Question

6. How many unpaired electrons does an atom of phosphorous have in its (3) 3 (4) 5 its ground complete an atom 7. A possible set of quantum numbers for the last electron added to come ground state is mg A 4 0 + 2 B 3 0 +1 - C. 4 1 -1 +2 D. 3 -72 (1) A (2) B (3) C (4) D 8. A possible set of quantum numbers to describe an m. = + (2) n = 4, = 0, m = 0, m = -% (3) n = 4, { = 5, m =0 m, a 5s subshells (1) n=512 = 5, m = 0, m, = +% (4135.0 de in the 6p orbitals? 4 What is the maximum total number of electrons possible in the portal 10. A ground-state atom of manganese has unpaired electrons and is (2) 6. diamagnetic (3) 5, paramagnetic (4)3, paramagnetic (1)6 (2)4 3) w we take away two electrons from the outer shell of strontium, it would have the same elect configuration as what element? (1) krypton (2) germanium (3) argon (4) ma 12. The orbital diagram for a ground-state nitrogen atom is 1s 2s 2p 1. I 1 I I 2. Il a al 3. 1 1 1 1 4. Il II 1 (4) 4 -1) 1 (2) 2 (3) 3 NOUNCY of 3.4 x 1011 3 -1? (1m = 109

Answer 1

6) Phosphorous & atomic number = 15 electronic configuration = 15² 25² 2p 62 5² p3 So, number of unpaired electrons = 3ph 3 7 silicon atomic number = 14 electronic configuration : 15²25² 226 35² 3p2 as last election is filled in 3p subshell so na3 ; P = 1 , mi= tl and ms = -1/2 8) 5s subshell n.5. s=0 mi: 0 ms = +/ 9 6p orbitals [11/11/16 , PP, In each portitals maximum z electroms can be filled So, in 3 porbitals maximum 6 es can be filled. so In op, maximum 6es can be filled. 10) manganese - atomic number = 25 electronic configuration - 15²25²20625²3p6 45²3d5 So, number of on paired e = 5 as there are 5 un paired electron and element with unpaired electron are called para magnetic

11) strontium & atomic number = 38 when its outer two election is removed then number of e left = 38-2 = 36 é the element with 36e in its ground state is krypton. 12) Nitrogen atomic no 7 electronic configurations 152 25² 2p3 [11] is 11111 2P 25 option I is correct