Question

1. answer the following questions regarding atoms of selenium (se) and bromine (br).

a. *for each of the atoms below, list the number of each subatomic particle in the space provided.

b. **write the full electron configuration of the selenium atom.
c. write the abbreviated (noble gas) electron configuration of a neutral atom of bromine.
d. **calculate the effective nuclear charge of selenium. work must be shown to earn credit.
e. **in terms of atomic structure, explain why the first ionization energy of selenium (se) is less than that of bromine (atomic number 35). (ap chemistry 2000, q7)

Explanation:

a.

$_{34}^{80}\text{Se}$:

Step1: Determine protons

Protons = atomic number. For Se, atomic number is 34.

Step2: Determine neutrons

Neutrons = mass number - atomic number. Mass number is 80, atomic number 34. So, \(80 - 34=46\).

Step3: Determine electrons

In a neutral atom, electrons = protons. So, electrons = 34.

$_{35}^{79}\text{Br}^{-1}$:

Step1: Determine protons

Protons = atomic number. For Br, atomic number is 35.

Step2: Determine neutrons

Neutrons = mass number - atomic number. Mass number is 79, atomic number 35. So, \(79 - 35 = 44\).

Step3: Determine electrons

For \(\text{Br}^{-1}\), electrons = protons+1. So, \(35 + 1=36\).

Selenium (Se) has atomic number 34. Using Aufbau principle (\(1s<2s<2p<3s<3p<4s<3d<4p\)).
\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{4}\)

Bromine (Br) has atomic number 35. The noble - gas core for Br is [Ar] (atomic number 18). Then add the remaining electrons.
\([\text{Ar}]4s^{2}3d^{10}4p^{5}\)

Answer:

For \(_{34}^{80}\text{Se}\): Protons: 34, Neutrons: 46, Electrons: 34.
For \(_{35}^{79}\text{Br}^{-1}\): Protons: 35, Neutrons: 44, Electrons: 36.

b.