Zadanie 5b
Wzorowane na: Chemistry HL Paper 2, May 2024, TZ1, pytanie 5(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 5a-5b.
The table lists the successive ionization energies of an element, labelled E here as a code letter rather than its real symbol.
| Number of ionization energy | first | second | third | fourth | fifth | sixth | seventh |
|---|---|---|---|---|---|---|---|
| IE / kJ mol−1 | 1010 | 1900 | 2910 | 4960 | 6270 | 21 270 | 25 430 |
Polecenie
Element E forms the oxoanion EO43−. Two possible Lewis structures of the ion are shown, with lone pairs drawn as dots.
(i) Use formal charges to show which of the two structures is the better representation of the ion.
(ii) The oxide E4O10 reacts with water to form the acid H3EO4. Write the balanced equation for this reaction.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) structure 1: E «5 − 0 − 4 =» +1 AND each O «6 − 6 − 1 =» −1 ✔
structure 2: E «5 − 0 − 5 =» 0; the O with the double bond «6 − 4 − 2 =» 0; each of the other three O «6 − 6 − 1 =» −1 ✔
structure 2 is more likely because its formal charges are closer to zero / fewer atoms carry a formal charge ✔
(ii) E4O10 + 6 H2O → 4 H3EO4 ✔
Schemat punktowania
(i) M1 and M2: formal charges of both structures, accept the values written on the diagram. M3 is earned only when the conclusion is justified by formal charges. (ii) [1]; ignore state symbols.