Chapter 28, Problem 24P

(a)

To determine

The full electron configuration for selenium.

Answer to Problem 24P

Fill in accordance with the periodic table: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{4}}$

Explanation of Solution

Given info:

Element: selenium

Formula used:

Arrangement or structure of electrons in the energy levels in the region of an atomic nucleus is known as Electronic structure or electronic configuration. In electronic configuration, $\text{spdf}$ demotes to the shape of any orbital. The $\text{spdf}$ notation employs numbers to designate principal shell and also, letters $\left(s,p,d,f\right)$ to categorize a sub shell. In $\text{spdf}$ notation, subscript indicates the numeral of electrons in a designated sub shell.

Calculation:

If $l$ is an angular quantum number of sub shell then max electrons it can hold: $\text{N=}2\left(2l+1\right)$

Thus, in $\text{spdf}$ maximum electrons are $\text{2,}\text{6,}\text{10,}\text{14 }$ respectively.

It is known that atomic number of Selenium: $\text{Z=34}$

For $\text{Z=34}$ , electronic configuration is

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{4}}$

(b)

To determine

The full electron configuration for gold.

Answer to Problem 24P

Fill in accordance with the periodic table: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{4f}}^{\text{14}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{1}}$

Explanation of Solution

Given info:

Element: gold

Formula used:

Arrangement or structure of electrons in the energy levels in the region of an atomic nucleus is known as Electronic structure or electronic configuration. In electronic configuration, $\text{spdf}$ demotes to the shape of any orbital. The $\text{spdf}$ notation employs numbers to designate principal shell and also, letters $\left(s,p,d,f\right)$ to categorize a sub shell. In $\text{spdf}$ notation, subscript indicates the numeral of electrons in a designated sub shell.

Calculation:

If $l$ is an angular quantum number of sub shell then max electrons it can hold: $\text{N=}2\left(2l+1\right)$

Thus, in $\text{spdf}$ maximum electrons are $\text{2,}\text{6,}\text{10,}\text{14 }$ respectively.

It is known that atomic number of gold: $\text{Z=79}$

For $\text{Z=79}$ , electronic configuration is

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{4f}}^{\text{14}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{1}}$

(c)

To determine

The full electron configuration for radium.

Answer to Problem 24P

Fill in accordance with the periodic table: ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{4f}}^{\text{14}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}$

Explanation of Solution

Given info:

Element: radium

Formula used:

Arrangement or structure of electrons in the energy levels in the region of an atomic nucleus is known as Electronic structure or electronic configuration. In electronic configuration, $\text{spdf}$ demotes to the shape of any orbital. The $\text{spdf}$ notation employs numbers to designate principal shell and also, letters $\left(s,p,d,f\right)$ to categorize a sub shell. In $\text{spdf}$ notation, subscript indicates the numeral of electrons in a designated sub shell.

Calculation:

If $l$ is an angular quantum number of sub shell then max electrons it can hold: $\text{N=}2\left(2l+1\right)$

Thus, in $\text{spdf}$ maximum electrons are $\text{2,}\text{6,}\text{10,}\text{14 }$ respectively.

It is known that atomic number of gold: $\text{Z=88}$

For $\text{Z=88}$ , electronic configuration is

  ${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{6}}{\text{3s}}^{\text{2}}{\text{3p}}^{\text{6}}{\text{3d}}^{\text{10}}{\text{4s}}^{\text{2}}{\text{4p}}^{\text{6}}{\text{4d}}^{\text{10}}{\text{4f}}^{\text{14}}{\text{5s}}^{\text{2}}{\text{5p}}^{\text{6}}{\text{5d}}^{\text{10}}{\text{6s}}^{\text{2}}{\text{6p}}^{\text{6}}{\text{7s}}^{\text{2}}$