2.2 · Atomic StructurePractice · 原子结构

One photon carries E = hc/λ = $6.626 \times 10^{-34} \times 3.00 \times 10^{8}$ ÷ ($5.32 \times 10^{-7}$) = $3.736 \times 10^{-19}$ J. The laser emits $0.0050$ J each second, so photons per second = $0.0050$ ÷ $3.736 \times 10^{-19}$ = $1.3 \times 10^{16}$. The power has only 2 significant figures.

Ionizing means moving the electron to n = ∞ (E = 0). One atom needs 0 − $E_2$ = $\dfrac{2.18 \times 10^{-18}}{2^2}$ = $5.45 \times 10^{-19}$ J. One mole needs $5.45 \times 10^{-19}$ × $6.022 \times 10^{23}$ = 3.28 × 10⁵ J = 328 kJ, a quarter of the energy to ionize from the ground state (about 1312 kJ/mol).

λ = hc/ΔE, so the smallest energy gap gives the longest wavelength. The levels crowd together higher up, so 4 → 3 has the smallest gap. Calculated: 2 → 1: 122 nm, 3 → 1: 103 nm, 4 → 2: 486 nm, 4 → 3: 1876 nm. The 4 → 3 line is in the infrared.

1s²2s¹2p¹ has 4 electrons, so it is Be. Ground-state Be is 1s²2s²; here one electron sits in the higher-energy 2p while 2s is not full, so this is an excited state. The other three are the ground states of N, Al, and K.

Ni is [Ar]4s²3d⁸. Cations lose the outer 4s electrons first, not 3d: $\ce{Ni^2+}$ is [Ar]3d⁸. Put 8 electrons into five 3d orbitals: one in each (5), then 3 more pair up, leaving 5 − 3 = 2 unpaired.

Valence 5s²5p³ means period 5, group 15 (with N, P, and As): Sb, element 51. Count: 36 + 2 + 10 + 3 = 51. As is in period 4 ([Ar]4s²3d¹⁰4p³), Bi is in period 6, and Sn has only 2 electrons in 5p.

Carbon has 2 electrons in 2p. Hund’s rule: electrons spread out into separate orbitals with parallel spins, giving [↑][↑][ ]. B crowds both into one orbital and C gives them opposite spins, both violating Hund’s rule; D puts two same-spin electrons in one orbital, violating the Pauli principle.

The photon carries E = hc/λ = $6.626 \times 10^{-34} \times 3.00 \times 10^{8}$ ÷ ($9.72 \times 10^{-8}$) = $2.05 \times 10^{-18}$ J, which equals $E_n - E_1$: $2.18 \times 10^{-18}\left(1 - \dfrac{1}{n^2}\right)$ = $2.05 \times 10^{-18}$. So $1 - \dfrac{1}{n^2}$ = 0.938, $n^2$ ≈ 16.2, and n = 4.

Transition metals lose their 4s electrons first. $\ce{Fe^2+}$ [Ar]3d⁶, $\ce{Mn^2+}$ [Ar]3d⁵, $\ce{Cr^3+}$ [Ar]3d³, $\ce{Cu+}$ [Ar]3d¹⁰. Only $\ce{Fe^2+}$ (Fe is [Ar]4s²3d⁶; remove both 4s electrons) is [Ar]3d⁶.

Se is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴. The p electrons are in 2p (6), 3p (6), and 4p (4), for 16 in all. Counting only the outer 4p misses the inner p electrons.