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1.	Which of the following species will have the largest and the smallest size Mg, Mg2+, Al, Al3+?.
Answer» Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic ions, the one with the large positive nuclear charge will have a smaller radius. Hence the largest species is Mg; the smallest one is Al³⁺ The size of an anion greater while that of the cation is smaller than that of its parent atom, e.g. F- (=1.36 Å)>F(=0.72 Å); Cl- (=1.81 Å)>Cl(=0.99Å); Na⁺ (=0.95Å)<Na(=1.90Å); Ca²⁺(=0.99 Å)<Ca(=1.97Å). Explanation: Let us consider the radii of Na, Na⁺ , Cl and Cl^- . The reason of the fact that Na+ ion is smaller than Na atom is that Na⁺ ion has 10 electrons (Na⁺ →1s² ,2s² p⁶ ) while Na atom has 11 electrons (Na →1s² ,2s² p⁶ ,3s¹ ). The nuclear charge (charge on the nucleus) in each case is the same, i.e. equal to +11 (atomic number of Na). This nuclear charge of +11 can pull 10 electrons of Na⁺ ion inward more effectively than it can pull a greater number of 11 electrons of Na atom. Thus Na⁺ ion is smaller than Na atom. The reason why Cl^- ion is bigger than Cl atom can also be explained on a similar basis. The Cl^- ion has 18 electrons (Cl^- →1s² ,2s² p⁶ ,3s² p⁶ ) while Cl atom has only 17 electrons (Cl→1s² ,2s² p⁶ ,3s² p⁵ ). The nuclear charge in each case is +17, which cannot pull 18 electrons of Cl^- ion as effectively as it can pull 17 electrons of Cl atom inward. Thus Cl^- ion is bigger than Cl atom.

Which of the following species will have the largest and the smallest size Mg, Mg2+, Al, Al3+?.

Answer»

Atomic radii decrease across a period. Cations are smaller than their parent atoms. Among isoelectronic ions, the one with the large positive nuclear charge will have a smaller radius.

Hence the largest species is Mg; the smallest one is Al³⁺

The size of an anion greater while that of the cation is smaller than that of its parent atom, e.g. F- (=1.36 Å)>F(=0.72 Å); Cl- (=1.81 Å)>Cl(=0.99Å); Na⁺ (=0.95Å)<Na(=1.90Å); Ca²⁺(=0.99 Å)<Ca(=1.97Å).

Explanation:

Let us consider the radii of Na, Na⁺ , Cl and Cl^- . The reason of the fact that Na+ ion is smaller than Na atom is that Na⁺ ion has 10 electrons (Na⁺ →1s² ,2s² p⁶ ) while Na atom has 11 electrons (Na →1s² ,2s² p⁶ ,3s¹ ). The nuclear charge (charge on the nucleus) in each case is the same, i.e. equal to +11 (atomic number of Na). This nuclear charge of +11 can pull 10 electrons of Na⁺ ion inward more effectively than it can pull a greater number of 11 electrons of Na atom. Thus Na⁺ ion is smaller than Na atom.

The reason why Cl^- ion is bigger than Cl atom can also be explained on a similar basis. The Cl^- ion has 18 electrons (Cl^- →1s² ,2s² p⁶ ,3s² p⁶ ) while Cl atom has only 17 electrons (Cl→1s² ,2s² p⁶ ,3s² p⁵ ). The nuclear charge in each case is +17, which cannot pull 18 electrons of Cl^- ion as effectively as it can pull 17 electrons of Cl atom inward. Thus Cl^- ion is bigger than Cl atom.

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