"Classification of Elements and Periodicity" — Practice

Question 1

Arrange $\mathrm{N^{3-}}$, $\mathrm{O^{2-}}$, $\mathrm{F^-}$, $\mathrm{Na^+}$ and $\mathrm{Mg^{2+}}$ in decreasing order of ionic radius, with reasoning.

Accepted Answer

All five are isoelectronic with ten electrons, so only nuclear charge distinguishes them: 7, 8, 9, 11 and 12 respectively. More protons holding the same ten electrons gives a smaller ion, so the decreasing order is $\mathrm{N^{3-} > O^{2-} > F^- > Na^+ > Mg^{2+}}$. The reasoning takes one line, and the trap is reaching for left-to-right periodic position instead, which is irrelevant once the species are isoelectronic.

Question 2

Why is the first ionisation enthalpy of beryllium greater than that of boron?

Accepted Answer

Beryllium is $1s^2\,2s^2$, so the electron removed comes from a completely filled 2s subshell, which is unusually stable and penetrates close to the nucleus. Boron is $1s^2\,2s^2\,2p^1$, and its outermost electron sits in a 2p orbital that is higher in energy and partly shielded by the 2s pair. The 2p electron is therefore easier to remove despite boron's larger nuclear charge, giving 801 against beryllium's 899 kJ mol$^{-1}$. The same reasoning puts magnesium above aluminium.

Question 3

An element shows successive ionisation enthalpies of 578, 1817, 2745 and 11577 kJ mol$^{-1}$. Identify the group it belongs to.

Accepted Answer

The jump comes between the third and fourth values, where the enthalpy rises by a factor of about four. Three electrons are therefore removed comparatively easily before the noble gas core is reached, so the element has three valence electrons and belongs to group 13; the values are those of aluminium. The general method is to count the removals preceding the large jump, and to read only its position, since a ratio of four here signals a core just as clearly as sodium's ratio of nine does.

Question 4

Explain why the electron gain enthalpy of chlorine is more negative than that of fluorine, contrary to the group trend.

Accepted Answer

Fluorine's valence electrons occupy the very compact 2p subshell, so an incoming electron is forced into a small region already holding seven electrons and meets strong interelectronic repulsion, which cancels much of the energy released on adding it. Chlorine's 3p subshell is substantially larger, so the same addition costs far less repulsion, giving $-349$ against fluorine's $-328$ kJ mol$^{-1}$. Below chlorine the normal trend resumes, with bromine at $-325$ and iodine at $-295$.

Question 5

Why do noble gases have positive electron gain enthalpies?

Accepted Answer

Their valence shells are complete, so an added electron cannot enter the existing shell and must begin a new one at a considerably greater distance from the nucleus. It is poorly attracted there and strongly repelled by the filled shell beneath, so energy must be supplied rather than released. A positive electron gain enthalpy means the anion is less stable than the neutral atom plus a free electron, which is why noble gas anions are not encountered.

Question 6

An element has the configuration $[\mathrm{Ar}]\,3d^{10}\,4s^2\,4p^4$. Give its period, group and block, and predict whether its oxide is acidic or basic.

Accepted Answer

The outermost principal quantum number is 4, so it lies in period 4. Its valence electrons are $4s^2\,4p^4$, six in all, placing it in group 16, and the last electron enters a p subshell, making it a p-block element. It is selenium, with $Z = 34$. Being a non-metal, it gives an acidic oxide, $\mathrm{SeO_2}$, which dissolves in water to give selenous acid. Note that the filled $3d^{10}$ is core, not valence, and must not be counted towards the group.

Question 7

Why were cobalt and nickel, and tellurium and iodine, placed out of atomic mass order in Mendeleev's table, and how did Moseley resolve this?

Accepted Answer

Chemical similarity demanded that cobalt precede nickel and tellurium precede iodine, but their atomic masses run the other way, with cobalt at 58.9 against nickel's 58.7 and tellurium at 127.6 against iodine's 126.9. Moseley measured the X-ray frequencies of the elements in 1913 and found they varied regularly with a whole number, the atomic number. Ordering by atomic number puts every pair in the chemically correct sequence, so the anomalies are anomalies of mass alone and vanish entirely.

Question 8

Explain why period 4 contains 18 elements while period 3 contains 8.

Accepted Answer

Period length equals twice the number of orbitals filled in that round. Period 3 fills 3s and 3p, that is $1 + 3 = 4$ orbitals, giving eight electrons. Period 4 fills 4s, then 3d, then 4p, that is $1 + 5 + 3 = 9$ orbitals, giving eighteen. The 3d subshell becomes energetically accessible only after 4s by the $(n+l)$ rule, which is why the first transition series appears in period 4 and not earlier, and why the table widens exactly there.

Question 9

Why does lithium resemble magnesium more closely than it resembles sodium, its own group neighbour?

Accepted Answer

This is the diagonal relationship. Going down a group increases atomic size while going across a period decreases it, and for the first two periods the two effects roughly cancel along a diagonal, leaving lithium and magnesium with similar sizes and similar charge-to-radius ratios, which is what governs polarising power and hence covalent character. Both form nitrides directly with nitrogen, both give the normal oxide rather than a peroxide on burning, and both have carbonates that decompose on heating, unlike sodium's.

Question 10

Why can nitrogen not form $\mathrm{NCl_5}$ while phosphorus readily forms $\mathrm{PCl_5}$?

Accepted Answer

Nitrogen is a second-period element, so its valence shell contains only 2s and 2p orbitals and there are no 2d orbitals to use. It cannot accommodate more than eight valence electrons and is therefore limited to a covalency of four. Phosphorus lies in period 3, where 3d orbitals are energetically accessible, so its octet can expand to hold five bonding pairs. The same restriction limits oxygen to a covalency of two and fluorine to one, while sulphur reaches six and chlorine seven.

Question 11

Arrange $\mathrm{Na_2O}$, $\mathrm{MgO}$, $\mathrm{Al_2O_3}$, $\mathrm{P_4O_{10}}$ and $\mathrm{Cl_2O_7}$ in order of increasing acidic character and justify the trend.

Accepted Answer

The order is $\mathrm{Na_2O < MgO < Al_2O_3 < P_4O_{10} < Cl_2O_7}$. Acidic character rises with the electronegativity of the element bonded to oxygen, which rises across the period as $Z_{	ext{eff}}$ climbs. $\mathrm{Na_2O}$ and $\mathrm{MgO}$ are basic metal oxides, $\mathrm{Al_2O_3}$ is amphoteric and reacts with both acids and alkalis, and the last two are acidic non-metal oxides giving phosphoric and perchloric acid with water. Reading oxide behaviour is the quickest route to placing an unfamiliar element.

Question 12

Zinc lies in the d-block but is not classed as a transition element. Explain.

Accepted Answer

A transition element is defined as one having a partly filled d subshell either in the free atom or in one of its common oxidation states. Zinc is $3d^{10}\,4s^2$ as an atom and $3d^{10}$ as the $\mathrm{Zn^{2+}}$ ion, so the d subshell is complete in every state it adopts. It therefore shows none of the characteristic transition properties: no variable oxidation states, no coloured ions and no significant catalytic behaviour. Cadmium and mercury are excluded for exactly the same reason.

Practice — "Classification of Elements and Periodicity"

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