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Question

Which of the following options are incorrect in accordance with the mentioned properties?


I) |IP1 of ion M2+|>|EA1 of M3+|
(IP = ionization potential, and EA = electron affinity)

II) S>Se>Te>O (order of EA)

III) Li<Be<B<C (order of electronegativity)

IV) Al3+<Mg2+<Na+<F−(order of ionic size)

V) Li+>Na+<K+ (order of hydrated size)

VI) NaCl>MgCl2>AlCl3(order of lattice energy)



A
(i) and (v)
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B
(i), (v) and (ii)
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C
(i), (v), and (vi)
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D
(i) and (iii)
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Solution

The correct option is C (i), (v), and (vi)
I) IP1 of ion M2+=EA1 of M3+
(IP = ionization potential, and EA = electron affinity)
I) IP1 of ion M2+ is the reverse of EA1 of M3+

II) S>Se>Te>O (order of EA)
Due to the small size oxygen has the poorest EA in the group

III) Li<Be<B<C (order of electronegativity)
As we move from left to rigth electronegativity increases due to increase in effective nuclear charge.

IV) Al3+<Mg2+<Na+<F (order of ionic size)
Greater the positive charge, greater the effective nuclear charge and hence smaller the radius.

V) Li+>Na+>K+ (order of hydrated size)
Smaller the size of the cation greater the hydrated size


VI) NaCl<MgCl2<AlCl3 (order of lattice energy)
greater the charge, greater the lattice energy


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