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Question

Electronegativity decreases in the following order:

A
N>P>As>Bi>Sb
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B
N>P>As>Sb>Bi
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C
N>P>As>Sb=Bi
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D
N>P>Sb>As>Bi
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Solution

The correct option is B N>P>As>Sb>Bi
This one is pretty simple isn’t it? We know that the number of valence electrons is the same for each of the pnictogens. As we go down the group, owing to the increase in the atomic size, progressively more electrons are there between the nucleus and the valence electrons. There are two things at play here: 1. Because of the increasing atomic size, the separation between the nucleus and the valence electrons increases. 2. There are progressively more electrons in between the valence electrons and the nucleus. These enveloped electrons shield the valence electrons from being held tightly by the nucleus. Thus, as we go down the group, the valence electrons are more loosely held. Hence electronegativity decreases.

Having reasoned using the theory, let us look at the actual Pauling electronegativity values shall we?
N (3.04) > P (2.20) > As (2.19) > Sb (2.05) > Bi (2.02). Do note that if we round off the values to one decimal, we can pretty much say that approximately, the electronegativities of P and As are the same. Also, using such approximation, Sb and Bi also have the same electronegativities.

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