An oxygen atom has a total of \[8\] electrons. How do you write the \[4\] quantum numbers for each of the \[8\] electrons in the ground state?
Answer
537.3k+ views
Hint: We have to know that the quantum is described as the set of numbers which is used to determine the position and energy of an electron present in the atom. There are a total of four quantum numbers. Principal quantum number, which is denoted as n, orbital angular quantum number (l), magnetic quantum number (\[{m_I}\]) and electron spin quantum number \[\left( {{m_s}} \right)\].
Complete answer:
The oxygen is a chemical compound having the atomic number,\[8\]and having the symbol, $'O'$. And the electronic configuration of oxygen is, \[1{s^2}2{s^2}2{p^4}\] and we can be represented as,
The \[4\]quantum numbers for each of the\[8\]electrons in the ground state can be written as,
The first energy level;
\[n = 1,l = 0,{m_l} = 0,{m_s} = - \dfrac{1}{2} \to \]The electrons located in the 1s orbital, spin – down
\[n = 1,l = 0,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the 1s orbital, spin – up
The second energy level;
\[n = 2,l = 0,{m_l} = 0,{m_s} = - \dfrac{1}{2} \to \] The electrons located in the 2s orbital, spin – down
\[n = 2,l = 0,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the 2s orbital, spin – up
\[n = 2,l = 1,{m_l} = - 1,{m_s} = - \dfrac{1}{2} \to \] The electrons located in the \[2{p_x}\] orbital, spin – down
\[n = 2,l = 1,{m_l} = - 1,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_x}\] orbital, spin – up
\[n = 2,l = 1,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_y}\] orbital, spin – up
\[n = 2,l = 1,{m_l} = 1,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_z}\] orbital, spin – up
Note:
We have to know that the principal quantum number determines the distance between the electrons and nucleus and it is directly proportional to the atomic size. The possible values of the principal quantum number may be \[1,2,3,4,...\] . Azimuthal quantum number also known as orbital angular momentum quantum numbers and the possible values of l is \[0,1,2,3,4,..,\left( {n - 1} \right)\]. The value of magnetic quantum \[{m_I}\]is equal to \[ - l,..., - 1,0,1...,l\]. And the possible values of electron spin quantum number\[\left( {{m_s}} \right)\]is \[ + \dfrac{1}{2}, - \dfrac{1}{2}\].
Complete answer:
The oxygen is a chemical compound having the atomic number,\[8\]and having the symbol, $'O'$. And the electronic configuration of oxygen is, \[1{s^2}2{s^2}2{p^4}\] and we can be represented as,
The \[4\]quantum numbers for each of the\[8\]electrons in the ground state can be written as,
The first energy level;
\[n = 1,l = 0,{m_l} = 0,{m_s} = - \dfrac{1}{2} \to \]The electrons located in the 1s orbital, spin – down
\[n = 1,l = 0,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the 1s orbital, spin – up
The second energy level;
\[n = 2,l = 0,{m_l} = 0,{m_s} = - \dfrac{1}{2} \to \] The electrons located in the 2s orbital, spin – down
\[n = 2,l = 0,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the 2s orbital, spin – up
\[n = 2,l = 1,{m_l} = - 1,{m_s} = - \dfrac{1}{2} \to \] The electrons located in the \[2{p_x}\] orbital, spin – down
\[n = 2,l = 1,{m_l} = - 1,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_x}\] orbital, spin – up
\[n = 2,l = 1,{m_l} = 0,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_y}\] orbital, spin – up
\[n = 2,l = 1,{m_l} = 1,{m_s} = + \dfrac{1}{2} \to \] The electrons located in the \[2{p_z}\] orbital, spin – up
Note:
We have to know that the principal quantum number determines the distance between the electrons and nucleus and it is directly proportional to the atomic size. The possible values of the principal quantum number may be \[1,2,3,4,...\] . Azimuthal quantum number also known as orbital angular momentum quantum numbers and the possible values of l is \[0,1,2,3,4,..,\left( {n - 1} \right)\]. The value of magnetic quantum \[{m_I}\]is equal to \[ - l,..., - 1,0,1...,l\]. And the possible values of electron spin quantum number\[\left( {{m_s}} \right)\]is \[ + \dfrac{1}{2}, - \dfrac{1}{2}\].
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Name the Largest and the Smallest Cell in the Human Body ?

Draw a diagram showing the external features of fish class 11 biology CBSE

What organs are located on the left side of your body class 11 biology CBSE

Difference between physical and chemical change class 11 chemistry CBSE

