Question

The number of p electrons in bromine atom is:
A. $17$
B. $7$
C. $15$
D. $13$

Answer 1

Bromine occurs in the group $7$ of the periodic table. They are known as halogens. They are generally non-metals and have electronic configurations just one electron short of the nearest noble gas. They form negative ions with a $- 1$ charge.

Complete step by step solution:
The halogens exist as diatomic molecules at room temperature, ${\text{2}}{{\text{5}}^ \circ }{\text{C}}$. Halogens include fluorine, chlorine, bromine, iodine and astatine. On moving down the group, atomic size increases. They have $7$ valence electrons in the outermost shell, but one electron is deficient to the nearest noble gas configuration. They can easily attain a full and complete outer shell by gaining one electron. They all gain an electron in reactions, thus forming a negative ion with $- 1$ charge.
Fluorine is the most electronegative element in the periodic table and it has no d orbitals in its valence shell. So it cannot expand its valence shell. Chlorine, bromine and iodine can expand its valence shell to hold as many as $14$ valence electrons.
The electronic configuration of ${\text{F,Cl,Br}}$ are given below:
${\text{F - 1}}{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^5}$
${\text{Cl - 1}}{{\text{s}}^2}{\text{2}}{{\text{s}}^2}{\text{2}}{{\text{p}}^6}3{{\text{s}}^2}3{{\text{p}}^5}$
${\text{Br - 1}}{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{{\text{s}}^2}3{{\text{p}}^6}3{{\text{d}}^{10}}4{{\text{s}}^2}4{{\text{p}}^5}$
From the above electronic configuration, to calculate the number of p electrons, we should consider${\text{2}}{{\text{p}}^6}{\text{3}}{{\text{p}}^6}{\text{4}}{{\text{p}}^5}$.
So by adding the total number of electrons, we get $6 + 6 + 5 = 17$ electrons.
Thus there are $17$ electrons in p orbital.

Hence the correct option is A.

Note: Down the group, the atomic size increases. Bromine has atomic number $35$. The electronic configuration of bromine atoms can be expressed with respect to the nearest noble gas configuration. Thus it will be $\left[ {{\text{Ar}}} \right]3{{\text{d}}^{10}}4{{\text{s}}^2}4{{\text{p}}^5}$. The electrons of bromine atoms in each shell can be represented as $2,8,18,7$.