5p - 3 = 15
Add 3 to each side:
5p = 18
Divide each side by 5:
p = 18/5 = 3.6
[Kr] 4d10 5s2 5p3
[Kr]4d105s25p3
[Kr]4d105s25p3
The electronic configuration of stibium (antimony) is: [Kr]4d10 5s2 5p3.
[Kr] 4d10 5s2 5p3
There are 5P3 = 5!/2! = 5*4*3 = 60 permutations.
Sb (antimony) has 51 electrons. Its noble gas notation is [Kr] 4d10 5s2 5p3
there are 51 electrons in all. So the atomic number of the element is 51 and the element is antimony.
Sb (antimony) has 51 electrons. Its noble gas notation is [Kr] 4d10 5s2 5p3
60 if you allow permutations, 20 if not. 60 is found by suing the permutation function 5P3, and we get 20 simply by dividing by 3 as there will be 3 permutations from a group of 3 numbers while keeping the order the same e.g. (1,2,3) , (2,3,1) and (3,1,2).
Assuming this is your original formula: -3p3+ 5p + -2p2 + -4 - -12p + 5 - -8p3You combine like terms, where the p exponent is the same, to produce:5p3 - 2p2 + 17p + 1
Group 1- as the configuration is ns1. The Kr in the question indicates the Krypton core. The 5s1 is the clue - the period number is the same as the principal quantum number (in this case 5) so the element is in group 1 and period 5 - so it is rubidium.