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Review Constants 1 Periodic Table Part A If the following elements were to form an ionic compound, which noble-gas configurat
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Li = lithium => no of electrons are 3 and it can removes 1 electron and remains 2 electrons and forms the nearest noble gas configuration => He having 2 electrons

Mg = magnesium => no of electrons are 12 and it can removes 2 electron and remains 10 electrons and forms the nearest noble gas configuration => Ne having 10 electrons

O = oxygen => no of electrons are 8 and it can gains 2 electron and having the total 10 electrons and forms the nearest noble gas configuration => Ne having 10 electrons

P = phosphorous => no of electrons are 15 and it can gains 3 electron and having the total 18 electrons and forms the nearest noble gas configuration => Ar having 18 electrons

K = potassium => no of electrons are 19 and it can removes 1 electron and remains 18 electrons and forms the nearest noble gas configuration => Ar having 18 electrons

Se = selenium => no of electrons are 34 and it can gains 2 electron and having the total 36 electrons and forms the nearest noble gas configuration => Kr having 36 electrons

Sr = strontium => no of electrons are 38 and it can removes 2 electrons and remains 36 electrons and forms the nearest noble gas configuration => Kr having 36 electrons

answers =>

He => Li

Ne => Mg , O

Ar => P , K

Kr => Se , Sr

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