Corrected exercises: metals, nonmetals and metalloids (202-SN1-RE)

Sodium: a metal, group 1, far to the left. Silicon: a metalloid, it sits on the staircase. Sulfur: a nonmetal. Germanium: a metalloid, just below silicon. Copper: a transition metal. Chlorine: a nonmetal, group 17. The reading rule is purely geographic, which is what makes it reliable: find the staircase, then see which side you are on.

Boron, silicon, germanium, arsenic, antimony, tellurium. These are the six elements whose properties hesitate between the two families: they have the luster of a metal and the brittleness of a nonmetal, and above all an intermediate conductivity on which the whole electronics industry rests. Polonium and astatine are sometimes added, depending on the author, but these six are the ones to remember.

Hydrogen: gas. Bromine: liquid. Iodine: solid; it sublimes easily, but it is a solid. Mercury: liquid. Iron: solid. Neon: gas. Notice what happens down group 17: chlorine is a gas, bromine a liquid, iodine a solid. London dispersion forces increase with the size of the molecule, and that is enough to change the state three times within a single column.

Mercury, which is a metal, and bromine, which is a nonmetal. One on each side of the staircase. Two near-liquids deserve a mention because they melt in the hand without being liquid at $25$ °C: gallium melts at $29.8$ °C and cesium at $28.5$ °C. The question is therefore always asked at a given temperature, never in the absolute.

The five nonmetals $H$, $N$, $O$, $F$, $Cl$ and the six noble gases $He$, $Ne$, $Ar$, $Kr$, $Xe$, $Rn$, eleven in all. Conclusion: they are ALL nonmetals, and no metal is a gas at $25$ °C, the most volatile still being mercury, which stays liquid. The reason lies in the nature of the bonding: a metal is a lattice of cations in a sea of electrons, a collective structure that takes a great deal of energy to break up, whereas a nonmetal forms small discrete molecules that simple London dispersion forces are enough to separate.

In a metal, the valence electrons no longer belong to any one atom in particular: they are delocalized over the whole crystal and move as soon as a potential difference is applied. In solid sodium chloride, the charge carriers do exist, namely the $Na^{+}$ and $Cl^{-}$ ions, but they are LOCKED at the lattice sites. Hence the distinction to remember: solid salt does not conduct, whereas molten or dissolved salt conducts very well, because its ions are then free to move. It is therefore not the presence of charges that decides, but their mobility.

The electrons are already free and already fixed in number: heating does not release any more. On the other hand, thermal agitation makes the cations of the lattice vibrate harder, and they become so many obstacles in the path of the electrons. The hotter it gets, the more frequent the collisions, and so the more the resistance increases. This is exactly the opposite of the behaviour of a semiconductor, where heating releases new carriers: this contrast will come back in exercise 7, and it is the surest test for telling a metal from a metalloid.

Metallic bonding is not DIRECTIONAL: a cation is attracted by the sea of electrons in all directions at once. If one plane of atoms slides over its neighbour, each cation finds exactly the same environment again, and the metal deforms without breaking: this is malleability and ductility. In an ionic crystal, by contrast, each ion is surrounded by ions of OPPOSITE charge. A slip of half a unit suddenly brings ions of the same charge face to face, repulsion takes over, and the crystal cleaves cleanly. A metal bends, a salt breaks, and the difference rests entirely on this alternation of charges.

What holds a metal together is the attraction between the cations and the sea of electrons, so two things matter: the charge of the cation and the number of electrons contributed to the cloud. Sodium gives up only one electron and leaves only an $Na^{+}$: the attraction is weak, it melts at $98$ °C, so low that it can be cut with a knife. Magnesium gives up two and leaves an $Mg^{2+}$: the attraction is far stronger, $650$ °C. Aluminum gives up three, $660$ °C. The enormous jump from sodium to magnesium, then the plateau from magnesium to aluminum, show that the charge counts for more than the bare count of electrons, the aluminum cation also being smaller and more deeply buried.

Sodium: $1$ delocalized electron per atom, cation $Na^{+}$. Magnesium: $2$ electrons, cation $Mg^{2+}$. Aluminum: $3$ electrons, cation $Al^{3+}$. The number of electrons given up is simply the number of valence electrons, hence the group number for the first columns: $1$, $2$ and $13$, which gives $3$.

Sodium $Na^{+}$, magnesium $Mg^{2+}$, aluminum $Al^{3+}$, potassium $K^{+}$, calcium $Ca^{2+}$. The charge is the number of valence electrons, hence the group number for columns 1, 2 and 13. Electrons are lost, never gained: that is the very definition of a metal.

Magnesium is $1s^{2}2s^{2}2p^{6}3s^{2}$. By losing its two $3s$ electrons, the ion becomes $1s^{2}2s^{2}2p^{6}$, exactly the configuration of neon, with $10$ electrons for $12$ protons. Beware of the most common writing error: electrons are removed from the OUTER shell, the $3s$, not from an inner shell.

Iron is $[Ar]3d^{6}4s^{2}$. The two $4s$ electrons are removed first, because they are the outermost despite the filling order, which gives $Fe^{2+}$ as $[Ar]3d^{6}$. One more $3d$ electron gives $Fe^{3+}$ as $[Ar]3d^{5}$. Its particular stability comes from this $d$ subshell being exactly HALF filled, five orbitals each holding one unpaired electron of the same spin: it is the same reason that explains the anomaly of chromium in the table of configurations, and it is why iron(III) salts are so common.

$\dfrac{4562}{496}=9.20$. The first electron removed is the lone $3s$ electron, alone in its shell and loosely held: that costs little. The second has to be removed from the neon core, a complete shell very close to the nucleus: sodium would then have to break up a noble-gas structure, which costs more than nine times as much. This sudden jump IS the reason why sodium stops at $+1$ and never exists as $Na^{2+}$: chemistry follows the point where the energy blows up.

Aluminum and oxygen: $Al^{3+}$ and $O^{2-}$. It takes $2$ aluminums for $3$ oxygens to reach $+6$ and $-6$, hence $Al_{2}O_{3}$. Potassium and sulfur: $K^{+}$ and $S^{2-}$, two potassiums are needed, hence $K_{2}S$. The method is always the same: find the charge from the column, cross over, simplify if possible. Here no simplification is needed, but it would be for calcium and oxygen, where $Ca_{2}O_{2}$ must become $CaO$.

Fluorine $F^{-}$, oxygen $O^{2-}$, nitrogen $N^{3-}$, chlorine $Cl^{-}$, sulfur $S^{2-}$. The charge is the number of electrons that are MISSING to reach eight valence electrons, that is $18$ minus the group number. Group 17 gains one electron, group 16 gains two, group 15 gains three.

Oxygen has $8$ electrons and gains $2$: the oxide ion carries $10$. Nitrogen has $7$ and gains $3$: the nitride ion carries $10$ as well. Both therefore have the configuration of neon, $1s^{2}2s^{2}2p^{6}$, the same as $F^{-}$, $Na^{+}$, $Mg^{2+}$ and $Al^{3+}$. Six different ions, a single electron arrangement: that is the definition of an isoelectronic series.

Because the number of PROTONS, for its part, does change. $N^{3-}$ has $7$ protons for $10$ electrons: the nucleus is short-handed, it holds a cloud that is too large for it poorly, and the cloud swells. $Al^{3+}$ has $13$ protons for the same $10$ electrons: the nucleus pulls much harder and tightens the cloud. Between the two, the radius decreases steadily as the atomic number rises. The rule to remember: in an isoelectronic series, the radius decreases as $Z$ increases, and it is the only case in which radii are compared without worrying about the period or the group.

$\dfrac{1.46}{0.50}=2.92$. For a rigorously identical electron arrangement, the radius therefore varies by almost a factor of three depending on the charge of the nucleus. This says two things: an anion is always larger than the atom it comes from, a cation always smaller, and the difference is far from marginal. It is also what allows a small cation such as $Al^{3+}$ to slip between large anions in a crystal lattice.

Magnesium and nitrogen: $Mg^{2+}$ and $N^{3-}$. It takes $3$ magnesiums for $2$ nitrogens to reach $+6$ and $-6$, hence $Mg_{3}N_{2}$, magnesium nitride, which incidentally forms along with the oxide when magnesium is burned in air. Calcium and chlorine: $Ca^{2+}$ and $Cl^{-}$, hence $CaCl_{2}$.

$Zn+CuSO_{4}$: YES, zinc is above copper, so it displaces it. $Cu+ZnSO_{4}$: NO, this is exactly the reverse reaction, and it does not occur. $Mg+HCl$: YES, magnesium is above hydrogen, so it attacks acids. $Cu+HCl$: NO, copper is below hydrogen, which is why a copper wire does nothing in hydrochloric acid whereas an iron nail fizzes in it.

$Zn+CuSO_{4}\rightarrow ZnSO_{4}+Cu$: zinc goes from $0$ to $+2$, copper from $+2$ to $0$, two electrons are exchanged, and the sulfate is a spectator. $Mg+2\,HCl\rightarrow MgCl_{2}+H_{2}$: magnesium goes from $0$ to $+2$, hydrogen from $+1$ to $0$, and two acid molecules are needed because each hydrogen gains only one electron.

Iron is above silver, so it displaces it: $Fe+2\,AgNO_{3}\rightarrow Fe(NO_{3})_{2}+2\,Ag$. The coefficient $2$ comes from the electron balance, iron giving up two electrons and each silver ion gaining only one. The grey deposit that grows in needles on the strip is metallic silver, and the solution turns the pale green of iron(II).

$n(Fe)=\dfrac{2.00}{55.85}=3.581\times 10^{-2}$ mol. The equation gives $2$ silvers per iron: $n(Ag)=2\times 3.581\times 10^{-2}=7.162\times 10^{-2}$ mol. Mass: $7.162\times 10^{-2}\times 107.87=7.73$ g. The deposit therefore weighs nearly four times the mass of iron consumed, which is not a paradox: moles are exchanged, not grams, and silver is twice as heavy per mole while being twice as numerous.

Because gold is at the very bottom of the series: it is such a poor reducing agent that no oxidizing agent present in nature, neither the oxygen of the air, nor water, nor ordinary acids, manages to strip its electrons away. It therefore stays at oxidation number $0$, that is, in the metallic state, since it was formed. Aluminum is at the opposite end, at the very top: it gives up its electrons at the slightest contact with oxygen, and it exists on Earth only as an oxide or a silicate. What is called a noble metal is nothing other than a metal that sits too low in this series, and exercise 10 will show what it costs to bring aluminum back up.

$Cl_{2}+2\,KBr$: YES, chlorine is above bromine, so it displaces it. The equation is $Cl_{2}+2\,KBr\rightarrow 2\,KCl+Br_{2}$: chlorine goes from $0$ to $-1$, bromine from $-1$ to $0$, two electrons are exchanged, and potassium remains a spectator at $+1$. $Br_{2}+2\,KCl$: NO, this is the reverse, and it does not occur.

In water, the bromine formed gives an orange to yellow-brown tint, deeper the more concentrated it is. The decisive test is the addition of cyclohexane, a nonpolar solvent that is not miscible with water: shake, let the layers settle, and the halogen moves into the upper organic phase because it is nonpolar too. Bromine gives a bright orange there, iodine a very characteristic violet, and it is this violet that identifies iodine unambiguously. A nonpolar compound dissolves in a nonpolar solvent: the solubility rule does all the analytical work here.

Fluorine goes from $0$ in $F_{2}$ to $-1$ in $HF$: it is REDUCED, so it is the OXIDIZING AGENT. Oxygen goes from $-2$ in water to $0$ in $O_{2}$: it is OXIDIZED, so it is the REDUCING AGENT, which is remarkable since oxygen is ordinarily the standard oxidizing agent. Balance: $4$ fluorine atoms each gain $1$ electron, so $4$ electrons are gained; $2$ oxygen atoms each lose $2$ electrons, so $4$ electrons are given up. $4=4$, the balance closes. Fluorine is the only element able to make the oxygen of water play this role, and that is what makes it the most powerful oxidizing agent in all of chemistry.

Oxidizing power is the tendency to gain an electron, and the ability to do so depends on how strongly the nucleus can attract that extra electron. Going down the group, one shell is added at each period: the electron that is gained sits farther and farther from the nucleus, and it is more and more SHIELDED by the inner shells. The effective attraction therefore weakens, and the oxidizing power with it. This is exactly the reasoning that makes electron affinity and electronegativity decrease down the same column: a single cause, three trends.

The order is reversed: $I^{-}>Br^{-}>Cl^{-}>F^{-}$. This is a logical consequence and not an extra fact to memorize: if fluorine gains electrons eagerly, the fluoride ion gives them up with equal reluctance; the strongest oxidizing agent has the weakest conjugate reducing agent. Iodide is therefore the best reducing agent of the four, and the only one that the iron(III) ion can oxidize, according to $2\,Fe^{3+}+2\,I^{-}\rightarrow 2\,Fe^{2+}+I_{2}$. The brown colour that then appears is the iodine formed, and this test is used to determine iron(III) in the laboratory.

Boron: $3$ valence electrons. Silicon: $4$. Phosphorus: $5$. For the $s$ and $p$ blocks, the number of valence electrons is read directly from the group number, minus $10$ from group 13 onwards. In the crystal, each silicon forms $4$ covalent bonds, one per valence electron, which exactly completes its octet and those of its neighbours. No electron is left free to move: that is why pure silicon is a very poor conductor when cold, and why it is so hard and melts so high.

The two solids do not face the same problem. In copper, the electrons are ALREADY free and fixed in number: heating adds none, but makes the cations vibrate, and they hinder the electrons' passage, so the conductivity falls. In silicon, all the electrons are engaged in bonds: they have to be given enough energy to cross the band gap and become free. Thermal agitation does exactly that, and more and more of it as the temperature rises. The number of carriers then increases much faster than the hindrance, and the conductivity goes up. This contrast is the decisive test for deciding between a metal and a metalloid, and it is also what makes electronic components sensitive to temperature.

Phosphorus supplies $5$ valence electrons where the lattice expects only $4$. Four go into the bonds, the FIFTH is left without a job and is released very easily: it becomes a NEGATIVE charge carrier. This is called $n$-type doping, and phosphorus is called a donor. One phosphorus atom for every million silicon atoms is enough to change the conductivity by several orders of magnitude, which explains the extreme purity requirements of the industry.

Boron supplies only $3$ valence electrons for $4$ bonds to form: one electron is missing, and this absence, called a HOLE, behaves like a POSITIVE charge carrier. When a neighbouring electron fills the hole, it leaves one behind it, and step by step the hole moves through the crystal as a positive charge would. This is called $p$-type doping, and boron is called an acceptor. The junction between an $n$ region and a $p$ region is the basic component of all electronics.

Conductor: copper, a metal; its electrons are already free, and its conductivity DECREASES as the temperature rises. Semiconductor: silicon, a metalloid; its band gap is narrow, and its conductivity INCREASES with temperature, strongly. Insulator: sulfur or diamond, nonmetals; their band gap is too wide for thermal agitation to change anything, and the conductivity stays practically ZERO. The three families of the periodic table therefore give the three electrical behaviours, and the metalloid staircase is exactly the border between the first and the third.

FALSE. Mercury is a liquid at $25$ °C, and that is precisely what earned it its uses in thermometry and barometry. Two other metals melt only just above that, gallium at $29.8$ °C and cesium at $28.5$ °C: gallium literally melts in the hand. The correct sentence: all metals are solids at $25$ °C except mercury, which is a liquid.

TRUE. It is even the signature of a semiconductor. Its electrons are engaged in covalent bonds and must cross a narrow band gap to become mobile; thermal agitation gives them the means, more and more as the temperature rises. A metal does exactly the opposite, since its electrons are already free and heating only hinders them. Measuring the conductivity at two temperatures is therefore enough to classify an unknown sample.

FALSE, and the direction is reversed. A metal displaces those that are BELOW it in the series, that is, those that are weaker reducing agents than itself. Zinc displaces copper, copper does not displace zinc. The way never to go wrong is to remember the physical meaning: the strongest reducing agent is the one that holds its electrons least tightly, so it is the one that gives them to the other. The correct sentence: a metal displaces from its compounds any metal located below it in the activity series.

FALSE as to the conclusion, true as to the premise. They do all have ten electrons, which makes them an isoelectronic series, but they do not have the same number of PROTONS: seven for nitrogen, eight for oxygen, nine for fluorine, eleven for sodium, twelve for magnesium. The more highly charged the nucleus, the tighter it pulls the same cloud of ten electrons, so the radius DECREASES steadily from $N^{3-}$ to $Mg^{2+}$, by a ratio of more than two. The correct sentence: for an identical electron arrangement, the radius decreases as the atomic number increases.

TRUE, and it is stronger than a mere family record. Fluorine is the most powerful oxidizing agent in all of ordinary chemistry, to the point of oxidizing the oxygen of water according to $2\,F_{2}+2\,H_{2}O\rightarrow 4\,HF+O_{2}$, a reaction in which oxygen, an oxidizing agent everywhere else, plays the role of reducing agent. The cause fits in one line: fluorine is the smallest of the halogens and the most electronegative of the elements, so its nucleus attracts the electron it gains from closer than any other.

$A$ is SODIUM: soft, can be cut with a knife, reacts violently with water releasing hydrogen gas, and gives a basic oxide; these are the four marks of an alkali metal, and the only one in the third period is sodium. $B$ is SILICON: conduction that appears only when hot is the mark of a semiconductor, hence a metalloid, and its use in microelectronics removes the last doubt. $C$ is CHLORINE: a yellow-green gas with bleaching properties, group 17. $D$ is SULFUR: a yellow solid with a blue flame, and the gas with the pungent odour is sulfur dioxide.

Electronegativity increases from left to right across a period, and the four elements fall in the order sodium, silicon, sulfur, chlorine. So sodium $0.93$, silicon $1.90$, sulfur $2.58$, chlorine $3.16$. Increasing order: $Na<Si<S<Cl$. The only point to watch is not to trust the order of the statement, which lists the elements out of order.

$A$ and $C$: $Na^{+}$ and $Cl^{-}$, hence $NaCl$, sodium chloride. $A$ and $D$: $Na^{+}$ and $S^{2-}$, two sodiums are needed, hence $Na_{2}S$, sodium sulfide. Both are ionic compounds, which is consistent with the electronegativity difference: $3.16-0.93=2.23$ for the first and $2.58-0.93=1.65$ for the second. The first lies well above the threshold of about $1.7$ that separates ionic from polar covalent; the second sits right at that threshold, just below it, a reminder that the threshold is only a rule of thumb, and that a compound of a metal with a nonmetal such as this one is classed as ionic.

$NaCl$: $M=22.99+35.45=58.44$ g/mol, so $\dfrac{22.99}{58.44}\times 100=39.34\%$. $Na_{2}S$: $M=2(22.99)+32.07=45.98+32.07=78.05$ g/mol, so $\dfrac{45.98}{78.05}\times 100=58.91\%$. The sulfide is much richer in sodium, which is no surprise since it contains two per formula unit while sulfur is lighter than chlorine.

$2\,Na+2\,H_{2}O\rightarrow 2\,NaOH+H_{2}$. It is a redox reaction, and two oxidation numbers prove it: sodium goes from $0$ to $+1$, losing one electron per atom, so $2$ for the equation; the hydrogen of water goes from $+1$ to $0$ in the hydrogen gas, gaining one electron per atom, so $2$ for the two atoms of the molecule formed. Balance $2=2$. Oxygen stays at $-2$ on both sides: it takes no part. The sodium hydroxide formed is what makes the solution basic, and the hydrogen released is what ignites in contact with air, which finally confirms the identification of $A$.

Aluminum is at the very top of the activity series, among the metals that hold their valence electrons least tightly. It gives them up to the first oxidizing agent that comes along, and the oxygen of the air is more than enough. Any exposed metallic aluminum therefore oxidizes immediately, and it exists on Earth only at oxidation number $+3$, as an oxide or a silicate. Gold, at the other end of the series, does exactly the opposite: no natural oxidizing agent attacks it, it stays at $0$, and it is found in nuggets. The abundance of an element and the ease of obtaining it in the metallic state are two unrelated questions.

Iron goes from $+3$ in $Fe_{2}O_{3}$ to $0$ in the metal: it is REDUCED, so iron oxide is the OXIDIZING AGENT. Carbon goes from $+2$ in $CO$ to $+4$ in $CO_{2}$: it is OXIDIZED, so carbon monoxide is the REDUCING AGENT. Balance: $2$ irons each gain $3$ electrons, so $6$ electrons are gained; $3$ carbons each lose $2$ electrons, so $6$ electrons are given up. $6=6$, the balance closes, which validates the coefficients. The whole blast furnace fits in this sentence: carbon is a strong enough reducing agent to pull iron out of its oxide.

Because aluminum is ABOVE carbon in the activity series, whereas iron is below it. A reducing agent can strip oxygen only from a metal that is a weaker reducing agent than itself: carbon manages it with iron, never with aluminum. The reaction $Al_{2}O_{3}+3\,C$ does not go the way we would hope at the temperatures of a blast furnace: the bond between aluminum and oxygen is stronger than the one carbon can offer, and aluminum can itself reduce oxides that carbon leaves intact, which is what thermite exploits. A reducing agent more powerful than any chemical reagent was therefore needed, electric current, with the Hall-Héroult process of 1886: it is precisely this date that makes the price of aluminum collapse.

$M(Al_{2}O_{3})=2(26.98)+3(16.00)=53.96+48.00=101.96$ g/mol. Mass percent of aluminum: $\dfrac{53.96}{101.96}\times 100=52.92\%$. For $1.00$ tonne of aluminum, $\dfrac{1.00}{0.5292}=1.89$ tonnes of alumina are needed, and since bauxite contains only half of it, $\dfrac{1.89}{0.500}=3.78$ tonnes of ore. Almost four tonnes mined for one tonne of metal, and this ratio takes into account neither the energy nor the red mud of the Bayer process.

Because the metal to be recycled is ALREADY at oxidation number $0$. All the expense of primary production goes into giving three electrons back to each aluminum atom, which is at $+3$ in its ore, and that is work that only electricity can do, in large quantity. The used metal, for its part, has no electrons left to receive: it only has to be MELTED, at $660$ °C, which requires only an ordinary latent heat. A change of state is paid for instead of a reduction, hence the factor of twenty. This is the conclusion of the whole set: the position of an element in the periodic table, then in the activity series, decides the price of its metal as much as its chemistry.