Outermost electronic configurations of four e | vedclass

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(B) The elements are identified as:$A: 3s^{2} \rightarrow Mg$$B: 3s^{2} 3p^{1} \rightarrow Al$$C: 3s^{2} 3p^{3} \rightarrow P$$D: 3s^{2} 3p^{4} \right

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$B < A < D < C$

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(B) The elements are identified as:$A: 3s^{2} \rightarrow Mg$$B: 3s^{2} 3p^{1} \rightarrow Al$$C: 3s^{2} 3p^{3} \rightarrow P$$D: 3s^{2} 3p^{4} \rightarrow S$Ionization enthalpy generally increases across a period from left to right.However,there are exceptions due to stable electronic configurations:$1$. $Mg$ $(3s^{2})$ has a fully filled $s$-orbital,making it more stable than $Al$ $(3s^{2} 3p^{1})$.$2$. $P$ $(3s^{2} 3p^{3})$ has a half-filled $p$-orbital,making it more stable than $S$ $(3s^{2} 3p^{4})$.Comparing the values:$Al Substituting the labels:$B < A < D < C$

Outermost electronic configurations of four elements $A, B, C, D$ are given below:
$A: 3s^{2}$
$B: 3s^{2} 3p^{1}$
$C: 3s^{2} 3p^{3}$
$D: 3s^{2} 3p^{4}$
The correct order of first ionization enthalpy for them is:

Similar Questions

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If first ionization enthalpies of element $X$ and $Y$ are $419 \ kJ \ mol^{-1}$ and $590 \ kJ \ mol^{-1}$,respectively and second ionization enthalpies of $X$ and $Y$ are $3069 \ kJ \ mol^{-1}$ and $1145 \ kJ \ mol^{-1}$,respectively. Then the correct statement is :-

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Which of the following orders is correct for the property mentioned in brackets?

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Outermost electronic configurations of four elements $A, B, C, D$ are given below:$A: 3s^{2}$$B: 3s^{2} 3p^{1}$$C: 3s^{2} 3p^{3}$$D: 3s^{2} 3p^{4}$The correct order of first ionization enthalpy for them is: