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Structure of Atom question

2024 · 9 Apr · Shift 1 · Q17
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Structure of Atom question

2024 · 9 Apr · Shift 1 · Q17

JEE MainChemistryStructure of AtomMCQ+4 / −1
Compare the energies of following sets of quantum numbers for multielectron system. (A) n=4,1=1\mathrm{n}=4,1=1n=4,1=1(B) n=4,1=2\mathrm{n}=4,1=2n=4,1=2(C) n=3,l=1\mathrm{n}=3, \mathrm{l}=1n=3,l=1(D) n=3,1=2\mathrm{n}=3,1=2n=3,1=2(E) n=4,1=0\mathrm{n}=4,1=0n=4,1=0 Choose the correct answer from the options given below :
  1. A
    (E)>(C)>(A)>(D)>(B)(\mathrm{E})\gt (\mathrm{C})\gt (\mathrm{A})\gt (\mathrm{D})\gt (\mathrm{B})(E)>(C)>(A)>(D)>(B)
  2. B
    (B)>(A)>(C)>(E)>(D)(\mathrm{B})\gt (\mathrm{A})\gt (\mathrm{C})\gt (\mathrm{E})\gt (\mathrm{D})(B)>(A)>(C)>(E)>(D)
  3. C
    (C)<(E)<(D)<(A)<(B)(\mathrm{C})\lt (\mathrm{E})\lt (\mathrm{D})\lt (\mathrm{A})\lt (\mathrm{B})(C)<(E)<(D)<(A)<(B)
  4. D
    (E)<(C)<(D)<(A)<(B)(\mathrm{E})\lt (\mathrm{C})\lt (\mathrm{D})\lt (\mathrm{A})\lt (\mathrm{B})(E)<(C)<(D)<(A)<(B)
View written solutionFree

Correct answer: C

  1. Identify the orbitals from the given quantum numbers

Given:

  • (A) n=4, l=1⇒4pn=4,\ l=1 \Rightarrow 4pn=4, l=1⇒4p
  • (B) n=4, l=2⇒4dn=4,\ l=2 \Rightarrow 4dn=4, l=2⇒4d
  • (C) n=3, l=1⇒3pn=3,\ l=1 \Rightarrow 3pn=3, l=1⇒3p
  • (D) n=3, l=2⇒3dn=3,\ l=2 \Rightarrow 3dn=3, l=2⇒3d
  • (E) n=4, l=0⇒4sn=4,\ l=0 \Rightarrow 4sn=4, l=0⇒4s
  1. Use the (n+l)(n+l)(n+l) rule for multielectron atoms

For multielectron systems, orbital energy increases with increasing value of (n+l)(n+l)(n+l). If two orbitals have the same (n+l)(n+l)(n+l) value, then the orbital with smaller nnn has lower energy.

Now calculate n+ln+ln+l for each:

  • (A) 4p4p4p: n+l=4+1=5n+l=4+1=5n+l=4+1=5
  • (B) 4d4d4d: n+l=4+2=6n+l=4+2=6n+l=4+2=6
  • (C) 3p3p3p: n+l=3+1=4n+l=3+1=4n+l=3+1=4
  • (D) 3d3d3d: n+l=3+2=5n+l=3+2=5n+l=3+2=5
  • (E) 4s4s4s: n+l=4+0=4n+l=4+0=4n+l=4+0=4
  1. Compare energies

Lowest (n+l)=4(n+l)=4(n+l)=4 group:

  • (C) 3p3p3p
  • (E) 4s4s4s

Since both have same n+l=4n+l=4n+l=4, lower nnn means lower energy: 3p<4s⇒(C)<(E)3p < 4s \Rightarrow (C) < (E)3p<4s⇒(C)<(E)

Next (n+l)=5(n+l)=5(n+l)=5 group:

  • (D) 3d3d3d
  • (A) 4p4p4p

Again, same n+l=5n+l=5n+l=5, lower nnn means lower energy: 3d<4p⇒(D)<(A)3d < 4p \Rightarrow (D) < (A)3d<4p⇒(D)<(A)

Highest:

  • (B) 4d4d4d with n+l=6n+l=6n+l=6

So overall increasing order of energy is: (C)<(E)<(D)<(A)<(B)(C) < (E) < (D) < (A) < (B)(C)<(E)<(D)<(A)<(B)

  1. Match with options

This corresponds to Option C.

  1. Final answer

(C)<(E)<(D)<(A)<(B)\boxed{(C)<(E)<(D)<(A)<(B)}(C)<(E)<(D)<(A)<(B)​ So, the correct option is C.

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