The electronegativity order of the following elements is
The electronegativity order of the following elements is
$\mathrm{N}>\mathrm{S}>\mathrm{Te}>1$
$\mathrm{I}>\mathrm{N}>\mathrm{S}>\mathrm{Te}$
$\mathrm{N}>\mathrm{I}>\mathrm{S}>\mathrm{Te}$
$\mathrm{N}>\mathrm{S}>\mathrm{I}>\mathrm{Te}$
Solution
Electronegativity is inversely proportional to the atomic size. Smaller the size, greater is the electronegativity. Nitrogen is placed in second period, so it has smallest size and highest electronegativity. Then, iodine belongs to halogen family which are electronegative elements. For, S and Te, on moving top to bottom, atomic size increases so electronegative character decreases. So, the correct order is $\mathrm{N}>\mathrm{I}>\mathrm{S}>\mathrm{Te}$.
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