Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureQuantum NumbersMedium2 minQB

ChemistryMediumsingle choice

The given orbital is $Ψ_{(r)} = k_{1} e^{- r / k_{2}} (r^{2} - 5 k_{3} r + 6 k_{3}^{2})$ .

For the above orbital, match column-I with column-II (assuming $k_{3} = 1$ ).

Codes:

P Q R S

Options:

Answer:

Solution:

$ψ_{r} = k \times [$ polynomial of r having degree $(n - l - 1)] \times r^{l} \times e^{\frac{- Z r}{n a _{0}}} \Rightarrow ψ_{r} = k_{1} e^{- r / k_{2}} (r^{2} - 5 k_{3} r + 6 k_{3}^{2})$ , has $l = 0$ and $n - l - 1 = 2 \Rightarrow n = 3$ and orbital is 3 s .

s-orbital has zero orbital angular momentum.

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Quantum Numbers

⏱ 2mℹ️ Source: QB

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