Interleaved — Phase 2

Chemistry interleaved practice

printable — key stays hidden on paper

Instructions: Work each problem on its own. These are deliberately mixed — decide the correct concept/method before computing. Use $h = 6.626\times10^{-34}\,\text{J·s}$ , $c = 3.00\times10^8\,\text{m/s}$ , $m_e = 9.11\times10^{-31}\,\text{kg}$ , $N_A = 6.022\times10^{23}$ . Show all working. Total: 50 marks.

1. (5 marks) A metal has work function $\phi = 3.20\times10^{-19}\,\text{J}$ . Light of wavelength $400\,\text{nm}$ strikes it. Calculate (a) whether electrons are ejected, and (b) the maximum kinetic energy of the ejected electrons in joules.

2. (5 marks) Write the ground-state electronic configuration of $\text{Cr}$ ( $Z=24$ ) and $\text{Cu}$ ( $Z=29$ ). Explain in one sentence each why these deviate from the naive Aufbau prediction.

3. (6 marks) Using Slater's rules, calculate the effective nuclear charge $Z_{\text{eff}}$ experienced by a $3p$ electron in a phosphorus atom ( $Z=15$ ).

4. (5 marks) An electron is confined to a region of width $\Delta x = 1.0\times10^{-10}\,\text{m}$ . Calculate the minimum uncertainty in its velocity. Use $\hbar = 1.055\times10^{-34}\,\text{J·s}$ .

5. (4 marks) Arrange the following isoelectronic species in order of increasing ionic radius, and justify: $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ .

6. (5 marks) Calculate the de Broglie wavelength of an electron accelerated so that it moves at $2.19\times10^6\,\text{m/s}$ .

7. (4 marks) State the set of four quantum numbers $(n, l, m_l, m_s)$ for the last electron added to a nitrogen atom ( $Z=7$ ). Which rule dictates its spin choice, and why?

8. (5 marks) Explain, using Fajan's rules, why $\text{AlCl}_3$ is more covalent than $\text{AlF}_3$ , and why $\text{LiI}$ is more covalent than $\text{LiF}$ .

9. (3 marks) The first ionization energy of boron ( $Z=5$ ) is lower than that of beryllium ( $Z=4$ ). Explain this anomaly.

10. (3 marks) Draw the Lewis structure of the sulfate ion $\text{SO}_4^{2-}$ and compute the formal charge on sulfur in the structure with four S–O single bonds versus the structure with two S=O double bonds. Which is preferred?

Answer keyMark scheme & solutions

1. (Photoelectric effect — 2.1.2) Photon energy $E = hc/\lambda = \dfrac{(6.626\times10^{-34})(3.00\times10^8)}{400\times10^{-9}} = 4.97\times10^{-19}\,\text{J}$ . Since $E > \phi$ , electrons are ejected. $KE_{\max} = E - \phi = 4.97\times10^{-19} - 3.20\times10^{-19} = 1.77\times10^{-19}\,\text{J}$ . Why this method: wavelength → photon energy first; the threshold check is $E$ vs $\phi$ , not $\lambda$ .

2. (Config exceptions Cr, Cu — 2.1.10 / 2.1.11) $\text{Cr}: [\text{Ar}]\,3d^5\,4s^1$ ; $\text{Cu}: [\text{Ar}]\,3d^{10}\,4s^1$ . A half-filled ( $d^5$ ) and fully-filled ( $d^{10}$ ) subshell gives extra exchange-energy stability, so one $4s$ electron shifts to $3d$ .

3. (Slater's rules — 2.2.1) P config: $1s^2\,2s^2 2p^6\,3s^2 3p^3$ . For a $3p$ electron:

Other electrons in $(3s,3p)$ group: $4$ electrons × $0.35 = 1.40$
$n=2$ shell (8 electrons) × $0.85 = 6.80$
$n=1$ shell (2 electrons) × $1.00 = 2.00$ Total screening $S = 1.40 + 6.80 + 2.00 = 10.20$ . $Z_{\text{eff}} = 15 - 10.20 = \mathbf{4.80}$ .

4. (Uncertainty principle — 2.1.4) $\Delta p \ge \dfrac{\hbar}{2\Delta x} = \dfrac{1.055\times10^{-34}}{2(1.0\times10^{-10})} = 5.28\times10^{-25}\,\text{kg·m/s}$ . $\Delta v = \Delta p/m_e = \dfrac{5.28\times10^{-25}}{9.11\times10^{-31}} = \mathbf{5.79\times10^{5}\,\text{m/s}}$ . Why: "confined to width" → position uncertainty → use $\Delta x\,\Delta p \ge \hbar/2$ .

5. (Isoelectronic series — 2.2.3) All have 10 electrons. Radius decreases as nuclear charge increases: $\text{Mg}^{2+}(12) < \text{Na}^+(11) < \text{F}^-(9) < \text{O}^{2-}(8)$ Increasing radius: $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-}$ .

6. (de Broglie — 2.1.3) $\lambda = \dfrac{h}{mv} = \dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(2.19\times10^6)} = \mathbf{3.32\times10^{-10}\,\text{m}}$ (≈ 332 pm). Why: "moving particle, wavelength" → $\lambda = h/p$ , distinct from photon $E=hc/\lambda$ .

7. (Quantum numbers + Hund — 2.1.5 / 2.1.9) N: $1s^2 2s^2 2p^3$ . Last electron enters a third, separate $2p$ orbital. One valid set: $n=2,\ l=1,\ m_l=+1,\ m_s=+\tfrac12$ . Hund's rule dictates the same spin ( $+\tfrac12$ ) as the other two $2p$ electrons to maximize multiplicity (minimize pairing repulsion).

8. (Fajan's rules — 2.3.4) Covalent character rises with (i) high cation charge/small size (polarizing power) and (ii) large, polarizable anion.

$\text{AlCl}_3$ vs $\text{AlF}_3$ : $\text{Cl}^-$ is larger/more polarizable than $\text{F}^-$ → $\text{AlCl}_3$ more covalent.
$\text{LiI}$ vs $\text{LiF}$ : $\text{I}^-$ far more polarizable than $\text{F}^-$ → $\text{LiI}$ more covalent.

9. (IE anomaly B < Be — 2.2.4) Be is $2s^2$ (stable filled subshell); B is $2s^2 2p^1$ . The $2p$ electron is higher in energy and better shielded, so it is easier to remove — hence $IE_1(\text{B}) < IE_1(\text{Be})$ .

10. (Formal charge / octet — 2.3.1 / 2.3.2) FC $= (\text{valence}) - (\text{lone}) - \tfrac12(\text{bonding})$ .

All single bonds: S has 4 bonds → FC $_S = 6 - 0 - \tfrac12(8) = +2$ .
Two double + two single bonds (expanded octet): S → FC $_S = 6 - 0 - \tfrac12(12) = 0$ . The structure with two S=O double bonds (FC $_S=0$ ) minimizes formal charges and is preferred.

[
  {"claim":"Photoelectric KE_max = 1.77e-19 J","code":"h=6.626e-34; c=3.00e8; lam=400e-9; phi=3.20e-19; E=h*c/lam; KE=E-phi; result = abs(KE-1.77e-19) < 2e-21"},
  {"claim":"Slater Z_eff for 3p in P = 4.80","code":"S = 4*0.35 + 8*0.85 + 2*1.00; Zeff = 15 - S; result = abs(Zeff-4.80) < 1e-9"},
  {"claim":"de Broglie lambda = 3.32e-10 m","code":"h=6.626e-34; m=9.11e-31; v=2.19e6; lam=h/(m*v); result = abs(lam-3.32e-10) < 5e-12"}
]