The number of degenerate orbitals in absence of electric field and magnetic field in n th excited state of X + y ion (Single e − specie) is four times the degeneracy value in 1 st excited state of He

Question

The number of degenerate orbitals in absence of electric field and magnetic field in $n^{th}$ excited state of $X^{+ y}$ ion (Single $e^{-}$ specie) is four times the degeneracy value in $1^{st}$ excited state of $He^{+}$ ion.

Z = atomic number of $X^{+ y}$ ion, element 'X' has diagonal relationship with 'Al'. Find the value of Z + n.

TestHub · Accepted Answer

Identify Element X: Since element X has a diagonal relationship with Aluminum (Al), X is Beryllium (Be), which gives an atomic number of $Z = 4$ .

Identify Ion X⁺ʸ: For Be to be a single-electron species, it must lose 3 electrons, making it Be³⁺ (where $y = 3$ ).

Calculate He⁺ Degeneracy: The 1ˢᵗ excited state of He⁺ corresponds to the shell level $n_{l e v e l} = 2$ , which has a degeneracy of $n_{l e v e l}^{2} = 2^{2} = 4$ orbitals.

Determine Excited State $n$ : The number of degenerate orbitals for X⁺ʸ is $4 \times 4 = 16$ , meaning its shell level is $16 = 4$ ; since the 4ᵗʰ level is the 3ʳᵈ excited state, we find $n = 3$ .

Final Value: Combining the terms gives $Z + n = 4 + 3 = 7$ .

Chemistry - Atomic Structure Question with Solution | TestHub

Doubts & Discussion