JEE MAIN - Chemistry (2022 - 27th July Evening Shift - No. 1)
The correct decreasing order of energy for the orbitals having, following set of quantum numbers :
(A) n = 3, l = 0, m = 0
(B) n = 4, l = 0, m = 0
(C) n = 3, l = 1, m = 0
(D) n = 3, l = 2, m = 1
is :
(D) > (B) > (C) > (A)
(B) > (D) > (C) > (A)
(C) > (B) > (D) > (A)
(B) > (C) > (D) > (A)
Explanation
(A) $\mathrm{n}+\ell=3+0=3$
(B) $\mathrm{n}+\ell=4+0=4$
(C) $\mathrm{n}+\ell=3+1=4$
(D) $\mathrm{n}+\ell=3+2=5$
Higher $\mathrm{n}+\ell$ valuc, higher the encrgy & if same $\mathrm{n}+\ell$ value, then higher $\mathrm{n}$ value, higher the energy.
Thus : $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$.
(B) $\mathrm{n}+\ell=4+0=4$
(C) $\mathrm{n}+\ell=3+1=4$
(D) $\mathrm{n}+\ell=3+2=5$
Higher $\mathrm{n}+\ell$ valuc, higher the encrgy & if same $\mathrm{n}+\ell$ value, then higher $\mathrm{n}$ value, higher the energy.
Thus : $\mathrm{D}>\mathrm{B}>\mathrm{C}>\mathrm{A}$.
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