Shortcut process of Electronic Configuration.pdf

Shortcut p rocess of Electronic Configuration Z X be an element with atomic number 𝑍 Steps: 1) Find the largest number among [0, 2, 10, 18, 36, 54, 86] which is smaller than 𝑍 2) According to your choices , the period of the element is : a. 0 ➔ 1 st b. 2 ➔ 2 nd c. 10 ➔ 3 rd d. 18 ➔ 4 th e. 36 ➔ 5 th f. 54 ➔ 6 th g. 86 ➔ 7 th Let it be 𝑛 3) Now, subtract the number you chose (say 𝑦 ) from 𝑍 . Let the result be 𝑥 4) Fill up 𝑛𝑠 , ( 𝑛 − 2 ) 𝑓 , ( 𝑛 − 1 ) 𝑑 , 𝑛𝑝 orbitals in the same order. Don’t write the orbital which doesn’t exist. 5) Let the result be 𝑛 𝑠 𝑎 , ( 𝑛 − 2 ) 𝑓 𝑏 , ( 𝑛 − 1 ) 𝑑 𝑐 , 𝑛 𝑝 𝑑 6) Identify the inert gas with atomic number 𝑦 according to the following table y Inert Gas 2 He 10 Ne 18 Ar 36 Kr 54 Xe 86 Rn 7) Let the inert gas be [ 𝑋 ] . Then, reqd configuration is [ 𝑋 ] 𝑛 𝑠 𝑎 , ( 𝑛 − 2 ) 𝑓 𝑏 , ( 𝑛 − 1 ) 𝑑 𝑐 , 𝑛 𝑝 𝑑 Example 1) 85 At Largest number is 54. Period is 6 85 - 54 = 31 6 𝑠 , 4 𝑓 , 5 𝑑 , 6 𝑝 all exist. Thus, we can fill them up with 31 electrons. 6 𝑠 2 , 4 𝑓 14 , 5 𝑑 10 , 6 𝑝 5 The inert gas with atomic number 54 is Xe. Ans : [ 𝑋𝑒 ] 6 𝑠 2 , 4 𝑓 14 , 5 𝑑 10 , 6 𝑝 5 2) 80 Hg Largest number is 54. Period is 6 80 - 54 = 2 6 6 𝑠 , 4 𝑓 , 5 𝑑 , 6 𝑝 all exist. Thus, we can fill them up with 26 electrons. 6 𝑠 2 , 4 𝑓 14 , 5 𝑑 10 , 6 𝑝 0 The inert gas with atomic number 54 is Xe. Ans : [ 𝑋𝑒 ] 6 𝑠 2 , 4 𝑓 14 , 5 𝑑 10 3) 70 Yb Largest number is 54. Period is 6 70 - 54 = 16 6 𝑠 , 4 𝑓 , 5 𝑑 , 6 𝑝 all exist. Thus, we can fill them up with 16 electrons. 6 𝑠 2 , 4 𝑓 14 , 5 𝑑 0 , 6 𝑝 0 The inert gas with atomic number 54 is Xe. Ans : [ 𝑋𝑒 ] 6 𝑠 2 , 4 𝑓 14 4) 47 Ag Largest number is 36. Period is 5 47 - 36 = 11 Among 5 𝑠 , 3 𝑓 , 4 𝑑 , 5 𝑝 , only 5 𝑠 , 4 𝑑 , 5 𝑝 exist. Thus, we can fill them up with 11 e lectrons. 5 𝑠 2 , 4 𝑑 9 , 5 𝑝 0 The inert gas with atomic number 36 is Kr. [ 𝐾𝑟 ] 5 𝑠 2 , 4 𝑑 9 Exception : since 4 𝑑 10 is much more stable than 4 𝑑 9 , so, one electron from 5 𝑠 moves to 4 𝑑 Ans : [ 𝐾𝑟 ] 5 𝑠 1 , 4 𝑑 10 5) 25 Mn Largest number is 18. Period is 4 25 - 18 = 7 Among 4 𝑠 , 2 𝑓 , 3 𝑑 , 4 𝑝 only 4 𝑠 , 3 𝑑 , 4 𝑝 exist. Thus, we can fill them up with 7 electrons. 4 𝑠 2 , 3 𝑑 5 , 4 𝑝 0 The inert gas with atomic number 54 is Xe. Ans : [ 𝑋𝑒 ] 4 𝑠 2 , 3 𝑑 5 Elements with exceptional electronic configuration Element Atomic number Actual configuration Expected configuration Cr 24 [Ar] 3d 5 4s 1 [Ar] 3d⁴4s² Cu 29 [Ar] 3d 10 4s 1 [Ar] 3d⁹4s² Nb 41 [Kr] 4d 4 5s 1 [Kr] 4d³5s² Mo 42 [Kr] 4d 5 5s 1 [Kr] 4d⁴5s² Ru 44 [Kr] 4d 7 5s 1 [Kr] 4d⁶5s² Rh 45 [Kr] 4d 8 5s 1 [Kr] 4d⁷5s² Pd 46 [Kr] 4d 10 [Kr] 4d⁸5s² Ag 47 [Kr] 4d 10 5s 1 [Kr] 4d⁹5s² La 57 [Xe] 5d 1 6s 2 [Xe] 4f¹6s² Ce 58 [Xe] 4f 1 5d 1 6s 2 [Xe] 4f²6s² Gd 64 [Xe] 4f 7 5d 1 6s 2 [Xe] 4f⁸6s² Pt 78 [Xe] 4f 14 5d 9 6s 1 [Xe] 4f¹⁴5d⁸6s²