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# The electronic configuration of the first artificial element:A. $\text{4}{{\text{f}}^{4}}\text{ 5}{{\text{d}}^{1}}\text{ 6}{{\text{s}}^{2}}$B. $\text{4}{{\text{f}}^{1}}\text{ 5}{{\text{d}}^{4}}\text{ 6}{{\text{s}}^{2}}$C. $\text{5}{{\text{f}}^{4}}\text{ 6}{{\text{d}}^{1}}\text{ 7}{{\text{s}}^{2}}$D. $\text{5}{{\text{f}}^{3}}\text{ 6}{{\text{d}}^{1}}\text{ 7}{{\text{s}}^{2}}$

Last updated date: 17th Jun 2024
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Hint: The first artificial element is placed at the right side to the uranium in the periodic table and at the left side of the actinide plutonium. The element has a silvery appearance and it is hard. It is also considered as radioactive.

- In the given question, we have to identify the correct electronic configuration of the first artificial element.
- Now, we know that the atomic number of uranium is 92 and the atomic number plutonium is 94.
- So, we know that the element is placed at the right of the uranium so it will have a greater atomic number that is 93.
- Now, to do the electronic configuration of the element with atomic number 93 we know that the noble gas radon gas has atomic number 86.
- So, with respect to the noble gas the electronic configuration of the element will be: $\text{(Rn) 5}{{\text{f}}^{4}}\text{ 6}{{\text{d}}^{1}}\text{ 7}{{\text{s}}^{2}}$
- The name of the element is neptunium and it is also known as a transuranic element because it comes after the uranium.
- It is represented by the 'Np'.
- Also, it is unstable and undergoes radioactive decay to form other elements.
Therefore, option C is the correct answer.

Note: The valence electrons or the electrons in the outermost shell of neptunium are 7. The element is highly reactive and has the most stable oxidation state of +5 in liquid solution whereas +4 is preferred in the solid neptunium.