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non-bonding electron pair to make both electrons in the new bond, so the resulting metal-ligand linkage is a dative covalent bond."},{"text":"A ligand donates a single electron to a metal ion to form a dative covalent bond.","correct":false,"explanation":"A ligand donates two electrons as a lone pair, not a single electron. Both bonding electrons come from the ligand, producing a dative covalent bond."},{"text":"A ligand donates a lone pair to a metal ion to form an ionic bond.","correct":false,"explanation":"Electron transfer would produce ions rather than the characteristic metal-ligand linkage. A ligand donates a lone pair to form a dative covalent bond."},{"text":"A ligand donates a bonding pair to a metal ion to form a non-polar covalent bond.","correct":false,"explanation":"A shared pair already involved in another covalent bond is not donated in ordinary ligand bonding. The ligand supplies an available lone pair to form a dative covalent bond."}]},{"id":"6591d7bc-3756-4ef4-903b-00b188c97f25","cardType":"cloze","points":null,"front":"Ligands are [ ] because they donate electron pairs; metal ions are [ ] because they accept them.","back":"Ligands are Lewis bases because they donate electron pairs; metal ions are Lewis acids because they accept them.","mcqOptions":[{"text":"Ligands are Lewis bases because they donate electron pairs; metal ions are Lewis acids because they accept them.","correct":true,"explanation":"Lewis bases donate electron pairs, whereas Lewis acids accept electron pairs. Ligands and metal ions therefore act as Lewis bases and Lewis acids respectively during complex formation."},{"text":"Ligands are Lewis acids because they donate electron pairs; metal ions are Lewis bases because they accept them.","correct":false,"explanation":"This reverses the Lewis definitions. Electron-pair donors are Lewis bases, so ligands are Lewis bases, while electron-pair-accepting metal ions are Lewis acids."},{"text":"Ligands are Brønsted-Lowry bases because they donate electron pairs; metal ions are Brønsted-Lowry acids because they accept them.","correct":false,"explanation":"Brønsted-Lowry theory concerns proton transfer, not electron-pair donation and acceptance. The stated behaviour identifies ligands as Lewis bases and metal ions as Lewis acids."},{"text":"Ligands are reducing agents because they donate electron pairs; metal ions are oxidising agents because they accept them.","correct":false,"explanation":"Oxidising and reducing agents are classified by electron transfer rather than electron-pair bonding. Ligands donate pairs as Lewis bases, and metal ions accept them as Lewis acids."}]},{"id":"229d63ba-2030-48e3-9093-5330e8d5c045","cardType":"cloze","points":null,"front":"In oxidation-number calculations, H₂O and NH₃ ligands are [ ], while Cl^-, OH^- and CN^- have charge [ ].","back":"In oxidation-number calculations, H₂O and NH₃ ligands are neutral, while Cl^-, OH^- and CN^- have charge -1.","mcqOptions":[{"text":"In oxidation-number calculations, H₂O and NH₃ ligands are neutral, while Cl^-, OH^- and CN^- have charge -1.","correct":true,"explanation":"Water and ammonia have no overall charge, whereas chloride, hydroxide and cyanide are all singly charged anions. Their ligand charges are therefore zero and -1 respectively."},{"text":"In oxidation-number calculations, H₂O and NH₃ ligands are neutral, while Cl^-, OH^- and CN^- have charge 0.","correct":false,"explanation":"Water and ammonia remain uncharged when acting as ligands, but chloride, hydroxide and cyanide each carry one negative charge. Their charges are not all zero."},{"text":"In oxidation-number calculations, H₂O and NH₃ ligands are positively charged, while Cl^-, OH^- and CN^- have charge -1.","correct":false,"explanation":"Coordinating to a metal does not make water and ammonia positively charged in these complexes. They are neutral ligands, while chloride, hydroxide and cyanide each have charge -1."},{"text":"In oxidation-number calculations, H₂O and NH₃ ligands are negatively charged, while Cl^-, OH^- and CN^- have charge +1.","correct":false,"explanation":"The negative charge symbols on chloride, hydroxide and cyanide must not be reversed. Water and ammonia are neutral, while each of those anionic ligands has charge -1."}]},{"id":"cbf38b7d-52a0-484f-a013-7ee889929e55","cardType":"cloze","points":null,"front":"The coordination number is the number of [ ], not always the number of [ ].","back":"The coordination number is the number of coordinate bonds, not always the number of ligand molecules.","mcqOptions":[{"text":"The coordination number is the number of coordinate bonds, not always the number of ligand molecules.","correct":true,"explanation":"Coordination number counts the coordinate bonds to the central metal ion. A multidentate ligand forms several such bonds, so ligand count and coordination number can differ."},{"text":"The coordination number is the number of ligand molecules, not always the number of coordinate bonds.","correct":false,"explanation":"This reverses the distinction: coordination number counts coordinate bonds, not ligand molecules. One multidentate ligand molecule can donate several lone pairs and make several bonds."},{"text":"The coordination number is the number of oxidation states, not always the number of ligand molecules.","correct":false,"explanation":"Oxidation number represents a formal charge assignment and does not count attachments to the metal. Coordination number counts coordinate bonds, which may exceed the number of ligand molecules."},{"text":"The coordination number is the number of metal ions, not always the number of donor atoms.","correct":false,"explanation":"A complex normally contains one central metal ion, so counting metal ions does not describe coordination. Coordination number instead counts coordinate bonds formed with ligand donor atoms."}]},{"id":"ae2ed0ec-bd57-42f0-a729-73689d83c213","cardType":"basic","points":1,"front":"What is the oxidation number of chromium in [Cr(H₂O)₄Cl₂]⁺?","back":"+3.","mcqOptions":[{"text":"+3.","correct":true,"explanation":"The four neutral water ligands contribute zero and two chloride ligands contribute -2. Since the complex is +1 overall, chromium must have oxidation number +3."},{"text":"+1.","correct":false,"explanation":"This treats the complex charge as chromium’s oxidation number and ignores the two chloride ligands. Accounting for their total -2 charge gives chromium an oxidation number of +3."},{"text":"+2.","correct":false,"explanation":"This typically results from counting only one chloride ion. Both chloride ligands contribute -1, so chromium must be +3 to produce the complex’s overall +1 charge."},{"text":"+4.","correct":false,"explanation":"This incorrectly treats the neutral water ligands as affecting the charge or overcounts chloride. With two -1 chlorides and an overall +1 charge, chromium is +3."}]},{"id":"3169fe02-624e-4282-9291-82543c3273c1","cardType":"basic","points":1,"front":"What is the coordination number of [Cr(H₂O)₄Cl₂]⁺?","back":"6.","mcqOptions":[{"text":"6.","correct":true,"explanation":"Each of the four monodentate water ligands and two monodentate chloride ligands forms one coordinate bond. The total number of coordinate bonds is therefore 6."},{"text":"4.","correct":false,"explanation":"This counts only the four water ligands and omits both chloride ligands. All six ligands are monodentate, so each contributes one to a coordination number of 6."},{"text":"2.","correct":false,"explanation":"This counts only the two chloride ligands and ignores the four water ligands. Every ligand shown forms one coordinate bond, giving a coordination number of 6."},{"text":"3.","correct":false,"explanation":"This likely confuses the coordination number with chromium’s +3 oxidation number. Coordination number counts the four water and two chloride coordinate bonds, giving 6."}]},{"id":"df132483-c8ae-4166-ac3f-8025d8003f87","cardType":"cloze","points":null,"front":"A coordination number of 6 usually gives an [ ] complex with bond angles of [ ].","back":"A coordination number of 6 usually gives an octahedral complex with bond angles of 90°.","mcqOptions":[{"text":"A coordination number of 6 usually gives an octahedral complex with bond angles of 90°.","correct":true,"explanation":"Six coordinate bonds usually arrange themselves towards the vertices of an octahedron. Adjacent bonds are at 90°, although opposite ligand-metal-ligand bonds are at 180°."},{"text":"A coordination number of 6 usually gives an octahedral complex with bond angles of 180°.","correct":false,"explanation":"An octahedral complex does contain 180° angles between opposite ligands, but its characteristic adjacent bond angle is 90°. Coordination number 6 therefore usually gives an octahedral, 90° arrangement."},{"text":"A coordination number of 6 usually gives a trigonal planar complex with bond angles of 120°.","correct":false,"explanation":"Trigonal planar geometry has coordination number 3 and bond angles of 120°, not coordination number 6. Six-coordinate complexes are usually octahedral with 90° angles."},{"text":"A coordination number of 6 usually gives a tetrahedral complex with bond angles of 109.5°.","correct":false,"explanation":"Tetrahedral geometry accommodates four coordinate bonds and has angles of 109.5°. A six-coordinate complex is usually octahedral, with adjacent bonds separated by 90°."}]},{"id":"cb173d27-da79-4722-9c6a-8575ad3483db","cardType":"cloze","points":null,"front":"A coordination number of 4 is often [ ] with bond angles of [ ].","back":"A coordination number of 4 is often tetrahedral with bond angles of 109.5°.","mcqOptions":[{"text":"A coordination number of 4 is often tetrahedral with bond angles of 109.5°.","correct":true,"explanation":"Four coordinate bonds often point towards the corners of a tetrahedron to minimise repulsion. The ideal angle between these bonds is approximately 109.5°."},{"text":"A coordination number of 4 is often tetrahedral with bond angles of 90°.","correct":false,"explanation":"A tetrahedral arrangement does not contain 90° bond angles; its ideal angle is 109.5°. Some four-coordinate complexes are square planar with 90° angles, but that is a different shape."},{"text":"A coordination number of 4 is often square planar with bond angles of 109.5°.","correct":false,"explanation":"Square planar complexes have 90° angles, not 109.5°. The usual four-coordinate geometry described here is tetrahedral, for which the ideal bond angle is 109.5°."},{"text":"A coordination number of 4 is often octahedral with bond angles of 90°.","correct":false,"explanation":"Octahedral geometry requires six coordinate bonds rather than four. Four-coordinate complexes are often tetrahedral, with ideal bond angles of approximately 109.5°."}]},{"id":"28126869-6a33-41d1-9b19-998c6788304e","cardType":"basic","points":1,"front":"Three bidentate ligands are bonded to one metal ion. What is the coordination number?","back":"6.","mcqOptions":[{"text":"6.","correct":true,"explanation":"Each bidentate ligand forms two coordinate bonds to the metal ion. Three such ligands therefore provide six coordinate bonds, giving a coordination number of 6."},{"text":"3.","correct":false,"explanation":"This counts the number of ligands rather than the number of coordinate bonds. Each of the three bidentate ligands forms two bonds, so the coordination number is 6."},{"text":"2.","correct":false,"explanation":"This is the denticity of each individual ligand, not the coordination number of the complex. Three bidentate ligands each form two bonds, giving a coordination number of 6."},{"text":"9.","correct":false,"explanation":"This results from incorrectly combining the number of ligands and their denticity by multiplication and addition. Three ligands forming two bonds each give a coordination number of 6."}]},{"id":"e6c9d52d-76d4-4209-a380-eea1ddce2b7c","cardType":"cloze","points":null,"front":"Ligand substitution is replacement of [ ] by [ ] in a complex ion.","back":"Ligand substitution is replacement of one ligand by another ligand in a complex ion.","mcqOptions":[{"text":"Ligand substitution is replacement of one ligand by another ligand in a complex ion.","correct":true,"explanation":"Ligand substitution occurs when a ligand attached to the central metal ion is displaced and another ligand takes its place, forming new coordinate bonds."},{"text":"Ligand substitution is replacement of one metal ion by another metal ion in a complex ion.","correct":false,"explanation":"Replacing the central metal ion would change the identity of the complex rather than constitute ligand substitution. Ligand substitution replaces one ligand with another ligand."},{"text":"Ligand substitution is replacement of one coordinate bond by another coordinate bond in a complex ion.","correct":false,"explanation":"Coordinate bonds are broken and formed during the process, but they are not the species being substituted. The definition is replacement of one ligand by another ligand."},{"text":"Ligand substitution is replacement of one complex ion by another complex ion in a complex ion.","correct":false,"explanation":"A new complex ion may be produced, but one complex ion is not contained and replaced within another. Ligand substitution specifically replaces one ligand with another ligand."}]},{"id":"b3417e99-df3b-4ed9-9add-49aaa047e023","cardType":"basic","points":1,"front":"What happens if a reversible substitution equilibrium is given a high concentration of the incoming ligand?","back":"Equilibrium shifts towards the substituted complex.","mcqOptions":[{"text":"Equilibrium shifts towards the substituted complex.","correct":true,"explanation":"Increasing the concentration of the incoming ligand causes the equilibrium to oppose that change by consuming it, so more of the substituted complex is formed."},{"text":"Equilibrium shifts towards the original complex.","correct":false,"explanation":"This reverses Le Chatelier’s principle. A high concentration of the incoming ligand is opposed by its consumption, so equilibrium shifts towards the substituted complex."},{"text":"Equilibrium shifts towards the free metal ion.","correct":false,"explanation":"Ligand substitution exchanges ligands without normally releasing the central metal as a free ion. Excess incoming ligand drives formation of the substituted complex."},{"text":"Equilibrium shifts towards the unbound ligands.","correct":false,"explanation":"A high incoming-ligand concentration does not favour further accumulation of unbound ligand. The equilibrium shifts to consume it by forming the substituted complex."}]},{"id":"1551e461-00ae-4c6a-9e31-2fc1ef551829","cardType":"basic","points":1,"front":"What complex forms when Cl^- replaces H₂O ligands in [Co(H₂O)₆]²⁺?","back":"[CoCl₄]^2-.","mcqOptions":[{"text":"[CoCl₄]^2-.","correct":true,"explanation":"Four chloride ligands coordinate to Co²⁺. Combining the +2 metal-ion charge with four -1 chloride charges gives the tetrahedral complex an overall charge of 2-."},{"text":"[CoCl₆]^4-.","correct":false,"explanation":"This incorrectly assumes that six chloride ions replace the six water molecules while preserving coordination number six. Co²⁺ forms the four-coordinate complex [CoCl₄]^2- under these conditions."},{"text":"[CoCl₄]²⁺.","correct":false,"explanation":"This ignores the negative charges of the chloride ligands. Co²⁺ plus four Cl^- ligands gives an overall charge of 2-, producing [CoCl₄]^2-."},{"text":"[CoCl₂]^2-.","correct":false,"explanation":"This uses too few chloride ligands and assigns the charge incorrectly. The product contains four chloride ligands and has the formula [CoCl₄]^2-."}]},{"id":"b5fac664-3d45-45ac-b0bc-2e5708612f76","cardType":"basic","points":1,"front":"What colour change occurs when concentrated Cl^- is added to [Co(H₂O)₆]²⁺?","back":"Pink to blue.","mcqOptions":[{"text":"Pink to blue.","correct":true,"explanation":"The pink hexaaquacobalt(II) complex is converted into blue [CoCl₄]^2- when a high chloride concentration shifts the ligand-substitution equilibrium towards the chloride complex."},{"text":"Blue to pink.","correct":false,"explanation":"This is the change observed when water is added to the blue chloride complex. Adding concentrated chloride instead converts pink [Co(H₂O)₆]²⁺ into blue [CoCl₄]^2-."},{"text":"Pink to green.","correct":false,"explanation":"Green may be seen in mixtures or confused with other transition-metal ions, but the chloride-substituted cobalt(II) complex is blue, so the stated change is pink to blue."},{"text":"Blue to green.","correct":false,"explanation":"The starting hexaaquacobalt(II) complex is pink rather than blue. Concentrated chloride forms the blue tetrachlorocobaltate(II) complex, so the overall change is pink to blue."}]},{"id":"0c5ba856-5c76-4ad8-b636-9cb68f71713e","cardType":"basic","points":1,"front":"What shape is [CoCl₄]^2-?","back":"Tetrahedral.","mcqOptions":[{"text":"Tetrahedral.","correct":true,"explanation":"The complex contains four chloride ligands around Co²⁺. With these ligands, the four-coordinate [CoCl₄]^2- ion adopts a tetrahedral arrangement."},{"text":"Octahedral.","correct":false,"explanation":"Octahedral geometry usually has coordination number six, as in [Co(H₂O)₆]²⁺. [CoCl₄]^2- has four ligands and is tetrahedral."},{"text":"Square planar.","correct":false,"explanation":"Although square-planar complexes also have coordination number four, this geometry is associated with particular metal ions and ligand fields. [CoCl₄]^2- is tetrahedral."},{"text":"Trigonal planar.","correct":false,"explanation":"Trigonal-planar geometry requires three bonding groups around the central ion. [CoCl₄]^2- has four chloride ligands, which arrange tetrahedrally."}]},{"id":"4d7b6203-0ecc-446f-8207-f9c90e7eab04","cardType":"basic","points":2,"front":"When water is added to the [Co(H₂O)₆]²⁺ / [CoCl₄]^2- equilibrium, which side is favoured?","back":"The left side - pink [Co(H₂O)₆]²⁺.","mcqOptions":[{"text":"The left side - pink [Co(H₂O)₆]²⁺.","correct":true,"explanation":"Adding water increases a reactant’s concentration on the left and dilutes chloride ions. Both effects favour formation of the pink hexaaquacobalt(II) complex."},{"text":"The right side - blue [CoCl₄]^2-.","correct":false,"explanation":"The blue chloride complex is favoured by concentrated chloride, not added water. Water shifts the equilibrium left towards pink [Co(H₂O)₆]²⁺."},{"text":"The left side - blue [CoCl₄]^2-.","correct":false,"explanation":"This combines the correct direction with the wrong species and colour. The left side contains pink [Co(H₂O)₆]²⁺, which is favoured when water is added."},{"text":"The right side - pink [Co(H₂O)₆]²⁺.","correct":false,"explanation":"This identifies the favoured species but places it on the wrong side. Added water favours the left side, where pink [Co(H₂O)₆]²⁺ is present."}]},{"id":"06e810b3-2902-49c1-9dbe-53ef087ff892","cardType":"basic","points":1,"front":"In excess NH₃, how many water ligands in [Cu(H₂O)₆]²⁺ are replaced?","back":"Four.","mcqOptions":[{"text":"Four.","correct":true,"explanation":"In excess ammonia, four water ligands are replaced to form [Cu(NH₃)₄(H₂O)₂]²⁺. Two water ligands remain coordinated to the copper(II) ion."},{"text":"Six.","correct":false,"explanation":"This assumes complete replacement of every water ligand. Excess ammonia replaces four, forming [Cu(NH₃)₄(H₂O)₂]²⁺, so two coordinated water ligands remain."},{"text":"Two.","correct":false,"explanation":"This confuses the number of water ligands remaining with the number replaced. Two water ligands remain, meaning four of the original six are replaced by ammonia."},{"text":"One.","correct":false,"explanation":"This understates the extent of substitution in excess ammonia. Four water ligands are replaced, producing [Cu(NH₃)₄(H₂O)₂]²⁺ with two water ligands remaining."}]},{"id":"4aa023fa-73a8-49d0-aeb2-7f4a3a05dbfe","cardType":"basic","points":1,"front":"What colour is [Cu(NH₃)₄(H₂O)₂]²⁺?","back":"Deep blue.","mcqOptions":[{"text":"Deep blue.","correct":true,"explanation":"Excess ammonia replaces four water ligands around Cu²⁺, forming the characteristic deep-blue tetraammine complex [Cu(NH₃)₄(H₂O)₂]²⁺."},{"text":"Pale blue.","correct":false,"explanation":"Pale blue is associated with the aqueous Cu²⁺ ion and its hydroxide precipitate. The tetraammine complex formed in excess ammonia is deep blue."},{"text":"Yellow.","correct":false,"explanation":"Yellow is associated with [CuCl₄]^2-, formed in concentrated chloride solution. The copper(II) tetraammine complex is instead deep blue."},{"text":"Green.","correct":false,"explanation":"Green may appear when blue and yellow copper(II) species coexist during chloride substitution. Pure [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue."}]},{"id":"a8408e04-a8e2-4c5f-9510-e1e2afd8c65a","cardType":"basic","points":2,"front":"Why can adding NH₃ dropwise to Cu²⁺ first form a pale blue precipitate?","back":"

NH₃ acts as a weak base.
OH^- deprotonates the aqua complex.

","mcqOptions":[{"text":"

NH₃ acts as a weak base.
OH^- deprotonates the aqua complex.

","correct":true,"explanation":"Ammonia partially reacts with water as a weak base, producing OH^-. The hydroxide ions remove protons from coordinated water, leading to a pale-blue hydroxide precipitate."},{"text":"

NH₃ acts as a strong acid.
H⁺ protonates the aqua complex.

","correct":false,"explanation":"Ammonia is not acting as a strong acid and does not supply H⁺. It acts as a weak base, producing OH^- that deprotonates the aqua complex."},{"text":"

NH₃ acts as a reducing agent.
Cu²⁺ gains two electrons.

","correct":false,"explanation":"Initial precipitation is an acid-base reaction, not reduction of Cu²⁺. Ammonia acts as a weak base, and the resulting OH^- deprotonates coordinated water."},{"text":"

NH₃ acts as a bidentate ligand.
NH₃ replaces the aqua ligands.

","correct":false,"explanation":"Ammonia does eventually act as a ligand in excess, but ligand substitution produces the soluble deep-blue complex. The initial precipitate forms because ammonia generates OH^-."}]},{"id":"b518952c-0e1d-429c-b554-8773cc3b0b54","cardType":"basic","points":2,"front":"Why is the observed colour often green when concentrated Cl^- is added to a Cu²⁺ aqua complex?","back":"A mixture of pale blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]^2- is present.","mcqOptions":[{"text":"A mixture of pale blue [Cu(H₂O)₆]²⁺ and yellow [CuCl₄]^2- is present.","correct":true,"explanation":"Chloride substitution is reversible, so both the pale-blue aqua complex and yellow tetrachlorocuprate(II) ions can coexist. Their combined observed colours make the solution appear green."},{"text":"A single complex of green [CuCl₆]^4- is present after complete ligand substitution.","correct":false,"explanation":"The green appearance is not the colour of a single hexachlorocopper(II) species. It results from a mixture of pale-blue aqua and yellow tetrachlorocuprate(II) complexes."},{"text":"A mixture of pale blue [Cu(H₂O)₆]²⁺ and red [CuCl₄]^2- is present.","correct":false,"explanation":"The aqua complex is pale blue, but [CuCl₄]^2- is yellow rather than red. Mixing the actual blue and yellow species accounts for the observed green."},{"text":"A suspension of blue Cu(OH)₂ and yellow [CuCl₄]^2- is present.","correct":false,"explanation":"Concentrated chloride favours ligand substitution rather than precipitation of copper(II) hydroxide. The green colour comes from coexisting pale-blue aqua and yellow chloride complexes."}]},{"id":"ef2d226d-3946-4121-ae7d-d0be90bcd974","cardType":"basic","points":1,"front":"What makes ethane-1,2-diamine a bidentate ligand?","back":"It has two nitrogen donor atoms, each with a lone pair.","mcqOptions":[{"text":"It has two nitrogen donor atoms, each with a lone pair.","correct":true,"explanation":"Each nitrogen atom in ethane-1,2-diamine donates one lone pair to the same metal ion. Two donor atoms therefore allow the ligand to form two coordinate bonds."},{"text":"It has four hydrogen donor atoms, each with a lone pair.","correct":false,"explanation":"The ligand contains four hydrogen atoms attached to nitrogen, but these hydrogens do not donate lone pairs. Bidentate bonding occurs through its two nitrogen donor atoms."},{"text":"It has two carbon donor atoms, each with a lone pair.","correct":false,"explanation":"The two carbon atoms form the ligand’s backbone and do not provide the relevant lone pairs. Its two nitrogen atoms are the sites that bond to the metal."},{"text":"It has one nitrogen donor atom, with two lone pairs.","correct":false,"explanation":"Ethane-1,2-diamine has two, not one, nitrogen donor atoms. Both nitrogens donate lone pairs to one metal ion, which is why the ligand is bidentate."}]},{"id":"d7e50cc6-0be5-45e4-b4a8-eb662e00bcdb","cardType":"basic","points":1,"front":"What denticity is EDTA^4-?","back":"Hexadentate - six coordinate bonds to one metal ion.","mcqOptions":[{"text":"Hexadentate - six coordinate bonds to one metal ion.","correct":true,"explanation":"EDTA^4- can donate lone pairs through two nitrogen atoms and four oxygen atoms. It therefore forms six coordinate bonds to one metal ion and is hexadentate."},{"text":"Bidentate - two coordinate bonds to one metal ion.","correct":false,"explanation":"Bidentate ligands form only two coordinate bonds to one metal ion. EDTA^4- uses six donor atoms, so it is classified as hexadentate."},{"text":"Tetradentate - four coordinate bonds to one metal ion.","correct":false,"explanation":"Tetradentate counts only four donor sites, commonly overlooking EDTA’s two nitrogen atoms. EDTA^4- has six donor atoms and is therefore hexadentate."},{"text":"Octadentate - eight coordinate bonds to one metal ion.","correct":false,"explanation":"Octadentate would require eight donor atoms binding to the same metal ion. EDTA^4- supplies six effective donor atoms, making it hexadentate."}]},{"id":"9a29dd35-a60b-4381-a207-c9641855c4fe","cardType":"basic","points":2,"front":"Why is substituting 3en for 6H₂O often thermodynamically favourable?","back":"Dissolved particles increase from 4 to 7, so entropy increases.","mcqOptions":[{"text":"Dissolved particles increase from 4 to 7, so entropy increases.","correct":true,"explanation":"The reactant side contains one complex ion and three en molecules, while the product side contains one complex ion and six released waters. Thus particles increase from 4 to 7."},{"text":"Dissolved particles decrease from 7 to 4, so entropy decreases.","correct":false,"explanation":"This reverses the particle count. Three en molecules and one aqua complex become one chelate complex plus six water molecules, so particles increase from 4 to 7."},{"text":"Dissolved particles decrease from 6 to 3, so entropy decreases.","correct":false,"explanation":"This counts only the exchanged ligands and omits the complex ions. Including all dissolved species gives four reactant particles and seven product particles, increasing entropy."},{"text":"Dissolved particles increase from 3 to 6, so entropy increases.","correct":false,"explanation":"Counting ligand molecules alone gives three en molecules and six waters, but the complex ion on each side must also be included. The total therefore rises from 4 to 7."}]},{"id":"3e33859f-3327-4666-af97-634e802e263f","cardType":"basic","points":1,"front":"For $\\text{M}^{n+} + x\\text{L} \\rightleftharpoons \\text{ML}_x^{n+}$, what is $K_\\text{stab}$?","back":"$K_\\text{stab}$ $= \\frac{[\\text{ML}_x^{n+}]}{[\\text{M}^{n+}][\\text{L}]^x}$","mcqOptions":[{"text":"$K_\\text{stab}$ $= \\frac{[\\text{ML}_x^{n+}]}{[\\text{M}^{n+}][\\text{L}]^x}$","correct":true,"explanation":"The stability constant is the equilibrium concentration of the complex divided by those of the free metal ion and ligands, with the ligand concentration raised to its coefficient $x$."},{"text":"$K_\\text{stab}$ $= \\frac{[\\text{M}^{n+}][\\text{L}]^x}{[\\text{ML}_x^{n+}]}$","correct":false,"explanation":"This is the reciprocal of the stability constant and instead represents dissociation of the complex. For formation, the complex concentration must be in the numerator."},{"text":"$K_\\text{stab}$ $= \\frac{[\\text{ML}_x^{n+}]}{x[\\text{M}^{n+}][\\text{L}]}$","correct":false,"explanation":"The stoichiometric coefficient $x$ becomes an exponent on the ligand concentration, not a multiplying factor. The correct denominator contains $[\\text{L}]^x$."},{"text":"$K_\\text{stab}$ $= \\frac{[\\text{ML}_x^{n+}]}{[\\text{M}^{n+}]^x[\\text{L}]}$","correct":false,"explanation":"Only the ligand has coefficient $x$ in the stated equation, so only its concentration is raised to that power. The metal-ion concentration remains to the first power."}]},{"id":"c7c862ef-79e4-4dd0-a12f-ec060a6ffdc0","cardType":"basic","points":1,"front":"What does a larger $K_\\text{stab}$ mean for a complex?","back":"The complex is more stable; equilibrium lies further right.","mcqOptions":[{"text":"The complex is more stable; equilibrium lies further right.","correct":true,"explanation":"A larger stability constant means the equilibrium concentration of the complex is greater relative to free metal ions and ligands. Formation is therefore more favourable and equilibrium lies further right."},{"text":"The complex is less stable; equilibrium lies further left.","correct":false,"explanation":"This reverses the meaning of the stability constant. A larger value indicates that complex formation is more favourable, so the complex is more stable and equilibrium lies further right."},{"text":"The complex is more labile; equilibrium is reached more quickly.","correct":false,"explanation":"A stability constant describes the equilibrium position, not the reaction rate or kinetic lability. A larger value means greater thermodynamic stability and a position further to the right."},{"text":"The complex is less soluble; equilibrium lies further right.","correct":false,"explanation":"Solubility cannot be inferred directly from a complex’s stability constant. A larger $K_\\text{stab}$ specifically means the formation equilibrium favours the more stable complex."}]},{"id":"186f16dd-b041-4de4-acc5-e64bfbe4eff1","cardType":"basic","points":2,"front":"Why can changing ligands change the observed colour of a transition-metal complex?","back":"

Ligands change the d-orbital energy gap.
Different wavelengths of visible light are absorbed.

","mcqOptions":[{"text":"

Ligands change the d-orbital energy gap.
Different wavelengths of visible light are absorbed.

","correct":true,"explanation":"Different ligands cause different amounts of splitting between the transition metal’s d-orbitals. The complex therefore absorbs different wavelengths of visible light, so the colour observed changes."},{"text":"

Ligands change the number of d-electrons.
Different frequencies of visible light are emitted.

","correct":false,"explanation":"Changing ligands usually does not change the number of d-electrons, which depends mainly on the metal ion’s oxidation state. Ligands alter the d-orbital energy gap, changing which wavelengths are absorbed."},{"text":"

Ligands change the number of protons in the metal ion.
Different intensities of visible light are absorbed.

","correct":false,"explanation":"The number of protons is fixed by the identity of the metal and cannot be changed by ligand substitution. Ligands change the d-orbital energy gap, so different wavelengths, rather than simply intensities, are absorbed."},{"text":"

Ligands change the charge on the metal nucleus.
Different amounts of visible light are reflected.

","correct":false,"explanation":"Ligand substitution does not alter the nuclear charge of the metal. It changes d-orbital splitting, causing selective absorption of different visible wavelengths; the remaining transmitted or reflected light determines the observed colour."}]}]}],"examCountdown":{"examDate":"2027-05-11","paperLabel":"Paper 1"},"heading":"Complex ions and ligand substitution","title":"Complex ions and ligand substitution Flashcards","topbar":{"backHref":"/as-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"aab06e47-78c5-5a2f-9091-35b7bc6f3f32","walled":"$L39"}],"$L3a","$L3b"]