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  • Outermost electronic configurations of four elements A, B, C, D are given below : (A) (B) <img al

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

 

Option: 1

(A)<(B)<(C)<(D)
 


Option: 2

(\mathrm{B})<(\mathrm{A})<(\mathrm{D})<(\mathrm{C})


Option: 3

(B)<(\mathrm{D})<(\mathrm{A})<(\mathrm{C})
 


Option: 4

(B)<(A)<(C)<(D)


Answers (1)

best_answer

\mathrm{(A) \: 3 s^{2} \rightarrow \mathrm{Mg}}
\mathrm{(B) \: 3s^{2} 3 p^{1} \rightarrow Al}
\mathrm{(C) \: 3 s^{2} 3 P^{3} \rightarrow P (Half \: filled) (\text{more stable than S }) }
\mathrm{(D) \: 3 s^{2} 3 p^{4} \rightarrow S(Partially\: filled) }

The penetration of a \mathrm{3s} -electron to the nucleus is more than that of a \mathrm{3p}-electron. of \mathrm{Al}.

Order of First ionization enthalpy will be -
\mathrm{(B)<(A)<(D)<(C) }

Option(2) is correct.

Posted by

manish painkra

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