electronegativity order of the following elements is:

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