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.
Sr has the above electron configuration. Sr has 38 electrons.
Yes, it is true.
See also the link below.
Period 5, Group 1
This would be Technetium.
This element is strontium (Sr).
silver
Xenon.
[Kr]4d105s25p3
[Kr]4d105s25p3
there are 51 electrons in all. So the atomic number of the element is 51 and the element is antimony.
[Kr] 4d10 5s2 5p3
[Kr] 4d10 5s2 5p3
The electronic configuration of stibium (antimony) is: [Kr]4d10 5s2 5p3.
[Kr]4d105s25p3
[Kr]4d105s25p3
there are 51 electrons in all. So the atomic number of the element is 51 and the element is antimony.
[Kr] 4d10 5s2 5p3
[Kr] 4d10 5s2 5p3
Sb (antimony) has 51 electrons. Its noble gas notation is [Kr] 4d10 5s2 5p3
Antimony Sb atom number 51, electronconfiguration:per shell: (K,L,M,N,O) = 2, 8, 18, 8+10, 5per orbital: [1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6]4d10, 5s2 5p3short writing: [Kr] 4d10 5s2 5p3
Sb (antimony) has 51 electrons. Its noble gas notation is [Kr] 4d10 5s2 5p3
It would be 3 electrons!Why?Antimony: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3 Nitrogen: 1s2,,2s2,2p3Phosphorus: 1s2 2s2 2p6 3s2 3p3Arsenic: 3d10 4s2 4p3Bismuth: 4f14 5d10 6s2 6p3so on..
There are 5P3 = 5!/2! = 5*4*3 = 60 permutations.
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).