1.

Four elements A, B, C and D form a series of compounds having the formulae AB, ` B_(2), CB_(3), DB_(2)` and `DB_(3)`. If the jumbled up atomic numbers of A, B, C and D are 13, 19 , 26 and 35 , What are the ordered atomic numbers of A, B C and C ?

Answer» ` 13 = Al = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(1)` (can lose `1 e ^(-))`
` 19 = K = 1ss^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 4s^(1) (can lose `1e ^(-))`
26 =` Fe = 1s^(2)2s^(2) 2p^(6) 3s^(2) 3p^(6)3d^(6) 4s^(2) ` (can lose ` 2e^(-) or 3 r^(-))`
35 ` = Br = 2s^(2) 2s^(2) 3p^(6) 3s^(2)3p^(6) 3d^(10)4s^(2) 4p_(x)^(2) 4p_(y)^(2) 4p_(z)^(1) ` (can gain or share ` 1 e ^(-)` )
Compunds formed will be ` KBr , Br_(2), AlBr_(3), Fe Br_(2) and FeBr_(3).`
Hence ,` A = K = 19 , B= Br = 35 , C = Al = 13 , D = Fe = 26.`


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