a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent. Solution a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent..