I.E of Bromine is higher than I.E of phosphorus. This is due to:

high

Z_{e f f}

of Br

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stable Electronic configuration of P

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small size of Br

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small size of P

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Solution

The correct option is A high $Z_{e f f}$ of Br
As we move from left to right in a periodic table the $Z_{e f f}$ increases and hence the energy required to remove electron from the last shell increases therefore, due to high $Z_{e f f}$ on Br the I.E of Bromine is higher than I.E of phosphorus.

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