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How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium ? |
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Answer» Solution :The electronic configuration of Na is `[Ne]3s^(1) " and that of Mg is " [Ne]3s^(2)`. The configuration of Mg is more stable (being completely filled) than of Na. Therefore, first ionization enthalpy of Mg is more than that of Na. After the LOSS of an electron from Na, it acquires the electronic configuration of noble gas, Ne i.e., `1s^(2) 2s^(2) 2P^(6)`. On the other hand in case of Mg atom, the electronic configuration becomes `[Ne]3s^(1)` . Thus the electronic configuration of `Na^(+)`is more stable than `Mg^(+)`and HENCE the second ionization enthalpy of Na is much larger than that of Mg. `Na([Ne]3s^(1)) TONA^(+)([Ne]) to Mg^(2+) ([Ne])` `Mg([Ne]3s^(2)) to Mg^(2+)([Ne]3s^(1))toMg^(2+)([Ne])` `IE_(1)(Na) lt IE_(1)(Mg) "" IE_(2)(Na) gt IE_(2)(Mg)` |
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