A)
\[{{S}^{2-}}>C{{l}^{-}}>\]\[{{K}^{+}}>C{{a}^{2+}}\] (Ionisation energy)
B)
\[C<N>F<O\] (\[{{2}^{nd}}\]Ionisation energy)
C)
\[B>AI>Ga>\] \[In>TI\] (Electronegativity)
D)
\[N{{a}^{+}}>L{{i}^{+}}>M{{g}^{2+}}>\]\[B{{e}^{2+}}>A{{l}^{3+}}\] (Ionic Radius)
Correct Answer: B
Solution :
[b]| [a] Correct order \[\to C{{a}^{2+}}>{{K}^{+}}>C{{l}^{-}}>{{S}^{2-}}\] (lonisation energy) |
| For Isoelectronic species \[(IE\propto {{Z}_{eff}})\] |
| [b] Correct order \[\to C<N<F<O\] \[({{2}^{nd}}IE)\] |
| Second electron removal from oxygen requires more energy as it acquires stable \[2{{s}^{2}}2{{p}^{3}}\] configuration after removal of one electron. |
| [c] Correct order \[\to B>TI>In>Ga>Al\] (Electronegativity) |
| In general EN increases in boron family from top to bottom due to increase in \[{{Z}_{eff}}\] on valence shell while Boron has highest E.N. due to its very small size. |
| [d] Correct order \[\to N{{a}^{+}}>L{{i}^{+}}>M{{g}^{2+}}\] \[>A{{l}^{3+}}>B{{e}^{2+}}\] (Ionic radius) Ionic radius depends on \[{{z}_{eff}}\] and number of shells. |
You need to login to perform this action.
You will be redirected in
3 sec
