Question

In Am$^{3+}$ (Z = 95), the number of f-electrons with ($n + \ell$) value equal to 8 is (where $n$, $\ell$ represent principal and azimuthal quantum numbers)

• A. 7
• B. 6
• C. 5
• D. 4

Hint:

Remember: $n + \ell$ rule is used to determine sublevel energies. For f-electrons, $\ell = 3$, so to get $n + \ell = 8$, $n$ must be 5 (5f).

Answer 1

The Correct Option is B

Atomic number of Am (Americium) is 95. The electron configuration of Am is [Rn]5f$^7$ 7s$^2$. For Am$^{3+}$, three electrons are removed, primarily from the 7s and 5f orbitals, resulting in [Rn]5f$^6$.
For f-electrons, $\ell = 3$. So, to get $n + \ell = 8$, $n = 5$ (since 5f: $n=5$, $\ell=3$). Hence, we count all the electrons in 5f orbitals.
Thus, 6 electrons exist in 5f orbital with $n + \ell = 5 + 3 = 8$.