Use the formula 2n2. In this case n = 3, so 2(32) = 18. The sublevels and number of electrons in each are 3s23p63d10, for a total of 18 electrons.
Three. s, p, and d.
2 core electrons and 8 valence electrons are there in N3- ion.
Three electrons more than the Nitrogen atom, hence then -3 charge.
Silicon
All of the electrons are paired. If you are asking how many lone pairs, there are 4.
Magnesium
2 core electrons and 8 valence electrons are there in N3- ion.
Three electrons more than the Nitrogen atom, hence then -3 charge.
7protons and 10 electrons.
The N atom is electrically neutral - the number of positively charged protons is equal to the number of negatively charged electrons. That is why there is no net charge on the atom. On the other hand the N3- anion (negatively charged ion) carries a charge of minus 3. It has gain 3 electrons, thereby incurring 3 negative charges. There are 3 more electrons than protons in N3-. N3- can combine with a cation to form an ionic compound.
Actually all the ions will have. Al3+, O2- and N3- ions will have the same number of electrons as neon (10 electrons), whereas Br- ion will have the same number of electrons as krypton (36 electrons)
N3- has 10 electrons and all of those are paired.
Silicon
All of the electrons are paired. If you are asking how many lone pairs, there are 4.
Let any number be n:- n3/n3 = n*n*n/n*n*n = 1 And in index form: n3/n3 = n3-3 = n0 = 1
The corresponding element is phosphorus.
I am checking the Wikipedia article on "quantum number", and don't find a quantum number "i" for the electron. If you mean "l", it seems that "l" can be between 0 and n-1. So, for n = 3, l can be between 0 and 2. If this is what you mean, I don't see any reason that would forbid this particular combination.
declare n1 number; n2 number; n3 number; begin n1:=3 n2 :=5 n3:= sum(n1,n2); dbms_output.put_line( n3); end