IB Chemistry S2.1: The ionic model worked solutions

There is one fewer negative charge and the nucleus is unchanged.

B changes the nucleus; C reverses the sign of electron loss; D removes a proton as well as an electron.

Adding two electrons fills the 3\mathrm{p} sublevel and gives \mathrm{S}^{2-}. Losing electrons gives a positive ion; changing proton number changes the element.

One ion of each type gives +2-2=0, so the empirical formula is \mathrm{SrO}. A has excess positive charge; C has excess negative charge; D is neutral but fails to reduce the ratio.

Two \mathrm{OH}^{-} ions balance \mathrm{Mg}^{2+}; two \mathrm{NH}_{4}^{+} ions balance \mathrm{CO}_{3}^{2-}. The subscript outside parentheses multiplies the whole ion; \mathrm{NH}_{3} is not ammonium.

Each chlorine is -1, so neutrality gives vanadium +3; the Roman numeral is III. [1] The letter V is the vanadium symbol and does not imply oxidation state +5.

Melting frees both types of ions to carry charge. The ions already exist in the solid; melting does not create a metallic sea of electrons.

Solvent–ion interactions must compete with the lattice attractions, so water solubility varies between ionic compounds. The tempting generalization that every salt is water-soluble is false.

The \mathrm{O{-}H} bond inside hydroxide is covalent. Lithium cations and hydroxide anions attract one another ionically. A exchanges the two locations; B overlooks the shared pair inside hydroxide; D overlooks the separate ions. [1]

(a) \mathrm{RbF}: 152+133=285\,\mathrm{pm} [1]. \mathrm{CsF}: 167+133=300\,\mathrm{pm} [1].

(b) \mathrm{RbF} has the larger value [1]. The charges are equal but the smaller separation gives stronger opposite-charge attraction, so more energy is needed to separate the ions [1].

(a) The extended lattice contains strong attractions between \mathrm{Ca}^{2+} and \mathrm{O}^{2-} ions [1]. A large energy input is needed to overcome enough of these attractions for the lattice to melt [1].

(b) Particles must overcome strong lattice attractions to leave the solid [1]. At ordinary temperatures very few have sufficient energy, giving low volatility [1].

(a) Ammonium is \ce{NH4+} [1]; phosphate is \ce{PO4^3-} [1]. (b) Two nitrate ions balance one \ce{Mg^2+}: +2+2(-1)=0, so \ce{Mg(NO3)2} [1]. Brackets preserve each nitrate group.

(a) P loses negative charge and becomes positive; Q gains negative charge and becomes negative [1 for both]. (b) The force is electrostatic attraction [1] between the positive and negative ions [1]. Transfer produces the ions; it does not define the force holding the solid together.

(a) Select ammonium chloride [1]. (b) Its constituent groups are \ce{NH4+} and \ce{Cl-} [1], which attract electrostatically [1]. N–H bonds within ammonium are covalent. Silicon dioxide is a covalent network and methanol consists of molecules.

(a) Three nitrate ions contribute 3(-1)=-3, so M must be 3+ [1]. Neutral water contributes zero charge [1]. (b) One \ce{M^3+} balances one \ce{PO4^3-}, giving \ce{MPO4} [1].

(a) \ce{CaF2(s) -> Ca^2+(g) + 2F-(g)}: correct one-to-two ion ratio [1] and solid-to-gaseous-ion states [1]. (b) The enthalpy change is positive because energy must be supplied against the attractions [1]. Aqueous ions would describe a different process.

(a) V has stronger attraction [1] because its oppositely charged ions have greater charge magnitudes [1]. (b) Hold centre-to-centre separation constant (or hold lattice arrangement constant) [1]. No exact melting-point multiplier follows from this qualitative comparison.

(a) The metal is in group 2 [1] and forms \mathrm{X}^{2+} [1]. The largest jump is from the second to third removal: after two outer electrons are removed, the next electron is from a more tightly bound inner shell [1].

(b) \mathrm{X}_{3}(\mathrm{PO}_{4})_{2} [1]. Three 2+ ions balance two 3- ions.

(c) It will be a poor conductor as a dry solid [1], because ions occupy fixed positions in its crystalline lattice and cannot migrate [1]. Neutrality check: 3(+2)+2(-3)=0.

(a) \mathrm{NH}_{4}^{+} [1]; \mathrm{HCO}_{3}^{-} [1].

(b) The nitrogen–hydrogen bonds are covalent [1]. Shared electron pairs are attracted to the positive nuclei of nitrogen and hydrogen [1].

(c) The polyatomic groups carry opposite charges [1]. Electrostatic attraction between these charged groups binds the crystal [1]. Internal covalent bonds do not cancel the overall ionic charges.

(a) R > P [1] and P > Q [1]. (b) At equal separation, R has greater ionic charges and stronger attraction than P [1]. At equal charges, P has a smaller separation and stronger attraction than Q [1]. (c) Both charge and separation change, and the arrangement changes too [1]. Without a quantitative lattice model or measured data, those competing effects do not establish a unique position [1].

(a) Three sulfate ions contribute -6; two M ions must contribute +6, so each is 3+ [1]. Each ion has 26-3=23 electrons [1]. (b) Proton number 26 identifies iron; three chloride ions balance each iron ion, giving \ce{FeCl3} [1], iron(III) chloride [1]. (c) Shifting lattice planes can place like charges together; repulsion causes fracture [1]. After melting, charged ions can migrate and carry current [1].

(a) M loses three electrons to give \ce{M^3+} [1]; oxygen gains two to give \ce{O^2-} [1]. (b) 2(+3)+3(-2)=0, hence \ce{M2O3} [1]. There are two moles of cations per mole of oxide: 2(0.120)=0.240\,\mathrm{mol} [1]. (c) Each formula unit involves transfer of six electrons [1], so the amount transferred is 6(0.120)=0.720\,\mathrm{mol} [1]. Count transferred electrons once, rather than adding the same loss and gain.

(a) The melt contains charged ions [1] that can migrate [1]; solid lattice positions prevent this migration. (b) At the cathode, \ce{Ca^2+ + 2e- -> Ca} [1]. At the anode, \ce{2Cl- -> Cl2 + 2e-} [1]. (c) Both equations involve two electrons per mole of product, so calcium and chlorine form in a 1:1 mole ratio [1]. Thus n(\ce{Cl2})=0.0500\,\mathrm{mol} [1].

(a) \ce{(NH4)2SO4} [1]; covalent bonds join atoms inside the polyatomic ions, whereas ionic attraction acts between the charged groups [1]. (b) In the solid, ions occupy fixed lattice positions [1]. In solution, dissolved ions migrate and carry current [1]. (c) The partially negative oxygen ends face ammonium cations, while partially positive hydrogen ends face sulfate anions [1 for both]. Attractive ion–dipole interactions stabilise separated ions [1]. This explains a contribution to dissolution, not a universal solubility rule.