MCQ
The decreasing order of the ionisation potential in the following elements is
- A$Ne > Cl > P > S > Al > Mg$
- ✓$Ne > Cl > P > S > Mg > Al$
- C$Ne > Cl > S > P > Mg > Al$
- D$Ne > Cl > S > P > Al > Mg$
In moving down the group from top to bottom the ionization potential decreases and on moving from left to right in a period it is increases.
$Ne$ is the noble element, it has completely filled octet hence, it has the highest ionization potential.
Half-filled $( P )$ and completely filled configuration $( Mg )$ are the cause of the higher value of $I.E.$
The outer electronic configuration of $Mg$ and $Al$ are $3 s ^2$ and $3 s ^2\, 3 p ^1$ respectively.
The decreasing order of the ionization potential of the elements is:
$Ne\, >\, Cl\, >\, P\, > \,S \,>\, Mg\, >\, Al$
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$(1)\,CH_3C \equiv C - CH_3$ $(2)\,CH_3CH_2 -CH_2 -CH_3$
$(3)\, CH_3CH_2C \equiv CH$ $(4)\, CH_3CH = CH_2$
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, \,\,\,\,K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2,\,\,\,\, K_2$
The equilibrium constant for the reaction
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be
$\to H-F = 565\, kJ/mol$
Calculate average bond energy of $Xe-F$ bond.....$ kJ/mol$