IB Chemistry SL The Periodic Table

Review IB Chemistry the periodic table through electron configurations, periodic trends, group reactions, oxidation states and transition chemistry.

Exam points

deduce electron configuration, period, group or oxidation state from compounds and ions
explain periodic trends and group reactions using nuclear charge, shielding and reactivity

Covered syllabus points

3.1.1—Organization 3.1.2—Electron configuration and position 3.1.3—Periodicity 3.1.4—Group trends 3.1.5—Metal/non-metal continuum 3.1.6—Oxidation states

[Maximum number: 1]

Lithium and boron are elements in period 2 of the periodic table. Lithium occurs in group 1 (the alkali metals) and boron occurs in group 3. Isotopes exist for both elements.

Distinguish between the terms group and period.

[Maximum number: 5]

Iron rusts in the presence of oxygen and water. Rusting is a redox process involving several steps that produces hydrated iron(III) oxide, $\mathrm{Fe}_{2} \mathrm{O}_{3} \bullet \mathrm{nH}_{2} \mathrm{O}$ , as the final product. The half-equations involved for the first step of rusting are given below.

Half-equation 1: $\quad \mathrm{Fe}(\mathrm{s}) \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-}$

Half-equation 2: $\quad \mathrm{O}_{2}(\mathrm{aq})+4 \mathrm{e}^{-}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq})$

(a)

Identify the oxidation number of each atom in the three species in half-equation 2.

Figure for Question (a) — IB Chemistry SL

[ 2 ]

0 | $\mathrm{I} \mid-\mathrm{II}$ - II -

Marking guidance:

Award [2] for five correct.
Award [1] for four correct.
Accept use of oxidation states ( 0,+1,-2,-2,+1 ) for oxidation numbers.
Penalize once for incorrect notation (eg, 2, 2-).

(b)

State the property that determines the order in which elements are arranged in the periodic table.

[ 1 ]

(c)

State the relationship between the electron arrangement of an element and its group and period in the periodic table.

[ 2 ]

[Maximum number: 5]

Chlorine undergoes many reactions.

(a)

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

[ 1 ]

$\mathrm{Cl}^{-}$ AND more «electron-electron» repulsion

Marking guidance:

Accept $\mathrm{Cl}^{-}$ AND has an extra electron.

(b)

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

[ 2 ]

Cl has a greater nuclear charge/number of protons/ $\mathrm{Z}_{\text {eff }}$ «causing a stronger pull on the outer electrons»

same number of shells

OR

same «outer» energy level

OR

similar shielding

(c)

2.67 g of manganese(IV) oxide was added to $200.0 \mathrm{~cm}^{3}$ of $2.00 \mathrm{moldm} \mathrm{m}^{-3} \mathrm{HCl}$ .

\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{MnCl}_{2}(\mathrm{aq})

[ 2 ]

(i)

State the oxidation state of manganese in $\mathrm{MnO}_{2}$ and $\mathrm{MnCl}_{2}$ .
$\mathrm{MnO}_{2}:$ $\mathrm{MnCl}_{2}$ :

[ 2 ]

$\mathrm{MnO}_{2}:+4$ $\mathrm{MnCl}_{2}:+2$

All question bank results loaded

Covered syllabus points

3.1 The periodic table question 1

3.1 The periodic table question 2

Question (a)

Question (b)

Question (c)

3.1 The periodic table question 3

Question (a)

Question (b)

Question (c)

Question (i)