Question

If the atomic number of element X is 7. The best electron-dot symbol for the element is:
(a)- $\cdot X$
(b)- $\cdot X\cdot $
(c)- $\cdot \overset{\cdot \cdot }{\mathop{\underset{\cdot }{\mathop{X\cdot }}\,}}\,$
(d)- $\cdot \overset{\cdot \cdot }{\mathop{\underset{\cdot \cdot }{\mathop{X\cdot }}\,}}\,\cdot $

Answer 1

Hint: To find the correct electron-dot structure, first, you have to find the atomic number, if the atomic number is given then, write its electronic configuration and find the valence shell and count the total number of electrons in the valence and draw the number dots around the symbol of the element.

Complete step-by-step answer:
Electron-dot structure can be found by knowing the valence electrons.The given atomic number is 7, so there are a total of 7 electrons in the element. Let us write its electronic configuration. The configuration will be:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$
There are 2 electrons in 1s, 2 electrons in 2s, and three electrons in 2p.
Next, we have to find the valence shell. The valence shell of this will be 2 because 2 is the last shell. And there are a total of 5 electrons in the last shell. Therefore, we have to draw 5 dots around the symbol of the electron.
The electron-dot structure is given below:
$\cdot \overset{\cdot \cdot }{\mathop{\underset{\cdot }{\mathop{X\cdot }}\,}}\,$

Therefore, the correct answer is option (c).

Note: Don't get confused that the number of valence shells is 3, i.e., in 2p. You have to see all the orbitals of the last shell. If the last shell is 3, then the number of electrons must be counted in 3s, 3p, and 3d.