Shib sankar dhar Asked a Question
July 15, 2020 10:51 pmpts 30 pts
(a) Decreasing 1onization potential for K, Ca & Ba is (a) Ba> K> Ca (b) Ca > Ba> K (c) K> Ba> Ca (d) K> Ca> Ba The ionization energy will be maximum for the process. (a) Ba > Ba Alkaline earth metals always fornm dipositive ions due to (a) 1E,-IE>11 eV (C)IE, -IE,<11 eV Amongst the following, the incorrect order is (a) TE, (Al) < IE, (Mg) (c)TE, (Mg) >IE, (Na) The correct order of decreasing first ionization energy is (a) C>B>Be> Li (c)B C>Be> Li Which of the following contiguration is expected to have maximum difference in second and third ionization energies a)(1s (2sP (2p? 42/ (b) Be > Be* (c) Cs Cs* (d) Li >Li* (b) IE, - IE, = 17 eV (d) None of these 5. (b) TE, (Na) < IE,(Mg) (d) IE, (Mg)> IE, (AI) (b) C>Be>B>Li (d) Be>Li>B>C V (b) (1s (2s)* (2p)° (3s)* (3p) (d) (1) (2s (2p) () (19 (2s)P (2p)° (3s)
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  • Priyanshu kumar Best Answer see this video shib for 42
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    Shib sankar dhar
    sir 47??
  • Dinesh khalmaniya thankyou
    42) option d
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    Dinesh khalmaniya
    got it??
  • Dinesh khalmaniya
    47) A because after removing one electron 2s2 configuration is obtained so it is fully filled and 3 electron is removed from fully filled orbital so it is difficult to remove so he...
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    Dinesh khalmaniya
    got it