1.9 Inorganic Chemistry Practice Problems – Chemistry GRE Subject Test – Flashcards

*+4* Solving this requires you to remember that the oxidation number of halogens is -1, and the sum of all oxidation numbers must be 0 in a neutral atom or equal to the charge in an ion. The ion charge here is -2, and so to find the oxidation number of Si we use: (Si#) + 6(-1) = -2 ----> (Si#) must be +4

*+6* Solving this requires you to remember that the oxidation number of the nitrate ion (NO3), is -1, and the sum of all oxidation numbers must be 0 in a neutral atom or equal to the charge in an ion. The ion charge here is +1, and so to find the oxidation number of Co we use: (Co#) + 5(-1) = +1 ----> (Co#) must be +6

*CN: 7 OX#: +6* The trick here is recognizing that Cs is an alkaline metal with a +1 ion charge. Since it is an ionic compound, XeF7- is the anion. *Cs is not coordinated with Xe*, as Cs is the cation. Only other atoms of the anion XeF7 contribute to the CN, and so and so the 7 fluorine atoms are the only atoms connected to xenon, yielding the coordination number 7. Determining the oxidation number in this case is the same as before (Xe#) + 1(+1) + 7(-1) = 0 ----> (Xe#) must be +6

*Nitrogen (-3 --> +2), oxygen (0 --> -2)* The key here is to remember that the oxidation number of a free element or atom in its elemental state (like O2) is 0. It is also important to note that H has an oxidation number of +1 in all compounds except metallic hydrides. From there, it is just a matter of comparing oxidation numbers on each side of the arrow to see how they've changed.

*(D) IEs increase from left to right across a period except for irregularities in atoms with three and six valence e-* In a period, all the valence electrons possess the same principle quantum number (n). As the number of electrons increases, so does the nuclear charge that binds them more strongly to the nucleus. This explains the increasing IEs from left to right across a period. In addition to this, there is a greater level of stability associated with half filled orbitals (configuration is np3 for a half filled p orbital, and the IE for np3 is greater than that of np4) as well as filled orbitals (ns2 and np6). Due to their greater stability, half filled and fully filled orbitals result in a higher IE for that atom. This explains the irregularities of atoms with 3 or 6 valence electrons when observing IE trends from left to right. *Elements with the highest IE values are in the top right corner of the periodic table*

*Atomic radii decrease with increasing Z from left to right across a period, but increase with increasing Z down a group.* Across a period, all of the valence electrons are at the same energy level (same value of n) and this energy level is pulled closer and closer to the nucleus as atomic number Z gets bigger, resulting in smaller radii. Down a group, however, energy levels are successively added, causing the valence electrons to be further and further from the nucleus as Z (and the value for n) increases, resulting in larger radii. *Elements with the greatest atomic radii are in the bottom left corner of the periodic table*

*(B) Tl+3* All of the ion choices are formed from group III elements, and since atomic and ionic radius increases moving down a group, Tl+3 is the largest. Ions follow essentially the same radii trend as atoms, but cations will usually be smaller than the neutral atom, while anions are usually larger. Here, they are all cations with the same charge so the trend is identical to that of atomic radii.

*(A) s-block metals* The strength of a base/acid formed by the reaction of an element's oxide with water is indicative of its metallic character. Highly metallic elements, like those in the s-block of the periodic table are likely to form strong bases: Li + 1/2 O2 --> Li2O followed by Li2O + H2O --> 2LI+ + 2OH- Nonmetals or weakly metallic elements, like those in the p block, are likely to form acids: SO3 + 3 H2O --> 2 H3O+ + SO4-2

*Fluorine* Nonmetals react by gaining electrons. Since fluorine is the most electronegative element, it has the strongest ability to gain electrons. This enables fluorine to react violently with elements that have very low electronegativities, such as Cs.

*Neutrons* These are two isotopes of polonium. Polonium always has the same atomic number, 84 (and thus always has 84 protons and 84 electrons). Isotopes share atomic number values, but differ in mass number A, which is the number of nucleons (protons + neutrons) in the nucleus. The isotopes have the same number of protons (we know that from the Z value and its identity) so any difference in mass number A must be due to neutrons, because A-Z = N.

*An increase of 1* During beta(-) decay, known as electron emission, one neutron decays into a proton, which stays in the nucleus, and a fast electron is ejected from the nucleus (called a beta particle). This causes the mass number A of the new element to stay the same as the parent, and the atomic number Z of the new element to be one more than the parent. *Beta(-) decay electron emission overall result: Daughter A = Parent A ; Daughter Z = Parent Z+1*

*Neutrons* During a nuclear fission reaction, one slow neutron is required to initiate the reaction, which produces three fast neutrons, perpetuating the chain reaction.

*False* Isotopes are atoms of the same element (meaning they have the same Z value) that have different values for A (not the same as seen in the question).

*1 Zn and 3 Cu* The key here is to realize that each unit cell does not have its own independent atoms, but shares its atoms with all the unit cells around it in a continuous structure. This means that for this fcc cell, the zinc occupying each of the eight corners belongs to eight adjacent cubic cells, thus each corner only has 1/8 of a zinc atom to itself. Each fcc cell has 8 corners, and 8 x 1/8 = 1, so there is one zinc atom. The copper atoms at each of the 6 faces in this unit cell each belong to two adjacent cubic cells, thus each face has 1/2 a copper atom to itself. Each fcc cell has 6 faces, and 6 x 1/2 = 3, so there are three copper atoms.

*I, II, and III* Increasing temperature (heat energy) excites electrons in a semiconductor, making them more mobile (and making more of them able to jump to the conduction band). In the same way, when exposed to light energy, electrons in a semiconductor may also be excited and thus able to jump into the conduction band. More electrons in the conduction band results in more electrical conductivity. The addition of impurities (doping) to a semiconductor is common practice to help increase electrical conductivity. This can be accomplished by an increase in the number of negative carriers (like the electron rich phosphorus added to silicon crystal) or an increase in the number of positive holes (like electron deficient boron added to silicon crystal).

*Q = S + I + 1/2 D - E -U* Q represents the sum of the heat produced in every step of the formation of NaF from solid sodium and gaseous fluorine atoms. The processes S (sublimation), I (ionization), and D (dissociation) are all endothermic, and so their energies are assigned a positive sign. Because only one fluorine is required, the value of D must be halved. Fluorine is the atom with the highest electron affinity and so taking on another electron is an exothermic process. *This is why E is assigned a negative sign above, however it should be noted that traditionally, electron affinity values are already negative*. This is probably a book error because the value of electron affinity should be subtracted when calculating Q if it is indeed exothermic, and assigning a negative sign to an already negative value is equivalent to saying the value should be added [ -(-EA) means +EA]. U is simply the energy released when the gaseous sodium ion binds to a gaseous fluorine ion. It is an exothermic reaction and is assigned a negative sign.

*(C) Sc* Sn, Si, C, and Ge all belong to group IV and have four valence electrons (ns2np2) thus allowing them to participate in four covalent bonds with four fluorine atoms Sc however is part of group IIIB (d block metals) and has only three valence electrons [Ar] 4s2 3d1 to participate in bonding, and thus cannot support covalent bonds to four fluorine atoms.

*Group II* This element has only two electrons in its highest (valence) electron shell (n = 3), therefore it must be a group II alkaline earth metal. Since we know it has 12 electrons, it is actually relatively easily to identify the element itself as Magnesium, which is in fact an alkaline earth metal.

*(A) form a 3+ ion* Elements in group IIIA have the electron configuration (ns2 np1), and are more likely to form cations by losing their electrons (as is typical of metals) than anions (it is the most direct path to an octet, after all). All elements in group IIIA are metals with the exception of boron, and they are all quite good conductors.

*(E) Gold* Gold offers very good resistance to acid and is known as a noble metal. The internet definition of this says that gold is a noble metal due to its resistance to corrosion and oxidation. It also mentioned that gold may be dissolved in aqua regia, a mixture of hydrochloric acid and nitric acid, so it would seem that gold is highly resistant to acid, but will react in some cases.

*Group I* This is behavior characteristic of the very reactive alkali metals, which readily react with water to form hydroxides as well as hydrogen: 2 Na (s) + 2 H2O (l) --> 2 NaOH (aq) + H2 (g)

*Group VIII* Noble gases have completely filled energy levels and are nonreactive under normal conditions. Under STP conditions, very weak interactive forces exist between individual atoms, and thus noble gases are gaseous at STP.

*Group V* Elements of group V are characterized by the electron configuration ns2np3. This is in accordance with Hund's Rule, which states that pairing of electrons may only occur after all individual orbitals in a given subshell have received one electron. Since np6 would indicate fully filled p orbitals, it is clear that np3 is indicative of half filled p orbitals, and all elements in group V have configuration ns2np3. (ns2 indicates a full s orbital).

*Group VI* Elements of group VI are characterized by the electron configuration ns2np4. This is in accordance with Hund's Rule, which states that pairing of electrons may only occur after all individual orbitals in a given subshell have received one electron. Since group V is characterized by half full p orbitals, and have configuration ns2np3, only one more electron is necessary to start pairing, and thus group VI with its ns2np4 configuration represents the first paired p orbitals.

*Group I* With their ns1 electronic configuration, all alkali metals have one valence electron, and the lowest ionization energies. They react very rapidly with nonmetals possessing highly negative electron affinity values.

*(D) Metal* Elemental lithium is an alkali metal (group I)

*(B) Nonpolar covalent compound* Although each of the four C-H bonds in methane are slightly polar themselves, the answer is in the symmetry of the molecule. The orientation of each of the bonds is such that the resultant dipole moment of the entire molecule is 0.

*(C) Polar covalent compound* The difference in electronegativity between the H and CL atoms is about 0.5, therefore, the diatomic HCl molecule (such as its gaseous form) is polar. HCl in aqueous solution however is considered a strong acid and dissociates completely.

*(E) Ionic substance* LiCl consists of an active metal and an active nonmetal. The difference in their electronegativities is about 2.5, characteristic of ionic substances.

*II only* Rb-Cl is ionic due to the large difference in electronegativity (~2.4) between the two atoms. This is visually obvious, because Rb and Cl are from opposite sides of the periodic table. *Be careful, since H (2.2) is also at the opposite side of the table from Cl (3.0), and thus H-Cl may seem ionic, but the actual difference in electronegativity between the two is only about 0.9, and thus the bond is not ionic but polar covalent.

*(D) A catalyst modifies the activation energy of a system*

*Strontium, Lithium, Boron, Chlorine* Electronegativity increases moving left to right or up a column on the periodic table. The most electronegative elements are located in the top right of the table.

Electronegativity is a measure of an atom's ability to attract bonding electrons in a molecule. Noble gases are generally nonreactive with other elements, they don't usually form molecules and thus do not need to be assigned an electronegativity value, which is molecule specific.

*(A) : O , (B) : Ar , (C) : Ca+2 , (D) : Br-* This is because atomic radii decrease with increasing Z from left to right across a period, but increase with increasing Z down a group.

*Barium, calcium, copper, sulfur, argon* This is because ionization energy increases moving left to right or up a column on the periodic table.

*Calcium, aluminum, carbon, oxygen, fluorine* Note that calcium is a member of the beryllium family and thus actually has a positive electron affinity. This means that it does not want to take on electrons at all, and to force it to do so takes energy (an endothermic reaction). This does not affect its position in this particular group.

*(A) H2O ; (B) NH3 ; (C) H2O* This is because basicity increases moving right to left or up a column on the periodic table.

*(B) Cl-* It is the only chemical species listed with 8 electrons, 7 valence and one from the negative charge. Don't be fooled by the Ar+, it has 8 valence electrons but has lost one from the positive charge.

There are no formal charges in SO2, all atoms are neutral. The geometry of SO2 is trigonal planar, but due to the lp, the shape of SO2 is bent. (Image)

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*(A) Ionic ; (B) Pure Covalent ; (C) Pure Covalent ; (D) Ionic ; (E) Ionic* In (A) and (D), the overall compounds are ionic, but the carbonate and ammonium (respectively) are bonded covalently internally. Multi-atom (polyatomic) ions like carbonate and ammonium are also called complex ions.

*(A) CaS and (B) Li2O* Since the ionic force between two ions is proportional to the charges of the ions, CaS (+2/-2) > NaI (+1/-1) and Li2O (+1/-2) > LiF (+1/-1).

*(A) and (B)* Hydrogen bonds exist only between a fluorine, oxygen, or a nitrogen atom and a hydrogen atom that is bound to a second fluorine, oxygen, or nitrogen atom (they don't have to be the same, just one of the three). In (A) the water is not actually hydrogen bonding with the glass, but the SiO2 is causing hydrogen bonding in the water due to the oxygens present in the glass.

*(A) CH3COOH * Acetic acid (CH3COOH) is the weakest acid because it has the smallest Ka. This is indicative of less dissociation in the solution. Another way to think of this is that, since a small Ka is related to a large pKa, and a large pKa is related to higher pH, this makes sense, because stronger acids have the lowest pHs.

*20 mL* This is a simple matter of recognizing that we can use the equation Mi x Vi = Mf x Vf to solve for our one missing variable, Vi. Rearrange the equation: Vi = Mf x Vf / Mi Solve: Vi = 0.1 M x 100 mL / 0.5 M Vi = 20 mL

*4* We know this because HCl is a strong acid, and thus we can assume that 100% of it has dissociated in the solution. That means that the [H+] of the solution is 10^-4 M. Since pH = -log[H+] ; pH = -log[10^-4M] ; pH = 4

*(A)* An equivalence point of pH = 7 on a titration curve is indicative of the titration of a strong acid with a strong base. HCl/NaOH are the only SA/SB pair listed. Equivalence points greater than 7 on a titration curve are indicative of a WA/SB pair while equivalence points less than 7 on a titration curve are indicative of a SA/WB pair.

*(B)* H2CO3, carbonic acid, is the only polyprotic acid listed. It is polyprotic because it has more than one ionizable proton. Polyprotic acids dissociate according to: H#A #H+ + A-# Ka = Ka1 + Ka2 +.... + Ka# where # = the number of ionizable protons. When # = 2, the polyprotic acid can be known as diprotic. H2CO3 has a # = 2 and dissociates as: 1. H2CO3 H+ + HCO3- Ka1 = 4.3 x 10^-7 2. HCO3- H+ + CO3-2 Ka2 = 4.8 x 10^-11

*It is reduced* The oxidizing agent is the species that causes another species to be oxidized. It causes other species to be oxidized by taking on their electrons, and so the oxidizing agent is reduced.