Periodic Trends

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Shortcut to Questions

Q: 1 2 3 4 5 6 7 8 9 10 11

1

Complete the following statements in terms of size, e.g. The lithium atom is smaller than the sodium atom


(i)The sodium ion is .................... the magnesium ion
(ii)The sulfide ion is .................... the oxygen atom
(iii) The magnesium ion is .................... the sulfide ion
(iv)The calcium ion is .................... the barium ion
(v) The nitride ion is .................... the phosphide ion

2

From the ions Be2+, F-, Li+, N3-, choose the one which has -

(a) the largest radius
(b)
the smallest radius

3

When the following atoms become their commonly found ions, does the size of the species increase or decrease?
(i) Na
(ii) Sr
(iii) P
(iv) Cl
(v) Cs
(vi) Sn
(vii) O
(viii) I

4

List the following in order of increasing radius
(i) Na, Na+, K, Mg2+
(ii) Na, Mg, Al
Give a brief explanation of your answers.

5

Classify the following oxides as basic, amphoteric, acidic or peroxides. CaO, PbO, SO2, SO3, P2O5, MgO, BaO, BaO2, Fe2O3, Al2O3, ZnO, SiO2

6

Arrange the following groups of elements in order of increasing first ionisation enthalpy.
(i) Li , K, Na, Cs
(ii) He, Rb, Na, Li
(iii) Ca, Br, As, K
(iv) Sn, Xe, Rb, I

7

Arrange the following groups of elements in order of increasing electronegativity.
(i) Li, F, O, C
(ii) Cl, I, F, Br
(iii) Cs, F, Na, K

8

Given that the electronegativity of hydrogen lies between the values for boron and carbon, predict whether
(a) the following bonds would be non-polar or polar
(b) the hydrogen atom would be the positive/negative end of the dipole?
Al - H, N - H, O - H, B - H

9

  From the elements B, He, H, Ba, O, Xe, F choose the one which has -
(a) the largest first ionization energy
(b) the smallest first ionization energy
(c) the highest electrical conductivity
(d) the highest electronegativity.

10

(a) Write the electronic configuration (in terms of sub-shells) for an atom of each of the following elements:
(i) Sn
(ii) Tl
(b) What are the usual oxidation states of tin and thallium in their compounds? (Each may have more than one).

11

(a) Sodium is element number 11 of the Periodic Table of elements. To which group of the table does it belong?
(b) Fluorine is element number 9 of the Periodic Table. To which group does it belong?
(c)The chemistry of the two elements is fundamentally different. For each element give a reaction which illustrates this.

Periodic Trends (Answers)  

1

The following generalizations apply to relating the sizes of atoms, ions and Periodic Table position of the elements involved.

  1. Down a Group: Atomic radius increases down any Periodic Table Group because in each successive period the outer electrons occupy an orbit that is further from the nucleus. This effect is enhanced through the increased screening of the outer electrons by those occupying orbits closer to the nucleus, thereby preventing the outer electrons from experiencing the full effect of the attraction by the nucleus.
  2. Across a period: Across any period of the Table from left to right, the atomic radius decreases. (Some small departures from this occur in the d-block.) This occurs because, for each additional proton added to the nucleus, the accompanying additional electron is allocated to the same set of outer orbitals which are being filled as that period is built up. The result is that as successive protons and electrons are added, all the outer electrons experience an increasing effective nuclear charge attracting them to the nucleus and hence the atomic radius decreases. (The "effective nuclear charge" is also referred to as "core charge".) When enough electrons have been added to attain the noble gas structure of
  3. ns2 (n-1)d10 np6 in Group 18, the next added electron must occupy an orbital with principal quantum number n increased by one, corresponding to an element from Group 1, so it is further out from the nucleus and the cycle starts again.

    Note that there is no comparable way of comparing the radii of the noble gases with those of other elements, so experimental values for their atomic radii are not normally listed.

  4. Cation radius: The cation of a given element has a smaller radius than the neutral atom due to the excess positive nuclear charge compared with the negative charge of the electrons. The greater attraction between the nucleus and the electrons draws the electrons closer to the nucleus. The larger the cationic charge, the greater will be this contraction of radius. Where multiple cations of an element exist, the cationic radius decreases greatly as the cation charge increases.
  5. E.g. Fe > Fe2+ > Fe3+

  6. Anion radius: Simple anions always have a larger radius than the neutral atom of the same element due to the excess negative charge from the electrons as compared with the positive nuclear charge. Repulsion between the electrons allows them to move further away from the nucleus. The larger the excess negative charge, the greater is the resulting anionic radius.

(i) The sodium ion (Na+) is larger than the magnesium ion (Mg2+) due to two effects. The elements sodium and magnesium are in the same Period, therefore outer electrons of the Mg atom experience greater effective nuclear charge and, more importantly, the magnesium cation, Mg2+, has a greater cationic charge than the Na+ cation. This second effect is by far the more significant.

(ii) The sulfide ion (S2-) is larger than the oxygen atom (O) due to two effects. Sulfur is in the same Group as oxygen and one Period lower, so its outer electrons occupy orbitals from the n = 3 level while oxygen's outer electrons occupy orbitals from the n = 2 level. Consequently for the neutral atoms, S would be larger than O. For the sulfide ion, reinforcing this effect is the large, excess 2- charge that greatly increases the radius of S2- as compared with the neutral S atom.

(iii) The magnesium ion (Mg2+) is smaller than the sulfide ion (S2-). If one were comparing just the neutral atoms Mg and S, the S atom would be smaller as both elements are in the same period and sulfur is more to the right. However, the presence of the excess 2+ charge on Mg2+ greatly reduces its size while the presence of the excess 2- charge on S2- greatly increases its radius. The effects of excess + or - charge on the radius of ions compared with their neutral atoms is far greater than the reduction observed in radius for neutral atoms that results from increasing effective nuclear charge from left to right across any Period.

(iv) The calcium ion (Ca2+) is smaller than the barium ion (Ba2+). The elements calcium and barium are both in the same Group 2 with calcium being higher, so for the neutral atoms, Ca would have a smaller radius, its outer electrons being in the n = 4 level as compared with the n = 6 level for the Ba atom. Both cations have a 2+ charge, so the relationship between their radii in the cationic state would be the same as for the neutral atoms.

(v) The nitride ion (N3-) is smaller than the phosphide ion (P3-). The elements nitrogen and phosphorus are in the same Group 15 with nitrogen being higher, so for the neutral atoms, N with its outer electrons in the n = 2 level would have a smaller radius than P with its outer electrons in the n = 3 level. Both anions have a 3- charge, so the relationship between their radii in the anionic state would be the same as for the neutral atoms.

2 See answer to the previous question for a summary of the principles involved.

(a) N3- has the largest radius. All the elements involved are in the same Period. Although for neutral atoms the largest would be that of the element furthest to the left, here the species are all ions. The effect of excess negative charge greatly overrides the reduction of radius found for neutral atoms from left to right in a given Period, so F- and N3- are the two largest species while Be2+ and Li+ are the two smallest. To choose which of F- or N3- is the larger, the principle involved is that the nitride ion has the greater excess negative charge, so N3- has the bigger radius.

(b) Following from the above, Be2+ has an excess charge of 2+ while Li+ has an excess charge of 1+, so Be2+ has the smaller radius.

3

See answer to Question 1 for a summary of the principles involved.

(i) Sodium is a Group 1 element so its only ionic state is Na+. Cations of a given element have a smaller radius than the neutral atom, so Na+ will decrease in size compared with the Na atom.

(ii) Strontium is a Group 2 element so its only ionic state is Sr2+. Cations of a given element have a smaller radius than the neutral atom, so Sr2+ will decrease in size compared with the Sr atom.

(iii) Phosphorus is a Group 15 element so its only ionic state is P3-. Anions of a given element have a larger radius than the neutral atom, so P3- will increase in size compared with the P atom.

(iv) Chlorine is a Group 17 element so its only ionic state is Cl-. Anions of a given element have a larger radius than the neutral atom, so Cl- will increase in size compared with the Cl atom.

(v) Caesium is a Group 1 element so its only ionic state is Cs+. Cations of a given element have a smaller radius than the neutral atom, so Cs+ will decrease in size compared with the Cs atom.

(vi) Tin (Sn) is a Group 14 element. It can have two ionic states, Sn2+ and Sn4+. Both are cations and cations of a given element always have a smaller radius than the neutral atom, so both Sn2+ and Sn4+. will decrease in size compared with the Sn atom. Note that the radius of Sn4+ would be much smaller than that of Sn2+ due to its extra 2+ charge.

(vii) Oxygen is a Group 16 element. Its most common ionic state is O2-. Anions of a given element have a larger radius than the neutral atom, so O2- will increase in size compared with the O atom.

(viii) Iodine is a Group 17 element so its only ionic state is I-. Anions of a given element have a larger radius than the neutral atom, so I- will increase in size compared with the I atom.

4

See answer to Question 1 for a summary of the principles involved.

(i) Mg2+ < Na+ < Na < K

Sodium and potassium are in the same Periodic Table Group (Group1) and potassium is lower down so it has the larger radius and and the ranking Na < K follows. Sodium and magnesium are in the same period and their cations will be much smaller than the neutral atoms, so the ranking (Na+, Mg2+) < Na < K follows. For neutral atoms, Na > Mg and due to the larger charge on the Mg2+ ion compared with the Na+ ion, this effect is greatly enhanced, so the combined ranking is Mg2+ < Na+ < Na < K

(ii) Al < Mg < Na

These elements are all in the same Period, so the consequence of increasing effective nuclear charge on the outer electrons causes a reduction in radius from left to right across the Period.

5

Basic oxides are those that react with H+ ion to form a salt and water. They can do this because basic oxides contain the O2- ion which reacts with the H+. Such compounds containing the O2- anion must also contain a cation. It is a defining property of metals that they form cations in compounds, the outer electrons being relatively easy to remove in reactions (i.e. they have low ionization energies). Thus if an element is a metal, it has a basic oxide.

e.g. Calcium is a Group 2 element, all of which are metals apart from beryllium. Hence CaO is a basic oxide and it reacts with H+ as follows:

CaO + 2H+ ® Ca2+ + H2O

The salt would be calcium chloride if the acid used were hydrochloric acid.

Other basic oxides in the list are BaO, MgO, Fe2O3.

Acidic oxides are those that react with OH- ion to form a salt and water. They can do this because the central atom is a non-metal and it can form covalent bonds to the O atom of the OH- ion. It is a defining property of non-metals that they can form covalent compounds with other non-metals. Thus if an element is a non-metal, it will have an acidic oxide.

e.g. CO2 + 2OH- ® CO32- + H2O

The salt would be sodium carbonate if the hydroxide used were sodium hydroxide.

Other acidic oxides in the list are SO2, SO3, P2O5, SiO3.

Amphoteric oxides are able to react with both H+ and OH- and such compounds are usually found as oxides of elements located near the middle of the periodic table. These elements typically exhibit some properties of metals as well as some of non-metals. Thus they have sufficient ionic character for their oxides to react with H+ but also the ability to form covalent bonds to the O atoms of OH- ions.

e.g. PbO + 2H+ ® Pb2+ + H2O

The salt would be lead(II) chloride if the acid used were hydrochloric acid.

PbO + H2O + 2OH- ® [Pb(OH)4]2-

The salt formed would be sodium tetrahydroxoplumbate(II) if sodium hydroxide were used.

Other amphoteric oxides in the list are Al2O3 and ZnO.

Peroxides of metals contain the O22- ion although there are covalent peroxides such as hydrogen peroxide, H2O2. The only peroxide in the list is BaO2.

6

The first ionization enthalpy (energy) of an element is defined as the energy required to remove the first electron from a gas phase atom of an element, and can be represented by the equation

X(g) ® X+(g) + e-

The following generalizations apply to trends in atom's first ionization energy with Periodic Table position.

Down a Group: As the outer electrons are progressively occupying orbits further from the nucleus and screened by increasing numbers of inner electrons, the attraction they experience to the nucleus decreases and therefore the energy needed to remove an electron (i.e. to ionize the atom) also decreases.

Across a Period: Recall that the reason that atomic radius decreases from left to right across any Period is the increasing effective nuclear charge experienced by the outer electrons as more protons are added to the nucleus while the accompanying additional electrons are placed in the same outer orbitals. The increased attraction to the nucleus that results not only decreases the atomic radius, but also leads to an increase in the amount of energy needed to remove an electron - i.e. to ionize the atom.

(i) These four elements are in the same Periodic Table Group (Group 1), so their first ionization energy will be greatest for the element highest in the group and least for the element lowest in the group.

Hence the ranking is Cs < K < Na < Li.

(ii) Of these four elements, sodium, lithium and rubidium are all in Group 1, so their first ionization energies will increase in the order Rb < Na < Li. The fourth element, helium, is in Group 18 and in the Period before lithium in the Table. There is always a decrease in first ionization energy at the start of a new Period, so He should have a higher first ionization energy than Li. [In fact, He has the highest first ionization energy of any element as its electrons are in the n = 1 level and He has no inner electrons to reduce the effective nuclear charge.]

Hence the overall ranking is Rb < Na < Li < He.

[Note however, that the ability to make predictions of first ionization energy involving elements which are in both different Periods and Groups is limited as the reduction that occurs down a Group is often more than cancelled out by the increase that occurs across a Period. For example, the first ionization energies of Li in Period 2 (526 kJ/mol) cf Mg in Period 3 (744 kJ/mol).]

(iii) These four elements are in the same Period of the Periodic Table so their first ionization energy will be greatest for the element furthest to the right in the Period and least for the element furthest to the left.

Hence the ranking is K < Ca < As < Br.

(iv) These four elements are all in the same Period, so their first ionization energies will increase from left to right across the period.

Hence the overall ranking is Rb < Sn < I < Xe.

7

The electronegativity of an atom is the power of that atom to attract electrons to itself. Electronegativity is not a directly measurable quantity like ionization energy or electron affinity but instead is calculated from measured bond energies. A number calculated from the bond energy is assigned as the electronegativity of the element. The larger the number, the greater is the electronegativity. The most electronegative atom is F with an electronegativity value of 3.99 while the least electronegative is Fr with a value of 0.7. It is useful to know that the order of electronegativities for the three most electronegative elements is F > O > Cl. The following general correlations are observed between Periodic Table position and electronegativity:

Down a Group. Electronegativity decreases down a Periodic Table Group as the outer electrons are further from the nucleus and screened from its charge.

Across a Period. Electronegativity increases from left to right across each Period as the effective nuclear charge increases.

(i) These four elements are in the same Period of the Table and their electronegativities will increase from left to right. Therefore the order is

Li < C < O < F.

(ii) All four elements are in the same Group of the Table, so their electronegativities will decrease down the Group. Therefore the order will be:

I < Br < Cl <F.

(iii) These elements are from different Groups and Periods. However, their electronegativity order can be deduced as follows: F is the most electronegative atom of all elements. Therefore it ranks highest. The remaining elements are all in the same Group (Group 1) and so their order will decrease down the Group -

i.e. Na > K > Cs.

Combining this with the knowledge that F is the most electronegative of all elements, the order is

Cs < K < Na < F.

8

(a) A covalent bond between any two atoms of different elements will generally be polar as each element will usually have at least a small difference in electronegativity from another element. This is confirmed for the atoms of boron, carbon and hydrogen in the data, while aluminium lies below boron in the same Group, so its electronegativity will be smaller than that of boron. The atoms of oxygen and nitrogen are in the same Period as boron and carbon and to the right of carbon, so will be more electronegative than carbon. Thus the order of electronegativities is

Al < B < H < C < N < O.

As all of these atoms have different electronegativities, the bonds Al-H, N-H, O-H and B-H will all be polar.

(b) From (a), H is more electronegative than B or Al, so it will be the negative end of the bond dipole in the Al-H and B-H bonds. In the N-H and O-H bonds, the H atom is the less electronegative and will therefore be the positive end these bond dipoles.

9

(a) Of these elements, helium is located in the extreme top, right hand corner of the Table and has the highest ionization energy of any element.

(b) Barium is located among the metals and of the elements listed, is located lowest in the Table. Consequently Ba would have the lowest first ionization energy.

(c) Electrical conductivity is highest for metals. Of the elements listed, only barium is a metal, so it would be expected to have the highest electrical conductivity.

(d) Fluorine has the largest electronegativity of all the elements. [Oxygen has the second largest electronegativity].

10

(a) (i) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2

(ii) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p1

(b) Sn +II and +IV

Tl +I and +III

11

(a) Group 1 IUPAC system

(b) Group 17 IUPAC system

(c) Sodium is a metal. Metals form cations in ionic compounds, have ionic hydrides containing the H- ion, do not form covalent bonds with nonmetals, have basic oxides containing the O2- ion.

Typical reactions include

2Na + H2 ® 2NaH .... ionic solid

Fluorine is a nonmetal. Nonmetals form anions in ionic compounds, have covalent hydrides, can form covalent bonds with non-metals, have acidic oxides involving covalent bonds to the O atoms.

F2 + H2 ® 2HF .... covalent gas