Question

Add charges, single electron dots, and/or pairs of dots as appropriate to show the Lewis symbols for the most stable ion of each element. Treat n=4 as the valenceshell in all cases.

To add formal charges, click the button below (which will turn yellow when activated) before clicking on the molecule. You may need to hit Enter on your keyboardafter typing the charge to make it register before clicking submit.
Place charges and electron dots as needed on each ion. To add a charge, click the +/- button, then click on an atom and type the desired charge.

As, Se, Br, K, Ca, Ga

Answer 1

Concepts and reason

The charge of any species is due to gain or loss of electrons from it. If it loses electrons, it gets positive charge and if it gains electrons, it gets negative charge.

Fundamentals

According to octet rule, an atom can lose or gain of electrons until it gets eight electrons in its valence shell. The Lewis symbols show the number of electrons in the valence shell.

The given elements are as follows:

K, Ca, Ga, As, Se, and Br.

Potassium (K) has one valence electron. Hence, to attain octet it loses one electron. Therefore, the formal charge will be +1.

Calcium (Ca) has 2 valence electrons. Hence, to attain octet it loses 2 electrons. Therefore, the formal charge will be +2.

Gallium (Ga) has 3 valence electrons. Hence, to attain octet it loses 3 electrons. Therefore, the formal charge will be +3.

Arsenic (As) has 5 valence electrons. Hence, to attain octet it gains 3 electrons. Therefore, the formal charge will be -3.

Selenium (Se) has 6 valence electrons. Hence, to attain octet it gains 2 electrons. Therefore, the formal charge will be -2.

Bromine (Br) has 7 valence electrons. Hence, to attain octet it gains 1 electron. Therefore, the formal charge will be -1.

All the elements belong to fourth period. Hence, the valence shell is fourth shell. For the ions ${{\rm{K}}^{\rm{ + }}}{\rm{,}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{C}}{{\rm{a}}^{{\rm{2 + }}}},{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{G}}{{\rm{a}}^{{\rm{3 + }}}}$ , there is no electrons in the valence shell. Hence, there should not be any electrons in their Lewis symbols.

For the ions ${\rm{A}}{{\rm{s}}^{{\rm{3 - }}}}{\rm{,}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S}}{{\rm{e}}^{{\rm{2}} - }},{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{B}}{{\rm{r}}^ - }$ , there is eight electrons in the valence shell. Hence, there should be eight valence electrons in their Lewis symbols.

Therefore, the Lewis symbols for the most stable ions of the given elements are as follows:

Ans:

The Lewis symbols for the most stable ions of the given elements are as follows: