JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 7)
Arrange the following orbitals in decreasing order of energy.
A. $\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0$
B. $\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0$
C. $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0$
D. $\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1$
The correct option for the order is :
A. $\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}=0$
B. $\mathrm{n}=4, \mathrm{l}=0, \mathrm{~m}=0$
C. $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=0$
D. $\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}=1$
The correct option for the order is :
$\mathrm{B}>\mathrm{D}>\mathrm{C}>\mathrm{A}$
$\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}$
$\mathrm{D}>\mathrm{B}>\mathrm{A}>\mathrm{C}$
$\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$
Explanation
(A) n = 3; l = 0; m = 0 ; 3s orbital
(B) n = 4; l = 0; m = 0 ; 4s orbital
(C) n = 3; l = 1; m = 0 ; 3p orbital
(D) n = 3; l = 2; m = 0 ; 3d orbital
As per Hund’s ruleo of energy is given by (n + l) value. If value of (n + l) remains same then energy is given by n only.
(B) n = 4; l = 0; m = 0 ; 4s orbital
(C) n = 3; l = 1; m = 0 ; 3p orbital
(D) n = 3; l = 2; m = 0 ; 3d orbital
As per Hund’s ruleo of energy is given by (n + l) value. If value of (n + l) remains same then energy is given by n only.
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