[{"data":1,"prerenderedAt":197},["ShallowReactive",2],{"topic-main-group-chemistry":3},{"topic":4,"prev":183,"next":189},{"id":5,"slug":6,"title":7,"shortTitle":7,"lesson":5,"category":8,"lastAlignmentAudit":9,"description":10,"metaDescription":11,"objectives":12,"conceptSections":28,"workedExample":86,"oerSources":95,"selfStudyQuestions":101,"relatedTopicSlugs":120,"relatedElements":124,"relatedReferences":135,"keyEquations":136,"howTo":153,"conceptProbe":161,"practiceTeaser":165,"gatedContent":173,"seoKeywords":176,"objectiveCount":182},29,"main-group-chemistry","Main Group Chemistry","advanced","2026-07-17","Predict the properties and reactions of the representative (main-group) elements from periodic trends, compare their allotropes, and learn the major industrial processes that make ammonia, nitric acid, sulfuric acid, and the metals.","Learn main group chemistry: periodic trends of the representative elements, alkali and alkaline-earth metal reactions, allotropes of carbon/phosphorus/oxygen/sulfur, hydrogen's reducing role, and the Haber, Ostwald, and contact processes. Free chemistry study guide.",[13,16,19,22,25],{"id":14,"text":15},"29.1","Predict the physical and chemical properties of a main-group element from its electron configuration and periodic-table position",{"id":17,"text":18},"29.2","Predict the products of, and balance the equations for, the characteristic reactions of alkali and alkaline-earth metals with water, oxygen, and the halogens",{"id":20,"text":21},"29.3","Compare the allotropes of carbon, oxygen, sulfur, and phosphorus and predict how their structural differences affect their properties",{"id":23,"text":24},"29.4","Predict the reactions of hydrogen and identify its bonding behavior and its role as a reducing agent",{"id":26,"text":27},"29.5","Identify the major industrial processes involving main-group elements (Haber-Bosch, Ostwald, contact process) and apply the chemical principles behind them",[29,34,38,47,56,65,74,82],{"heading":30,"content":31,"relatedObjectives":32},"Periodicity of the Representative Elements","\u003Cp>The \u003Cstrong>representative (main-group) elements\u003C/strong> are those in which the s and p orbitals are filling — groups 1, 2, and 13–18. Their chemistry follows directly from their position in the periodic table, so once you know where an element sits you can predict how it behaves.\u003C/p>\u003Cp>Two trends do most of the work. First, \u003Cstrong>metallic character increases down a group and decreases from left to right across a period\u003C/strong>: the alkali metals at the lower left are the most metallic and reactive, while the nonmetals at the upper right are the least metallic. Second, the \u003Cstrong>group number gives the characteristic ion charge\u003C/strong> — group 1 forms 1+ ions, group 2 forms 2+, group 13 forms 3+, group 16 forms 2−, and group 17 forms 1−. Combining a cation and an anion in the ratio that makes the compound neutral gives the formula (for example Na\u003Csup>+\u003C/sup> and O\u003Csup>2−\u003C/sup> give Na\u003Csub>2\u003C/sub>O; Al\u003Csup>3+\u003C/sup> and Se\u003Csup>2−\u003C/sup> give Al\u003Csub>2\u003C/sub>Se\u003Csub>3\u003C/sub>).\u003C/p>\u003Cp>The alkali metals are named for the fact that they and their oxides react with water to form strongly basic (alkaline) solutions; the alkaline-earth metals are named the same way for their heavier oxides.\u003C/p>",[33],291,{"heading":35,"content":36,"relatedObjectives":37},"Periodic Trends for Predicting and Ranking Properties","\u003Cp>To compare elements quantitatively, use the periodic trends in size and in how tightly an atom holds its electrons. \u003Cstrong>Atomic radius increases down a group and decreases across a period.\u003C/strong> \u003Cstrong>Ionization energy and electronegativity decrease down a group and increase across a period.\u003C/strong>\u003C/p>\u003Cp>These trends let you rank a set of elements or pick the “more” member of a pair. Within the halogens, for instance, fluorine (top of group 17) has the highest electronegativity and the highest first ionization energy. Reactivity of the metals runs the opposite way: a larger atom lower in a group holds its valence electron(s) more loosely, so it is oxidized more readily — rubidium reacts with water far more vigorously than sodium, and the alkali metals (group 1) are more reactive than the alkaline-earth metals (group 2) in the same period.\u003C/p>\u003Cp>The same reasoning predicts metallic versus covalent behavior: elements low and to the left are the most metallic, and those high and to the right form the most covalent, molecular substances.\u003C/p>\u003Cp>\u003Cstrong>Covalent character within a family of ionic compounds\u003C/strong> is decided by the \u003Cstrong>cation, not the element trend\u003C/strong>. A smaller, more highly charged cation (higher charge density) pulls on — \u003Cem>polarizes\u003C/em> — the electron cloud of its anion, sharing it partially and giving the bond covalent character. Among NaCl, CaCl₂, and BeCl₂, the tiny, doubly charged Be²⁺ is by far the most polarizing, so BeCl₂ is the most covalent of the three chlorides even though all three are metal–chloride compounds. Reach for this cation-polarizing-power reasoning whenever a question compares the covalency of salts of the \u003Cem>same\u003C/em> anion; the ordinary electronegativity trend compares elements, not cations.\u003C/p>",[33],{"heading":39,"content":40,"relatedObjectives":41,"deepDive":43},"Reactions of the Alkali and Alkaline-Earth Metals","\u003Cp>The active metals of groups 1 and 2 share three characteristic reactions.\u003C/p>\u003Cp>\u003Cstrong>With water\u003C/strong> they give the metal hydroxide and hydrogen gas: 2M + 2H\u003Csub>2\u003C/sub>O → 2MOH + H\u003Csub>2\u003C/sub> for group 1, and M + 2H\u003Csub>2\u003C/sub>O → M(OH)\u003Csub>2\u003C/sub> + H\u003Csub>2\u003C/sub> for group 2. The reaction is more vigorous for the larger, more reactive metals lower in each group.\u003C/p>\u003Cp>\u003Cstrong>With oxygen\u003C/strong> the group-1 product follows a trend set by cation size: lithium gives the simple oxide (Li\u003Csub>2\u003C/sub>O), sodium gives the peroxide (Na\u003Csub>2\u003C/sub>O\u003Csub>2\u003C/sub>), and potassium, rubidium, and cesium give the superoxide (for example KO\u003Csub>2\u003C/sub>). The alkaline-earth metals give the simple oxide MO.\u003C/p>\u003Cp>\u003Cstrong>With the halogens\u003C/strong> both groups give the ionic metal halide (for example 2Na + Cl\u003Csub>2\u003C/sub> → 2NaCl; Sr + Br\u003Csub>2\u003C/sub> → SrBr\u003Csub>2\u003C/sub>). In every case the metal is oxidized and the nonmetal is reduced; balancing the equation is just a matter of matching the ion charges and then conserving atoms.\u003C/p>",[42],292,[44],{"label":45,"body":46},"Why this matters: why sodium ships in a jar of oil","\u003Cp>Open a chemical stockroom and the Group 1 metals sit submerged in mineral oil: the water reaction on this page is why. Exposed sodium reacts with mere air humidity, and a pellet dropped into water dashes across the surface on a cushion of hydrogen gas that the reaction’s own heat can ignite. Cesium, at the bottom of the group where the trend peaks, shatters its container on contact with water.\u003C/p>\u003Cp>The oil is not fussiness; it is the storage protocol the group’s reactivity trend dictates. Reading down the group, the demonstrations get less like chemistry class and more like pyrotechnics, exactly as the increasing-activity trend predicts.\u003C/p>",{"heading":48,"content":49,"relatedObjectives":50,"deepDive":52},"Allotropes: How Structure Drives Properties","\u003Cp>\u003Cstrong>Allotropes\u003C/strong> are different structural forms of the same element, and the structural difference is what drives the difference in properties. The recurring theme is the \u003Cstrong>strength of sideways (π) overlap between p orbitals\u003C/strong>, which is strong for small period-2 atoms and weak for larger atoms below them.\u003C/p>\u003Cp>\u003Cstrong>Carbon:\u003C/strong> graphite is built from stacked sheets with a delocalized π system (soft, electrically conducting), while diamond is a three-dimensional σ network (very hard, insulating). Silicon, being larger, cannot sustain that π system and forms only the diamond-type network.\u003C/p>\u003Cp>\u003Cstrong>Nitrogen vs. phosphorus:\u003C/strong> small nitrogen makes a strong N≡N triple bond and exists as diatomic N\u003Csub>2\u003C/sub>; larger phosphorus has weak π overlap, so it forms single P–P bonds and builds the tetrahedral P\u003Csub>4\u003C/sub> molecule (reactive white phosphorus) or polymeric red phosphorus.\u003C/p>\u003Cp>\u003Cstrong>Oxygen vs. sulfur:\u003C/strong> oxygen forms the double-bonded diatomic O\u003Csub>2\u003C/sub> and the higher-energy allotrope ozone, O\u003Csub>3\u003C/sub>; larger sulfur cannot form strong π bonds and instead links into crown-shaped S\u003Csub>8\u003C/sub> rings.\u003C/p>",[51],293,[53],{"label":54,"body":55},"Why this matters: one element, guardian and pollutant","\u003Cp>Oxygen’s two allotropes have opposite reputations. O\u003Csub>2\u003C/sub> is the breathable workhorse; O\u003Csub>3\u003C/sub> (ozone) is a lung irritant and smog component at street level, yet twenty kilometers up, the stratospheric ozone layer absorbs the ultraviolet radiation that would otherwise sterilize the surface. Same element, and the extra atom changes everything, including which reputation applies at which altitude.\u003C/p>\u003Cp>Phosphorus makes the same point on a matchbox: white phosphorus is so reactive it ignites in air and is stored underwater, while its polymeric red allotrope waits patiently on the striker strip until friction asks. Structure, not identity, sets the personality.\u003C/p>",{"heading":57,"content":58,"relatedObjectives":59,"deepDive":61},"Hydrogen: Bonding Behavior and Its Role as a Reducing Agent","\u003Cp>Hydrogen sits by itself: its oxidation state depends on its partner. It is \u003Cstrong>+1 when bonded to a more electronegative nonmetal\u003C/strong> (as in H\u003Csub>2\u003C/sub>O or HCl) and \u003Cstrong>−1 in ionic hydrides\u003C/strong> formed with active metals (as in NaH or CaH\u003Csub>2\u003C/sub>).\u003C/p>\u003Cp>A defining part of hydrogen's chemistry is that it acts as a \u003Cstrong>reducing agent\u003C/strong>. Hydrogen gas reduces the oxides of less-active metals to the metal or a lower oxide while being oxidized to water, for example H\u003Csub>2\u003C/sub> + MnO\u003Csub>2\u003C/sub> → MnO + H\u003Csub>2\u003C/sub>O.\u003C/p>\u003Cp>An ionic hydride reacts with water to give the metal hydroxide plus hydrogen gas: CaH\u003Csub>2\u003C/sub> + 2H\u003Csub>2\u003C/sub>O → Ca(OH)\u003Csub>2\u003C/sub> + 2H\u003Csub>2\u003C/sub>. Here the hydride ion (H\u003Csup>−\u003C/sup>) is both the Lewis base (it donates the electron pair) and the reducing species (it is oxidized from −1 to 0), while water acts as the Lewis acid and the oxidizing agent — the two H atoms of each product H\u003Csub>2\u003C/sub> come from two different sources.\u003C/p>",[60],294,[62],{"label":63,"body":64},"Why this matters: hydrogen as the clean reducing agent","\u003Cp>The defining pattern of this section, hydrogen pulling oxygen away from other substances, is becoming climate infrastructure. Conventional steelmaking reduces iron ore with coke, exhaling CO\u003Csub>2\u003C/sub> by the ton; “green steel” pilot plants substitute H\u003Csub>2\u003C/sub> as the reducing agent, and the exhaust becomes water vapor. A fuel-cell vehicle runs the same clean oxidation electrochemically: hydrogen in, electricity out, water dripping from the tailpipe.\u003C/p>\u003Cp>Whether the hydrogen economy scales is an engineering and economics question, but its chemistry is exactly the reducing-agent behavior on this page.\u003C/p>",{"heading":66,"content":67,"relatedObjectives":68,"deepDive":70},"Industrial Processes: Haber-Bosch, Ostwald, and Contact","\u003Cp>Three named processes turn main-group elements into some of the most heavily produced industrial chemicals, and each illustrates how reaction chemistry, kinetics, and economics jointly determine a process.\u003C/p>\u003Cp>The \u003Cstrong>Haber-Bosch process\u003C/strong> makes ammonia: N\u003Csub>2\u003C/sub> + 3H\u003Csub>2\u003C/sub> ⇌ 2NH\u003Csub>3\u003C/sub> over an iron catalyst. The \u003Cstrong>Ostwald process\u003C/strong> turns that ammonia into nitric acid in stages: 4NH\u003Csub>3\u003C/sub> + 5O\u003Csub>2\u003C/sub> → 4NO + 6H\u003Csub>2\u003C/sub>O, then 2NO + O\u003Csub>2\u003C/sub> → 2NO\u003Csub>2\u003C/sub>, then 3NO\u003Csub>2\u003C/sub> + H\u003Csub>2\u003C/sub>O → 2HNO\u003Csub>3\u003C/sub> + NO. The \u003Cstrong>contact process\u003C/strong> makes sulfuric acid by oxidizing sulfur dioxide over a vanadium(V) oxide catalyst, 2SO\u003Csub>2\u003C/sub> + O\u003Csub>2\u003C/sub> ⇌ 2SO\u003Csub>3\u003C/sub>, then absorbing the SO\u003Csub>3\u003C/sub> into sulfuric acid to form oleum, which is diluted to H\u003Csub>2\u003C/sub>SO\u003Csub>4\u003C/sub>.\u003C/p>\u003Cp>The reversible product-forming steps in Haber and the contact process are both \u003Cstrong>exothermic and mole-reducing\u003C/strong>, so higher pressure and lower temperature favor the products at equilibrium. Their operating choices differ: Haber uses high pressure because the yield gain justifies compression, whereas the contact process operates near atmospheric pressure because SO\u003Csub>3\u003C/sub> conversion is already high and further compression is not economical. Both use a moderate temperature and a catalyst to obtain a workable rate.\u003C/p>",[69],295,[71],{"label":72,"body":73},"Sulfuric acid as an economic vital sign","\u003Cp>The Contact process’s product is no ordinary chemical: sulfuric acid has long been produced in greater tonnage than any other industrial chemical, feeding fertilizer manufacture above all, plus metal processing, batteries, and chemical synthesis. Economists once used a nation’s sulfuric acid consumption as a rough index of its industrialization, the way electricity use serves today.\u003C/p>\u003Cp>That is worth a pause: the Haber-Bosch, Ostwald, and contact processes quietly underwrite the world’s food supply and heavy industry. The equilibrium, rate, and catalytic reasoning you learned in earlier topics is not academic; it is load-bearing civilization.\u003C/p>",{"heading":75,"content":76,"relatedObjectives":77,"deepDive":78},"Metallurgy and the Manufacture of Phosphorus","\u003Cp>How a metal is won from its ore depends on how reactive it is. \u003Cstrong>Very active metals are obtained by electrolytic reduction of a molten salt\u003C/strong>, because no ordinary chemical reductant can reduce them: the Downs cell electrolyzes molten NaCl to give sodium (2NaCl → 2Na + Cl\u003Csub>2\u003C/sub>), and the Hall-Héroult cell electrolyzes alumina dissolved in molten cryolite to give aluminum (cathode: Al\u003Csup>3+\u003C/sup> + 3e\u003Csup>−\u003C/sup> → Al). \u003Cstrong>Less-active metals are obtained by chemical reduction with carbon or carbon monoxide\u003C/strong> (for example ZnO + C → Zn + CO; SnO\u003Csub>2\u003C/sub> + 2C → Sn + 2CO).\u003C/p>\u003Cp>Phosphorus is produced industrially by the same carbon-reduction idea applied to a nonmetal: phosphate rock is heated with sand and coke, 2Ca\u003Csub>3\u003C/sub>(PO\u003Csub>4\u003C/sub>)\u003Csub>2\u003C/sub> + 6SiO\u003Csub>2\u003C/sub> + 10C → 6CaSiO\u003Csub>3\u003C/sub> + 10CO + P\u003Csub>4\u003C/sub>, and the volatile P\u003Csub>4\u003C/sub> distills off. Burning that phosphorus gives the acidic oxide P\u003Csub>4\u003C/sub>O\u003Csub>10\u003C/sub> (P\u003Csub>4\u003C/sub> + 5O\u003Csub>2\u003C/sub> → P\u003Csub>4\u003C/sub>O\u003Csub>10\u003C/sub>), which reacts with water to give phosphoric acid: P\u003Csub>4\u003C/sub>O\u003Csub>10\u003C/sub> + 6H\u003Csub>2\u003C/sub>O → 4H\u003Csub>3\u003C/sub>PO\u003Csub>4\u003C/sub>.\u003C/p>\u003Cp>One more lever: \u003Cstrong>removing a product can drive an otherwise unfavorable reduction\u003C/strong>. Calcium normally cannot reduce Cs⁺ (cesium is the more reactive metal), yet Ca + 2CsCl → CaCl₂ + 2Cs works industrially because at the operating temperature the cesium is volatile and distills out of the melt — continuously removing a product pulls the equilibrium forward (Le Chatelier), just as the escape of volatile P₄ drives the phosphorus furnace.\u003C/p>",[69],[79],{"label":80,"body":81},"The Bronze Age was an activity-series fact","\u003Cp>Human history followed the reduction difficulty ranked on this page. Gold and silver, needing no reduction at all, were worked first. Copper and tin oxides surrender to charcoal at the modest furnace temperatures early metalworkers could reach: hence bronze, millennia before iron, whose ores demand hotter furnaces and better technique. Aluminum, despite being the most abundant metal in Earth’s crust, waited until 1886, because no chemical reductant handles it economically: it took electrolysis.\u003C/p>\u003Cp>Read the metallurgy methods in this section as a timeline and they tell the same story: the harder a metal grips its oxygen, the later humanity got to use it.\u003C/p>",{"heading":83,"content":84,"relatedObjectives":85},"Decision Framework and Common Mistakes","\u003Cp>A reliable way to work main-group problems:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Read the position.\u003C/strong> Use the group number for the ion charge and the up/down, left/right trends for metallic character, size, ionization energy, and electronegativity.\u003C/li>\u003Cli>\u003Cstrong>Match the reaction to the family.\u003C/strong> Active metal + water → hydroxide + H\u003Csub>2\u003C/sub>; metal + oxygen → the size-appropriate oxide/peroxide/superoxide; metal + halogen → ionic halide.\u003C/li>\u003Cli>\u003Cstrong>Let structure explain properties.\u003C/strong> When comparing allotropes, argue from the strength of π overlap (small atoms make strong multiple bonds; larger atoms make single-bonded networks or rings).\u003C/li>\u003Cli>\u003Cstrong>For industrial steps\u003C/strong>, name the process from its inputs and outputs, write the balanced equation, predict the equilibrium shift, and then distinguish that thermodynamic effect from the economic operating choice. Haber uses high pressure; the contact process operates near atmospheric pressure even though higher pressure would favor SO\u003Csub>3\u003C/sub>.\u003C/li>\u003C/ol>\u003Cp>Common mistakes: forgetting that the group-1 oxygen product changes from oxide to peroxide to superoxide down the group; assuming every element forms the same allotrope as the one above it (silicon has no graphite analog; phosphorus has no stable P\u003Csub>2\u003C/sub>); confusing the direction of an equilibrium shift with the most economical plant conditions; and trying to reduce sodium or aluminum chemically with carbon when they require electrolysis.\u003C/p>",[33,42,51,60,69],{"title":87,"problem":88,"steps":89,"answer":94},"Predicting and Balancing a Main-Group Reaction","Predict the product and write the balanced equation for the reaction of potassium with water, and separately for magnesium burning in oxygen.",[90,91,92,93],"Identify the family: potassium is a group-1 (alkali) metal and magnesium is a group-2 (alkaline-earth) metal, so both are active metals.","Potassium + water gives the metal hydroxide + hydrogen gas. Balance with the 1+ charge of K: 2K + 2H2O -> 2KOH + H2.","Magnesium + oxygen gives the simple oxide (group 2 always forms MO). Balance Mg2+ with O2-: 2Mg + O2 -> 2MgO.","Check each equation conserves atoms and charge: K (2=2), O and H balance in the first; Mg (2=2) and O (2=2) in the second.","2K + 2H2O -> 2KOH + H2 (potassium hydroxide + hydrogen gas), and 2Mg + O2 -> 2MgO (magnesium oxide). Both reactions oxidize the metal; the alkali-metal reaction with water is the more vigorous of the two.",[96,97,98,99,100],"OpenStax Chemistry 2e, Ch 18.1: Periodicity (CC BY 4.0)","OpenStax Chemistry 2e, Ch 18.2: Occurrence and Preparation of the Representative Metals (CC BY 4.0)","OpenStax Chemistry 2e, Ch 18.4: Structure and General Properties of the Nonmetals (CC BY 4.0)","OpenStax Chemistry 2e, Ch 18.5: Occurrence, Preparation, and Compounds of Hydrogen (CC BY 4.0)","OpenStax Chemistry 2e, Ch 18.7-18.10: Nitrogen, Phosphorus, Oxygen, and Sulfur (CC BY 4.0)",[102,104,107,109,111,114,116,118],{"question":103},"How does the periodic table position of a main-group element tell you its characteristic ion charge and how metallic it is?",{"question":105,"hint":106},"Rank a set of main-group elements by electronegativity, ionization energy, or atomic radius using the periodic trends.","Across a period vs. down a group — the two directions move size and ionization energy in opposite senses.",{"question":108},"What are the products when an alkali or alkaline-earth metal reacts with water, with oxygen, and with a halogen?",{"question":110},"Why does the group-1 oxygen product change from oxide (Li) to peroxide (Na) to superoxide (K, Rb, Cs)?",{"question":112,"hint":113},"Why does carbon form graphite but silicon does not, and why is nitrogen diatomic (N2) while phosphorus is tetrahedral (P4)?","Compare the strength of sideways pi overlap for small period-2 atoms versus larger atoms.",{"question":115},"What oxidation states does hydrogen take, and how does it act as a reducing agent?",{"question":117},"Name the Haber, Ostwald, and contact processes from their reactants and products. Then compare the pressure and temperature choices for Haber and the contact process, distinguishing the equilibrium effect from the industrial operating choice.",{"question":119},"When is a metal obtained by electrolysis versus by chemical reduction with carbon?",[121,122,123],"periodic-properties-and-trends","predicting-and-balancing-chemical-reactions","redox-reactions-and-redox-balancing",[125,126,127,128,129,130,131,132,133,134],"Na","K","Mg","Al","C","N","P","O","S","H",[],[137,141,145,149],{"label":138,"equation":139,"note":140},"Active metal + water","2M + 2H2O -> 2MOH + H2  (group 1);  M + 2H2O -> M(OH)2 + H2  (group 2)","M is the alkali or alkaline-earth metal; the metal is oxidized and hydrogen is released.",{"label":142,"equation":143,"note":144},"Haber-Bosch (ammonia)","N2 + 3H2 ⇌ 2NH3","Exothermic and mole-reducing; high pressure + moderate temperature + iron catalyst favor NH3.",{"label":146,"equation":147,"note":148},"Contact process (sulfuric acid)","2SO2 + O2 ⇌ 2SO3  (V2O5 catalyst), then SO3 -> oleum -> H2SO4","Higher pressure and lower temperature favor SO3 at equilibrium; plants use a V2O5 catalyst near 450 °C and near-atmospheric pressure because conversion is already high.",{"label":150,"equation":151,"note":152},"Phosphoric acid from phosphorus(V) oxide","P4 + 5O2 -> P4O10;  P4O10 + 6H2O -> 4H3PO4","An acidic (nonmetal) oxide plus water gives an oxyacid.",{"title":154,"steps":155},"How to Predict the Products of a Main-Group Reaction",[156,157,158,159,160],"Identify each element's group and read its characteristic ion charge (group 1 -> 1+, group 2 -> 2+, group 13 -> 3+, group 16 -> 2-, group 17 -> 1-).","Match the reaction to its family: active metal + water -> metal hydroxide + H2; metal + oxygen -> the size-appropriate oxide, peroxide, or superoxide; metal + halogen -> ionic metal halide.","Write the formula of each product by combining the ions in the ratio that makes the compound neutral.","Balance the equation by conserving atoms (the charges are already satisfied by the formulas).","For an industrial process, name it from its inputs and outputs, write the balanced step(s), and apply Le Chatelier to any reversible exothermic, mole-reducing step.",{"question":162,"answer":163,"type":164},"A student claims that, just as carbon forms graphite, silicon should also have a graphite-like allotrope, and that phosphorus should exist as P2 molecules just as nitrogen exists as N2. Explain why both claims are wrong.","Both claims ignore the effect of atom size on sideways (pi) overlap. Graphite and N2 depend on strong pi bonding between small period-2 atoms: carbon's small 2p orbitals overlap well to give graphite's delocalized pi sheets, and nitrogen's give a strong N triple bond, so nitrogen is diatomic. Silicon and phosphorus are larger (period 3), so their 3p orbitals overlap sideways too weakly to sustain that pi bonding. Silicon therefore forms only the diamond-type 3-D network (no graphite analog), and phosphorus forms single P-P bonds, building the tetrahedral P4 molecule instead of P2.","conceptual",[166,169],{"id":167,"problem":168,"type":164},"pt-29-1","Write the balanced equation for the reduction of cesium chloride by elemental calcium at high temperature, and explain why the reaction proceeds even though cesium is normally more reactive than calcium.",{"id":170,"problem":171,"type":172},"pt-29-2","How many kilograms of Ca3(PO4)2 are needed to prepare 5.0 kg of phosphorus by carbothermal reduction if the yield is 90%?","calculation",{"workedExampleCount":174,"hasWorksheets":175},10,true,[177,178,179,180,181],"main group chemistry","representative elements","alkali and alkaline earth metals","allotropes","Haber Ostwald contact process",5,{"id":184,"slug":185,"lesson":184,"title":186,"shortTitle":186,"description":187,"category":8,"objectiveCount":182,"problemCount":188},28,"nuclear-chemistry","Nuclear Chemistry","Explore radioactive decay, nuclear reactions, half-life calculations, fission, and fusion.",38,{"id":190,"slug":191,"lesson":190,"title":192,"shortTitle":193,"description":194,"category":8,"objectiveCount":195,"problemCount":196},30,"review-and-capstone","Review and Capstone","Review & Capstone","A comprehensive end-of-course review built from a 50-question practice final exam, organized into seven strands that span the whole general-chemistry sequence. Recall the key method for each strand, refresh the prerequisites, then test every skill on fresh practice problems.",7,50,1785108608238]