How many valence electrons does sulfer have available for bonding to other atoms?

Well, both magnesium and barium are $"alkaline earths...."$, the which give dications upon oxidation.... ...from Group 2 of . And given Group 2, there are TWO , which are readily...

The carbon atom has an electronic configuration of $1s^(2)2s^(2)2p^(2)$. The $2s$ and the $2p$ orbitals participate in ; a tetrahedral $sp^3$ hybridization scheme is commonly invoked in order to describe...

The for phosphorus is $1s^2 2s^2 2p^6 3s^2 3p^3$. The are usually the outermost s and p electrons and in this case, there are 2 3s electrons and 3 3p...

Look at the group number at the top of . The column number at the top is the group number. For example, the column with Be at the top is...

This may seem a bit confusing when one looks at , since chlorine is the seventh element in its row, but iodine is the seventeenth. However, in the case of...

In terms of actual , there is literally only one example. You can consider that their oxidation states have never gotten higher than $+8$, except for Iridium: In terms...

The group that the are in are equal to the number of electrons in their respective outer shells.

And the zinc cation is $Zn^(2+)$. When the ions form a salt, clearly they form an electrically neutral species; and thus the formulation of the salt is $Zn_3P_2$. Each phosphorus...

If we look at the of sulfur ($1s^2 2s^2 2p^6 3s^2 3p^4$). Since the third p orbital holds the maximum of 6 electrons and sulfur only have 4 this means...

We need a stoichiometic equation: $SO_2(g) + 1/2O_2(g) rarr SO_3(g)$ The equation unequivocally tells us that the reaction of $64*g$ $SO_2(g)$ with $16*g$ $O_2(g)$ gives $80*g$ $SO_3(g)$. The given masses...