Question

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


$I)|I{P}_1\text{of ion}{M}^{2+}|>|E{A}_{1}of{M}^{3+}|$
$\text{(IP = ionization potential, and EA = electron affinity)}$

$II)S>Se>Te>O\text{(order of EA)}$

$III)Li<Be<B<C\text{(order of electronegativity)}$

$IV)A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}\text{(order of ionic size)}$

$V)L{i}^{+}>N{a}^{+}<K^{+}\text{(order of hydrated size)}$

$VI)NaCl>MgC{l}_2>AlC{l}_3\text{(order of lattice energy)}$

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

Answer 1

Answer: (A), (C)

The correct options are
A (i) and (v)
C (i), (v), and (vi)

$I)I{P}_{1}ofion{M}^{2+}=E{A}_{1}of{M}^{3+}$
$\text{(IP = ionization potential, and EA = electron affinity)}$
$I)I{P}_{1}ofion{M}^{2+}$ is the reverse of $E{A}_{1}of{M}^{3+}$

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

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

$IV)A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}\text{(order of ionic size)}$
Greater the positive charge, greater the effective nuclear charge and hence smaller the radius.

$V)L{i}^{+}>N{a}^{+}>K^{+}\text{(order of hydrated size)}$
Smaller the size of the cation greater the hydrated size


$VI)NaCl<MgC{l}_2<AlC{l}_3\text{(order of lattice energy)}$
greater the charge, greater the lattice energy