A) \[Na<Mg<He\]
B) \[Mg<Na<He\]
C) \[Mg<He<Na\]
D) \[Na<He<Mg\]
Correct Answer: B
Solution :
\[\underset{(1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}},\text{ }3{{s}^{1}})}{\mathop{Na}}\,\xrightarrow[removal\text{ }of\text{ }3s{{e}^{-}}]{I{{E}_{1}}}\] \[\underset{unstable\text{ }configuration}{\mathop{\underset{(1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}})}{\mathop{N{{a}^{+}}}}\,}}\,\,\,\xrightarrow[removal\text{ }of\text{ }2p{{e}^{-}}]{I{{E}_{2}}}\underset{(1{{s}^{2}},2{{s}^{2}},2{{p}^{5}})}{\mathop{N{{a}^{2+}}}}\,\] \[\underset{\begin{smallmatrix} (1{{s}^{2}},\text{ }2{{s}^{2}},\text{ }2{{p}^{6}},\text{ }3{{s}^{2}}) \\ stable\text{ }configuration \end{smallmatrix}}{\mathop{Mg}}\,\xrightarrow[removal1\text{ }of\text{ }3s{{e}^{-}}]{I{{E}_{1}}}\] \[\underset{\underset{stable\text{ }configuration}{\mathop{(1{{s}^{2}},\text{ }2{{s}^{2}},2{{p}^{6}}3{{s}^{1}})}}\,}{\mathop{M{{g}^{+}}}}\,\xrightarrow[removal\text{ }of\text{ }second\text{ }3s{{e}^{-}}]{I{{E}_{2}}}\underset{({{1}^{2}},\,\,2{{s}^{2}},\,\,2{{p}^{6}})}{\mathop{M{{g}^{2+}}}}\,\] In case of sodium metal removal of 2nd electron takes place from 2p subshell which is more closer to nucleus arid thus more energy is required to remove the electron, whereas removal of 2nd electron from magnesium takes place from 3s subshell. \[\underset{(1\,{{s}^{2}})}{\mathop{He}}\,\xrightarrow[removal\,\,of\,\,1{{s}^{2}}{{e}^{-}}]{I{{E}_{1}}}\underset{(1{{s}^{1}})}{\mathop{H{{e}^{+}}}}\,\xrightarrow[removal\,\,1{{s}^{1}}{{e}^{-}}]{I{{E}_{2}}}\underset{(1{{s}^{0}})}{\mathop{H{{e}^{2+}}}}\,\] Noble gases have the highest ionisation energies in their respective periods. In helium, removal of 2nd electron takes place from \[1{{s}^{1}}\]subshell which is nearest to nucleus. Therefore, the increasing order of \[I{{E}_{2}}\] (in kJ\[mo{{l}^{-1}}\]) of these elements is as follows \[Mg<Na<He\]
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