IB Chemistry S1.3: Electron configurations worked solutions

At fixed light speed, c=\lambda f (B), so increasing f reduces \lambda (D). Since E=hf (B), photon energy increases with f. Choose A.

The second transition is 2\to1, whose endpoint is the hydrogen ground level. It is ultraviolet. Its energy gap exceeds the preceding visible gap ending at n=2, so its photon has greater energy. Choose A.

Add occupied and empty places to obtain capacity 36+14=50. Using C=2n^2 (R), solve n=\sqrt{50/2}=5 (D). Option C is correct. This counts hypothetical shell capacity rather than specifying an atom.

A p sublevel has three orbitals. Each holds up to two electrons with opposite spins, so N=2(3)=6 (D). The matching pair is three orbitals and six electrons: B.

The argon core contributes 18 electrons and the written sublevels contribute 10+2, giving 30. A neutral Z=32 atom needs two more (D). After 3d and 4s, these enter 4p, giving 4p^2. Choose A.

Use the observed configurations before counting. Chromium has five singly occupied 3d orbitals plus one 4s single, so u=6. Copper has five paired 3d orbitals and one 4s single, so u=1. This gives B.

An atom prepared in n=2 already stores excitation energy. The remaining upward gap to the unbound level is smaller than the ground-to-unbound gap. Multiplying by particles per mole does not change that comparison. This is an excited-state threshold, giving B.

The isolated large ratio is \mathrm{IE}_5/\mathrm{IE}_4=6.7. The fifth removal first reaches an inner shell, so the first four removals exhaust the outer shell. The outer count is four, option D.

(a) An atomic orbital describes the spatial probability distribution for an electron of specified energy in the stated atomic environment. It is not a path followed by the electron. (b) One orbital accommodates at most two electrons, with opposite spins. The two lobes of one p orbital do not double this maximum.

(a) Draw a sphere for s and two lobes sharing a central nucleus for p. Mark each nuclear origin. (b) The envelope encloses a chosen region with high probability of finding the electron; it does not map its motion. (c) The p envelopes point along three mutually perpendicular axes. (d) Seven f orbitals each hold two electrons, giving 14 for the sublevel (D); one f orbital holds two. The local solution diagram is supplied separately.

(a) N_e=Z-q (D) gives 30-2=28 for zinc and 31-3=28 for gallium. (b) Each ion is [\mathrm{Ar}]3d^{10}. (c) Zinc loses two 4s electrons; gallium loses its 4p electron and two 4s electrons. Removal empties the highest occupied principal shell before 3d. The neutral-atom filling ladder cannot be reversed to choose the ion removal order.

(a) Draw five horizontal levels, with n=1 lowest and progressively shrinking gaps. (b) Put UV beside the 3\to1 arrow, visible beside 3\to2, and IR beside 4\to3. (c) Draw a dashed unbound reference above the finite levels and an upward arrow from n=1 to that reference. The local vector solution diagram supplies these exact arrows.

(a) Period 3 group 17 gives D the arrangement [\mathrm{Ne}]3s^2 3p^5; period 3 group 2 gives E [\mathrm{Ne}]3s^2. (b) D fills its p sublevel by gaining one electron, making charge -1. E loses two outer electrons, making charge +2. (c) One E ion balances two D ions: 2+2(-1)=0 (D), so the formula using the given labels is \mathrm{ED}_2.

(a) Only photon energies matching allowed upward level gaps are absorbed, so selected wavelengths are depleted from the transmitted continuous beam. (b) Electrical energy first raises electrons to excited states. When they return to lower states, photons carry the positive energy differences, producing bright lines against the dark background. (c) The separate line positions imply discrete allowed energy differences. Compare the wavelengths of several characteristic lines from the samples under matched conditions. Shared positions support a common species; brightness alone does not identify it.

(a) With N_e=Z-q (D), oxygen gives 8-(-2)=10, sodium 11-1=10 and magnesium 12-2=10. They are isoelectronic. (b) Ten electrons fill 1s^2 2s^2 2p^6. (c) Neutral oxygen has eight electrons, 1s^2 2s^2 2p^4. Gaining two electrons completes 2p; its nucleus retains eight protons. Identical electron counts do not imply identical nuclei.

(a) Within shell 3 the sublevel ordering is 3s<3p<3d. (b) Place one parallel-spin electron in each p orbital before pairing: [↑↓][↑][↑]. The draft pairs while leaving an equal-energy orbital empty, contrary to Hund's rule. (c) The full configuration is 1s^2 2s^2 2p^6 3s^2 3p^4. Count occupied orbitals as 1+1+3+1+3=9 (D). The two singly occupied 3p orbitals contain the two unpaired electrons. Reversing all spin arrows gives an equivalent diagram.

(a) \mathrm{Al}^{+}(g)\to\mathrm{Al}^{2+}(g)+e^- (R). The second removal starts with a gaseous singly charged ion and removes one electron per ion; molar IE refers to a mole of these events. (b) Thirteen electrons give [\mathrm{Ne}]3s^2 3p^1. (c) The first three removals exhaust shell 3. The fourth removes a 2p core electron from closer to the nucleus with less shielding, so the large shell-boundary rise is from IE_3 to IE_4.

(a) The four positive gaps are 0.488, 4.578, 1.06 and 3.03, in 10^{-19}\,\mathrm{J} (R). The smallest gap, 5\to4, gives the longest wavelength. (b) \lambda=hc/E (D) gives (6.63\times10^{-34})(3.00\times10^8)/(0.488\times10^{-19})=4.07582\times10^{-6}\,\mathrm{m}, or 4.08\times10^{-6}\,\mathrm{m}. (c) The cascade releases (0.488+1.06+3.03)\times10^{-19}=4.578\times10^{-19}\,\mathrm{J} (D), equal to the direct endpoint difference. The cascade emits three photons, each with its own gap and wavelength; the direct route emits one photon with the combined gap.

(a) A normal emission line measures a difference between two bound energies. The limiting ground-series frequency measures the gap between ground and unbound states. (b) E=hf (B) gives (6.63\times10^{-34})(1.85\times10^{15})=1.22655\times10^{-18}\,\mathrm{J}, or 1.23\times10^{-18}\,\mathrm{J}. (c) Multiply the unrounded joules per atom by 6.02\times10^{23}\,\mathrm{mol^{-1}} and convert J to kJ (D). The result is 738\,\mathrm{kJ\,mol^{-1}}, to three significant figures. (d) The remaining escape gap is 1.22655\times10^{-18}-2.40\times10^{-19}=9.8655\times10^{-19}\,\mathrm{J} (D). Thus \lambda_{\max}=hc/E=2.01612\times10^{-7}\,\mathrm{m}=202\,\mathrm{nm} (D). The ground-state threshold is c/f=162\,\mathrm{nm}. Excitation reduces the required energy, so a longer wavelength is sufficient.

(a) The jump from 4510 to 21400 occurs at the fifth removal. Four electrons are removed before the inner shell is reached. (b) Four main-group outer electrons mean group 14; period 3 then identifies silicon. (c) Its condensed ground state is [\mathrm{Ne}]3s^2 3p^2. The listed energies are illustrative model inputs, not measured silicon ionization energies. (d) Sum the first four IEs: 820+1650+3300+4510=10280\,\mathrm{kJ\,mol^{-1}}. For the stated amount, Q=0.0150(10280)=154.2\,\mathrm{kJ}, or 154\,\mathrm{kJ} to three significant figures (D). This is only gaseous cation formation. Oxide formation also involves oxygen species, electron gain and lattice formation, so these removals alone cannot give its total energy change.

(a) Convert to energy per atom: E=906\times1000/(6.02\times10^{23})=1.50498\times10^{-18}\,\mathrm{J} (D). Then \lambda=hc/E=1.32161\times10^{-7}\,\mathrm{m}=132\,\mathrm{nm} (D). (b) A 170\,\mathrm{nm} photon has energy hc/\lambda=1.17\times10^{-18}\,\mathrm{J}, equivalent to 704\,\mathrm{kJ\,mol^{-1}} (D), below 906\,\mathrm{kJ\,mol^{-1}}. It cannot free a ground-state electron. Absorption into a bound state additionally requires a matching allowed gap. (c) Longer limiting wavelength means a smaller positive gap to the same unbound state. Its lower level is therefore above the ground state. First IE includes the missing ground-to-excited gap, so the excited-series limit alone is insufficient.

(a) All written totals are correct, but the neutral chromium ground state is the observed exception [\mathrm{Ar}]3d^5 4s^1. Both ion records are valid: Cr loses 4s first and then one 3d electron, whereas Cu loses its single 4s electron. (b) Distribute four Cr^{2+} d electrons separately with parallel spins, [↑][↑][↑][↑][\ ], giving four unpaired electrons. Cu^+ has [↑↓][↑↓][↑↓][↑↓][↑↓], so none are unpaired. (c) Capture reduces the charge to +1 and raises the electron count from 22 to 23. The relaxed Cr^+ ground state is [\mathrm{Ar}]3d^5: all five d orbitals are singly occupied, giving five unpaired electrons. It is the one-electron-loss product of the correct neutral Cr configuration; a neutral-atom initial filling ladder alone cannot choose its arrangement.

(a) Proton number is Z, neutron number is A-Z (D). Magnesium has 12 protons and 25-12=13 neutrons; fluorine has 9 protons and 19-9=10 neutrons. (b) Mg loses its two 3s electrons to form \mathrm{Mg}^{2+}; F gains one 2p electron to form \mathrm{F}^{-}. Both contain ten electrons, 1s^2 2s^2 2p^6. (c) One 2+ ion balances two 1- ions, giving \mathrm{MgF}_2 (D). On melting, the ions become mobile and carry charge through the liquid.

(a) E=hc/\lambda (D) gives (6.63\times10^{-34})(3.00\times10^8)/(480\times10^{-9})=4.14375\times10^{-19}\,\mathrm{J}. (b) N=nN_A (B) gives 0.0200(6.02\times10^{23})=1.204\times10^{22} photons. Total NE=4989.075\,\mathrm{J}, or 4.99\times10^3\,\mathrm{J} (D). (c) Absorbed heat is 0.650(4989.075)=3242.89875\,\mathrm{J}. Using Q=C\Delta T (D), \Delta T=3242.89875/18.0=180\,\mathrm{K} to three significant figures. (d) The atom ends at a lower energy: \Delta E_{\rm atom}=-E_\gamma (R). The positive photon energy is the energy transferred out.

(a) The dominant jump is between the second and third removals, 7420/1190=6.24 compared with 1190/610=1.95 (D). Two outer electrons are removed before reaching the core, so the model is group 2. (b) Period 3 group 2 gives [\mathrm{Ne}]3s^2. Its usual \mathrm{M}^{2+} ion balances one \mathrm{O}^{2-} ion, giving MO. (c) The next element is aluminium. Both \mathrm{Mg}^{2+} and \mathrm{Al}^{3+} have ten electrons. The aluminium ion has one more proton with comparable shielding, so the stronger attraction contracts its electron cloud and its radius is smaller.