Q. The attempt in ionic bond formation is to

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Q. Upon reaction of O₂ with metals of Period 3 elements, ionic bonds are formed if the

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Q. The ionic equation O⁻²(aq)+ H₂O(l) → 2OH⁻. This will result in an

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Q. The magnitude of ionization energy depends upon

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Q. The ionization energy between last element of one period and first element of next period receives a rapid

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Q. The nuclear charge increases in periodic table

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Q. The electronic configuration is evident by the successive

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Q. The force of attraction between nucleus and electrons increases across periods because

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Q. The force of repulsion between electron pairs causes to remove electron with less

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Q. There is less attraction between the fifth electron of Boron and nucleus, hence that electron resides in

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Q. The electronic configuration of electrons involves their addition starting from lowest energy level to

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Q. Electrons always reside in certain energy level outside

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Q. The number of lobes in p-orbital is

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Q. The nuclear charge down the group

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Q. The less amount of attraction between the nucleus and outer electrons is due to

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Q. The process by which electrons spin and pair up against each other is called

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Q. Boron has one more proton than

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Q. The energy level which is closest to nucleus has

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Q. The decrease in the force of attraction is due to the increase in the

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Q. The aqueous phase contains ions

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