
What would be the sum of all four quantum numbers associated with the unpaired electrons present in a nitrogen atom?
A. $10\dfrac{1}{2}$
B. $8\dfrac{1}{2}$
C. 7
D. 6
Answer
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Hint: The atomic number of nitrogen is 7 means nitrogen contains 7 electrons. The electronic configuration of nitrogen is $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$ . Means nitrogen contains 3 unpaired electrons in 2p orbital.
Complete step by step solution:
- In the question it is asked to find the sum of all four quantum numbers associated with the unpaired electrons present in a nitrogen atom.
- We are supposed to write the sum of all four quantum numbers associated with the unpaired electrons in 2p orbital.
- The principal quantum number (n) for 2p orbital is 2.
- The angular momentum quantum number (l) for 2p orbital is 1.
- The magnetic quantum number (m) for 2p orbital = -1, 0, +1
- The Spin quantum number (s) for unpaired electron is $\dfrac{1}{2}$
- We can write all the above quantum numbers individually as follows.
(1) n = 2, l = 1 , m = 1 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1+1+\dfrac{1}{2}=4\dfrac{1}{2}$
(2) n = 2, l = 1 , m = 0 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1+0+\dfrac{1}{2}=3\dfrac{1}{2}$
(3) n = 2, l = 1 , m = -1 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1-1+\dfrac{1}{2}=2\dfrac{1}{2}$
- The sum of all the four quantum numbers associated with the unpaired electrons present in a nitrogen atom = $4\dfrac{1}{2}+3\dfrac{1}{2}+2\dfrac{1}{2}=10\dfrac{1}{2}$ .
- Therefore the sum of all four quantum numbers associated with the unpaired electrons present in a nitrogen atom is $10\dfrac{1}{2}$ .
So, the correct answer is “Option A”.
Note: To calculate the sum of the all four quantum numbers associated with the unpaired electrons present in a nitrogen atom we should know about the quantum numbers related to 2p orbital which contains unpaired electrons. We are not supposed to consider the paired electrons.
Complete step by step solution:
- In the question it is asked to find the sum of all four quantum numbers associated with the unpaired electrons present in a nitrogen atom.
- We are supposed to write the sum of all four quantum numbers associated with the unpaired electrons in 2p orbital.
- The principal quantum number (n) for 2p orbital is 2.
- The angular momentum quantum number (l) for 2p orbital is 1.
- The magnetic quantum number (m) for 2p orbital = -1, 0, +1
- The Spin quantum number (s) for unpaired electron is $\dfrac{1}{2}$
- We can write all the above quantum numbers individually as follows.
(1) n = 2, l = 1 , m = 1 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1+1+\dfrac{1}{2}=4\dfrac{1}{2}$
(2) n = 2, l = 1 , m = 0 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1+0+\dfrac{1}{2}=3\dfrac{1}{2}$
(3) n = 2, l = 1 , m = -1 and s = $\dfrac{1}{2}$ , the sum of these four quantum numbers = $2+1-1+\dfrac{1}{2}=2\dfrac{1}{2}$
- The sum of all the four quantum numbers associated with the unpaired electrons present in a nitrogen atom = $4\dfrac{1}{2}+3\dfrac{1}{2}+2\dfrac{1}{2}=10\dfrac{1}{2}$ .
- Therefore the sum of all four quantum numbers associated with the unpaired electrons present in a nitrogen atom is $10\dfrac{1}{2}$ .
So, the correct answer is “Option A”.
Note: To calculate the sum of the all four quantum numbers associated with the unpaired electrons present in a nitrogen atom we should know about the quantum numbers related to 2p orbital which contains unpaired electrons. We are not supposed to consider the paired electrons.
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