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Thus, to find the number of electrons possible per shell. First, we look at the n=1 shell (the first shell). It has: The 1s orbital; An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons. The n=2 (second) shell has: The 2s orbital; The 2p orbitals; s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons.
The total number of electrons is 16. We'll distribute them in the following manner. $1s^2, 2s^2, 2p^6, 3s^2, 3p^4$, where the superscripts of each orbital represent the number of electrons in it. Now, we have a total of 6 electrons in the last (3rd) orbital. Therefore the valence number is 6.
6. You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table: The Group 1 atoms have 1 valence electron. The Group 2 atoms have 2 valence electrons. The Group 3 atoms have 3 valence electrons. The Group 4 atoms have 4 valence electrons. The Group 5 atoms have 5 valence electrons.
The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions.
Helium has two 1s electrons, therefore, if two He atoms form a bond, 4 electrons has to be placed into the molecular orbitals. Since every orbital can hold a maximum of 2 electrons, two electrons would be in the sigma orbital, and the other two would be in the sigma antiorbital. An orbital and antiorbital would effectively cancel out, therefore ...
$\begingroup$ Firstly, it depends on what you count as "valence electrons". If you say that, for all the d-block metals, the ns and (n-1)d electrons count as "valence electrons", then the answer is to just look at the group number. However, that obviously doesn't work for Zn, which effectively only has 2 valence electrons.
1. When determining the number of electrons transferred in a redox reaction is it the total in both half equations? For example: 2IX− +ZnX2+ IX2 +Zn 2 I X − + Z n X 2 + I X 2 + Z n. First we split it up into the two half-reactions and get the following. 2I IX2 +2eX− 2 I I X 2 + 2 e X −. ZnX2+ +2eX− Zn Z n X 2 + + 2 e X − Z n.
From the magnet deflections one can get a charge-to-mass ratios, and get back to number of protons (charge on a fully stripped ion) and number of neutrons (from number of protons and the mass). This is the heart of mass spectrometry (although one can use time-of-flight techniques rather than, or in addition to, magnetic separation).
The number of valance electrons counted divided by 8 will give the number of sigma bonds formed. This is just like counting the number of atoms which are getting complete octets, i.e. forming sigma bonds to get octet complete and becoming stable. The remainder is the no. of non bonded electrons, which when divided by 2 gives the number of lone ...
There are two ways for this to happen. 1) Nucleus gains an addition proton (+1 charge) 2) Atom loses an electron (- (-1) charge) Both of these cases result in the atom gaining +1 charge. Case 1 is very difficult to achieve, hence its not a good idea to think about in any chemical reaction (in fact it is not one at all, see comments!), instead ...