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Beyond Advanced · The p-Block Elements

About this Chemistry practice set on The p-Block Elements

This set contains 70 questions on The p-Block Elements, arranged to build skill gradually. Work through them in order, and return to any chapter where your accuracy is low. Try each question yourself before you submit, then open the step-by-step solution to compare your method.

In chemistry, separate the question into what must be recalled and what must be reasoned. Physical chemistry needs a clean setup with units, organic chemistry needs the mechanism written step by step, and inorganic chemistry rewards recalling the trend first and the exception second.

Read more in our guide to using previous year questions, study the topic in the concept notes, or browse all JEE preparation guides.

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1
Which of the following statements about the group 13 elements and their compounds is/are correct?
Not attempted
A
Gallium has a smaller atomic radius than aluminium and a slightly higher first ionisation enthalpy, because the ten 3d3d electrons shield the nuclear charge poorly.
B
TlI3\mathrm{TlI_3} is a salt made of Tl3+\mathrm{Tl^{3+}} and three I−\mathrm{I^-} ions, since Tl3+\mathrm{Tl^{3+}} is the stable state of thallium in its halides.
C
In Al2Cl6\mathrm{Al_2Cl_6} each of the two bridges Al−Cl−Al\mathrm{Al-Cl-Al} is a three-centre two-electron bond, exactly as the B−H−B\mathrm{B-H-B} bridges in B2H6\mathrm{B_2H_6}.
D
The first ionisation enthalpy of Tl is higher than that of In, because the 4f4f and 5d5d electrons poorly screen the 6s6s and 6p6p electrons.
2
A 1.98 g1.98\,\mathrm{g} sample of an alloy containing only Al, Mg and Si is analysed. With excess dilute HCl (Si does not react) it releases 1.792 L1.792\,\mathrm{L} of H2\mathrm{H_2} at STP. With excess hot concentrated NaOH (Mg does not react, and Al gives Na[Al(OH)4]\mathrm{Na[Al(OH)_4]} while Si gives Na2SiO3\mathrm{Na_2SiO_3}) it releases 2.016 L2.016\,\mathrm{L} of H2\mathrm{H_2} at STP. Take Al=27\mathrm{Al}=27, Mg=24\mathrm{Mg}=24, Si=28\mathrm{Si}=28 and a molar volume of 22.4 L mol−122.4\,\mathrm{L\,mol^{-1}}. What is the mass percentage of Si in the alloy?
Not attempted
Your answer: —
Correct answer: 21.21
3
Gas-phase measurements show that the heat released on forming the adduct X3B←NMe3\mathrm{X_3B\leftarrow NMe_3} becomes larger along BF3<BCl3<BBr3\mathrm{BF_3 < BCl_3 < BBr_3}, although fluorine is the most electronegative halogen. Which explanation correctly accounts for this order?
Not attempted
A
The B-F bond is the weakest of the boron-halogen bonds, so BF3\mathrm{BF_3} is the least stable acceptor and loses a fluoride most readily to the amine.
B
The inductive electron withdrawal is strongest for the larger halogens, so boron is most electron-deficient in BBr3\mathrm{BBr_3}.
C
On forming the adduct boron changes from planar sp2sp^2 to pyramidal sp3sp^3 and loses its pπp\pi-pπp\pi back bonding with the halogens; this loss, which must be paid for, is largest for the efficient 2p2p-2p2p overlap in BF3\mathrm{BF_3} and smallest for BBr3\mathrm{BBr_3}, so the net gain is smallest for BF3\mathrm{BF_3}.
D
BF3\mathrm{BF_3} is dimeric through fluorine bridges (as Al2Cl6\mathrm{Al_2Cl_6} is) and the dimer must be broken before the amine can bind, whereas BBr3\mathrm{BBr_3} is monomeric.
4
Gaseous aluminium chloride below about 450 K450\,\mathrm{K} has the molecular formula Al2Cl6\mathrm{Al_2Cl_6}. In it every Al atom is four-coordinate (each line in a diagram is an Al-Cl or Al-Al bond), there is no Al-Al bond, and every Cl atom is bonded to at least one Al atom. Which diagram (connectivity only, geometry ignored) satisfies all these conditions?
Not attempted
A
Diagram 1
AlAlClClClClClCl
B
Diagram 2
AlAlClClClClClCl
C
Diagram 3
AlAlClClClClClCl
D
Diagram 4
AlAlClClClClCl
5
Consider the real D2h_{2h} geometry of Al2Cl6\mathrm{Al_2Cl_6} (two edge-sharing tetrahedra). Write its Lewis structure so that each bridging chlorine forms one normal bond to one aluminium and donates one of its lone pairs to the other aluminium (dative bond). Let xx be the largest number of atoms of the molecule that can lie in a single plane in the real geometry, and yy the total number of lone pairs on the six chlorine atoms. What is x+yx+y?
Not attempted
A
2222
B
2020
C
2424
D
2121
6
Match List-I with List-II and select the correct combination. List-IList-IIP)Gallium1)dimeric, each metal four-coordinateQ)Boric acid in water2)strong oxidant, easily reduced to the +1 stateR)Tl3+(aq)3)liquid over a range of about 2370 KS)AlCl3 (vapour, below 450 K)4)accepts OH− from water, acts as a Lewis acid5)uses vacant d orbitals of the central atom to expand its octet\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \text{Gallium} & \text{1)} \text{dimeric, each metal four-coordinate} \\ \text{Q)} \text{Boric acid in water} & \text{2)} \text{strong oxidant, easily reduced to the +1 state} \\ \text{R)} \mathrm{Tl^{3+}(aq)} & \text{3)} \text{liquid over a range of about 2370 K} \\ \text{S)} \mathrm{AlCl_3\ (vapour,\ below\ 450\ K)} & \text{4)} \text{accepts }\mathrm{OH^-}\text{ from water, acts as a Lewis acid} \\ & \text{5)} \text{uses vacant }d\text{ orbitals of the central atom to expand its octet} \end{array}
Not attempted
A
P-1, Q-2, R-3, S-4
B
P-3, Q-4, R-1, S-2
C
P-4, Q-3, R-2, S-1
D
P-3, Q-4, R-2, S-1
7
Which plot of first ionisation enthalpy (vertical axis, arbitrary linear scale) against the element (B, Al, Ga, In, Tl along the horizontal axis) correctly shows the experimental trend for group 13?
Not attempted
A
Plot 1
BAlGaInTl
B
Plot 2
BAlGaInTl
C
Plot 3
BAlGaInTl
D
Plot 4
BAlGaInTl
8
A 2.1189 g2.1189\,\mathrm{g} sample of impure borax, Na2B4O7⋅10H2O\mathrm{Na_2B_4O_7\cdot 10H_2O} (M=381.4 g mol−1M=381.4\,\mathrm{g\,mol^{-1}}, 90.0%90.0\% pure by mass, impurities inert), is dissolved in water. The skeleton of the anion is shown in the figure (all bonds are single bonds; the borate carries a charge of 2−2-). Just enough HCl is added to convert the borate completely into B(OH)3\mathrm{B(OH)_3} (no other boron species remain). A large excess of mannitol is then added and the resulting solution is titrated with 0.100 M0.100\,\mathrm{M} NaOH to the phenolphthalein end point, treating B(OH)3\mathrm{B(OH)_3} as a monobasic acid in the presence of mannitol. What volume of NaOH, in mL, is needed for this final titration?
Not attempted
BOOBBOOOBOHOHOHOH
Your answer: —
Correct answer: 200
9
The anion in borax is [B4O5(OH)4]2−\mathrm{[B_4O_5(OH)_4]^{2-}}. Every boron atom carries exactly one OH group; each boron is either trigonal planar (sp2sp^2, three O neighbours, no formal charge) or tetrahedral (sp3sp^3, four O neighbours, formal charge −1-1); every other oxygen bridges exactly two boron atoms. How many B-O σ\sigma bonds does the anion contain?
Not attempted
A
1212
B
1414
C
1616
D
1818
10
The figure shows the structure of diborane. Which of the following statements is/are correct?
Not attempted
BBHHHHHHdashed: bridging B-H-B bonds
A
Excess NMe3\mathrm{NMe_3} cleaves B2H6\mathrm{B_2H_6} symmetrically to give two molecules of H3B⋅NMe3\mathrm{H_3B\cdot NMe_3}, in each of which boron is sp3sp^3 hybridised.
B
All six hydrogen atoms of B2H6\mathrm{B_2H_6} are chemically equivalent, since the molecule has only 1212 valence electrons and these are delocalised over all eight bonds.
C
The bridging B-H bonds are longer than the terminal B-H bonds, because each bridge is a three-centre two-electron bond in which two electrons hold three atoms together.
D
Heating the product of the unsymmetrical cleavage by NH3\mathrm{NH_3}, [BH2(NH3)2]+[BH4]−\mathrm{[BH_2(NH_3)_2]^+[BH_4]^-}, removes hydrogen and gives borazine in which every boron is sp2sp^2 hybridised.
11
Boric acid behaves as a very weak monobasic Lewis acid: B(OH)3+2H2O⇌[B(OH)4]−+H3O+\mathrm{B(OH)_3+2H_2O\rightleftharpoons[B(OH)_4]^-+H_3O^+}. In the presence of a large excess of mannitol, all boron-containing acid and base forms are in equilibrium with a single apparent dissociation constant Ka′=1.0×10−5K_a'=1.0\times10^{-5} (defined for total acid and total conjugate base). To 75.0 mL75.0\,\mathrm{mL} of 0.100 M0.100\,\mathrm{M} B(OH)3\mathrm{B(OH)_3} containing a large excess of mannitol is added 25.0 mL25.0\,\mathrm{mL} of 0.100 M0.100\,\mathrm{M} NaOH. Take log⁡2=0.301\log2=0.301. What is the pH of the resulting solution?
Not attempted
Your answer: —
Correct answer: 4.7
12
Solid boric acid consists of planar B(OH)3\mathrm{B(OH)_3} units arranged in flat layers. In a layer each unit donates its three O-H hydrogens to oxygen atoms of three neighbouring units and accepts three hydrogen bonds from three other neighbours (see the figure for the central unit). Counting every hydrogen bond only once, what is the average number of hydrogen bonds per H3BO3\mathrm{H_3BO_3} formula unit in a layer?
Not attempted
B(OH)₃B(OH)₃B(OH)₃B(OH)₃B(OH)₃B(OH)₃B(OH)₃arrow: O–H···O bond, donor → acceptor
A
66
B
33
C
99
D
1.51.5
13
A 1.000 g1.000\,\mathrm{g} sample of aluminium foil covered by a thin film of Al2O3\mathrm{Al_2O_3} (and nothing else) dissolves completely in excess hot aqueous NaOH, giving Na[Al(OH)4]\mathrm{Na[Al(OH)_4]} and 1.2096 L1.2096\,\mathrm{L} of H2\mathrm{H_2} at STP (22.4 L mol−122.4\,\mathrm{L\,mol^{-1}}). Take Al=27\mathrm{Al}=27 and O=16\mathrm{O}=16. What is the mass percentage of Al2O3\mathrm{Al_2O_3} in the foil?
Not attempted
A
2.88%2.88\%
B
1.32%1.32\%
C
27.1%27.1\%
D
2.80%2.80\%
14
Match List-I with List-II and select the correct combination. List-IList-IIP)B2H6+6H2O1)H3B⋅NMe3Q)B2H6 with excess NH3 at low temperature2)B3N3H6R)strong heating of the ionic product of Q3)[BH2(NH3)2]+[BH4]−S)B2H6 with excess NMe34)B(OH)3 and H25)LiBH4\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{B_2H_6+6H_2O} & \text{1)} \mathrm{H_3B\cdot NMe_3} \\ \text{Q)} \mathrm{B_2H_6}\text{ with excess }\mathrm{NH_3}\text{ at low temperature} & \text{2)} \mathrm{B_3N_3H_6} \\ \text{R)} \text{strong heating of the ionic product of Q} & \text{3)} \mathrm{[BH_2(NH_3)_2]^+[BH_4]^-} \\ \text{S)} \mathrm{B_2H_6}\text{ with excess }\mathrm{NMe_3} & \text{4)} \mathrm{B(OH)_3}\text{ and }\mathrm{H_2} \\ & \text{5)} \mathrm{LiBH_4} \end{array}
Not attempted
A
P-4, Q-3, R-2, S-1
B
P-3, Q-4, R-1, S-2
C
P-4, Q-2, R-3, S-1
D
P-1, Q-3, R-2, S-4
15
Borazine, B3N3H6\mathrm{B_3N_3H_6}, is called inorganic benzene. Which of the following statements is/are correct?
Not attempted
A
Because each B-N bond is polar, the planar D3hD_{3h} molecule has a large net dipole moment.
B
Borazine undergoes electrophilic substitution with HCl, replacing ring hydrogen by chlorine, because B-N is non-polar like C-C.
C
Three molecules of HCl add to borazine to give B3N3H9Cl3\mathrm{B_3N_3H_9Cl_3}, in which chlorine is bonded to boron and hydrogen to nitrogen.
D
The six π\pi electrons of the ring are supplied one by each ring atom, exactly as in benzene.
16
Which of the following statements about group 14 is/are correct?
Not attempted
A
Hydrolysis of CCl4\mathrm{CCl_4} to CO2\mathrm{CO_2} and HCl is thermodynamically favourable, yet CCl4\mathrm{CCl_4} is inert towards water; SiCl4\mathrm{SiCl_4} is hydrolysed at once.
B
In aqueous solution Ge(II) is more stable than Ge(IV), as predicted by the inert pair effect for the lighter members of the group.
C
PbO2\mathrm{PbO_2} oxidises concentrated HCl to Cl2\mathrm{Cl_2}, whereas SnO2\mathrm{SnO_2} does not, because the stability of the +2+2 state increases down the group.
D
[SiF6]2−\mathrm{[SiF_6]^{2-}} is known but [CF6]2−\mathrm{[CF_6]^{2-}} is not, because carbon has no dd orbitals and its covalency cannot exceed four.
17
A fullerene Cn\mathrm{C}_n is a closed polyhedral cage in which every carbon is sp2sp^2 hybridised and bonded to exactly three others, and the only faces are pentagons and hexagons. How many hexagonal faces does C128\mathrm{C}_{128} have?
Not attempted
A
6464
B
7474
C
5454
D
118118
18
Graphite has ABAB stacking of planar hexagonal carbon nets (layer B is displaced relative to layer A so that half the atoms of a layer lie directly above atoms of the layers adjacent to it, as in the figure). The C-C distance in a layer is 142.0 pm142.0\,\mathrm{pm}, the spacing between neighbouring layers is 340 pm340\,\mathrm{pm}. Take the hexagonal unit cell with a rhombic base (side aa, angle 120∘120^\circ) containing exactly one A layer and one B layer. Use 3=1.732\sqrt3=1.732, M(C)=12.01 g mol−1M(\mathrm{C})=12.01\,\mathrm{g\,mol^{-1}}, NA=6.022×1023 mol−1N_A=6.022\times10^{23}\,\mathrm{mol^{-1}}. What is the density of graphite in g cm−3\mathrm{g\,cm^{-3}}?
Not attempted
layer Alayer Blayer A340 pm340 pmside view of ABAB stacking (layers are hexagonal nets)
Your answer: —
Correct answer: 2.24
19
At 298 K298\,\mathrm{K} the standard Gibbs energy of formation of diamond from graphite is +2.85 kJ mol−1+2.85\,\mathrm{kJ\,mol^{-1}} (at 1 bar1\,\mathrm{bar}). The densities are 2.26 g cm−32.26\,\mathrm{g\,cm^{-3}} (graphite) and 3.51 g cm−33.51\,\mathrm{g\,cm^{-3}} (diamond), M(C)=12.01 g mol−1M(\mathrm{C})=12.01\,\mathrm{g\,mol^{-1}}. Treat both solids as incompressible and neglect the 1 bar1\,\mathrm{bar} standard pressure against the answer. At what pressure, in kbar (1 kbar=108 Pa1\,\mathrm{kbar}=10^{8}\,\mathrm{Pa}), do graphite and diamond become equally stable at 298 K298\,\mathrm{K}?
Not attempted
Your answer: —
Correct answer: 15.06
20
Red lead, Pb3O4\mathrm{Pb_3O_4}, is warmed with dilute nitric acid and the mixture is filtered. Which description of the residue and the filtrate is correct for the reaction of one mole of Pb3O4\mathrm{Pb_3O_4}?
Not attempted
A
All the lead dissolves as three moles of Pb(NO3)2\mathrm{Pb(NO_3)_2} and oxygen is evolved, since Pb4+\mathrm{Pb^{4+}} oxidises water.
B
A brown residue of PbO2\mathrm{PbO_2} (one mole) remains and the filtrate contains three moles of Pb(NO3)2\mathrm{Pb(NO_3)_2}.
C
All the lead dissolves as Pb(NO3)4\mathrm{Pb(NO_3)_4} and Pb(NO3)2\mathrm{Pb(NO_3)_2} in the ratio 1:21:2, because nitric acid is an oxidant that stabilises the +4+4 state.
D
A brown residue of PbO2\mathrm{PbO_2} (one mole) remains and the filtrate contains two moles of Pb(NO3)2\mathrm{Pb(NO_3)_2}; no oxidation-state change occurs in the reaction.
21
Match List-I with List-II and select the correct combination. List-IList-IIP)SnCl2(aq)1)inert towards water for kinetic reasonsQ)PbO22)fumes in moist air, hydrolysed to hydrated silica and HClR)CCl43)oxidises hot concentrated HCl to Cl2S)SiCl44)reduces HgCl2 to white Hg2Cl2, then grey Hg5)gas at room temperature\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{SnCl_2(aq)} & \text{1)} \text{inert towards water for kinetic reasons} \\ \text{Q)} \mathrm{PbO_2} & \text{2)} \text{fumes in moist air, hydrolysed to hydrated silica and HCl} \\ \text{R)} \mathrm{CCl_4} & \text{3)} \text{oxidises hot concentrated HCl to }\mathrm{Cl_2} \\ \text{S)} \mathrm{SiCl_4} & \text{4)} \text{reduces }\mathrm{HgCl_2}\text{ to white }\mathrm{Hg_2Cl_2}\text{, then grey Hg} \\ & \text{5)} \text{gas at room temperature} \end{array}
Not attempted
A
P-3, Q-4, R-2, S-1
B
P-4, Q-3, R-1, S-2
C
P-4, Q-3, R-2, S-1
D
P-1, Q-2, R-3, S-4
22
Take mean bond enthalpies (in kJ mol−1\mathrm{kJ\,mol^{-1}}): C=O\mathrm{C{=}O} in CO2\mathrm{CO_2} 799799, C−O\mathrm{C{-}O} 358358, Si=O\mathrm{Si{=}O} 640640, Si−O\mathrm{Si{-}O} 452452. Consider the conversion of a discrete molecule O=E=O\mathrm{O{=}E{=}O} into the network solid in which each E is bonded to four O by single bonds (each O shared between two E). Which of the following is/are correct?
Not attempted
A
For carbon the conversion to the network is endothermic by about 166 kJ166\,\mathrm{kJ} per mole of E, so CO2\mathrm{CO_2} is molecular.
B
For silicon the conversion is exothermic by about 528 kJ528\,\mathrm{kJ} per mole of Si, so silica is a network solid.
C
The π\pi contribution of the Si=O bond (D(Si=O)−D(Si−O)D(\mathrm{Si{=}O})-D(\mathrm{Si{-}O})) is larger than that of the C=O bond, which explains why silicon forms SiO2\mathrm{SiO_2} molecules.
D
If the C=O\mathrm{C{=}O} bond enthalpy were only 700 kJ mol−1700\,\mathrm{kJ\,mol^{-1}} the network form of carbon dioxide would be favoured on enthalpy grounds.
23
The silicate anion present in amphiboles has the empirical formula Si4O116−\mathrm{Si_4O_{11}^{6-}}; it is a double chain of corner-sharing SiO4\mathrm{SiO_4} tetrahedra (see the figure for a sketch). Each SiO4\mathrm{SiO_4} tetrahedron has four oxygen vertices, and every shared vertex belongs to exactly two tetrahedra. What is the average number of oxygen vertices shared per tetrahedron?
Not attempted
double chain (chains linked by shared O)
Your answer: —
Correct answer: 2.5
24
A branched silicone molecule is built from the following chlorosilanes: one CH3SiCl3\mathrm{CH_3SiCl_3} (branching unit T), three arms each made of five (CH3)2SiCl2\mathrm{(CH_3)_2SiCl_2} units, and three (CH3)3SiCl\mathrm{(CH_3)_3SiCl} molecules that cap the three chain ends (see the figure). All Si-Cl bonds are hydrolysed to Si-OH and then every Si-OH condenses to form Si-O-Si links, with no Si-OH left over. How many moles of water are consumed in total (hydrolysis plus condensation, net) in making one mole of this molecule?
Not attempted
TMMMT: branch unit; circles: (CH₃)₂SiO; M: (CH₃)₃Si end
A
3636
B
1919
C
1818
D
1515
25
Which of the following statements about carbon monoxide and carbon dioxide is/are correct?
Not attempted
A
In Ni(CO)4\mathrm{Ni(CO)_4} the metal donates electron density from filled dd orbitals into the empty π∗\pi^* orbitals of CO, which lowers the C-O bond order and the C-O stretching frequency relative to free CO.
B
In carboxyhaemoglobin CO is bonded to iron through its oxygen atom, whose lone pair is the better donor.
C
CO2\mathrm{CO_2} has no dipole moment because the C=O bonds are themselves non-polar.
D
CO dissolves in water to give formic acid, being its acid anhydride, so aqueous CO is acidic.
26
Coke is gasified in a blast of air (taken as 21% O221\%\ \mathrm{O_2}, 79% N279\%\ \mathrm{N_2} by volume) mixed with steam. The only reactions are 2C+O2→2CO2\mathrm{C}+\mathrm{O_2}\to2\mathrm{CO} (ΔH=−221 kJ\Delta H=-221\,\mathrm{kJ}) and C+H2O→CO+H2\mathrm{C}+\mathrm{H_2O}\to\mathrm{CO}+\mathrm{H_2} (ΔH=+131.3 kJ\Delta H=+131.3\,\mathrm{kJ}), and the steam flow is adjusted so that the gasifier is exactly thermoneutral (net heat change zero). All oxygen and steam are consumed. What is the mole percentage of H2\mathrm{H_2} in the exit gas (which contains CO\mathrm{CO}, H2\mathrm{H_2} and N2\mathrm{N_2})?
Not attempted
Your answer: —
Correct answer: 18.44
27
The silicate Si3O96−\mathrm{Si_3O_9^{6-}} (as found in benitoite) is a cyclic silicate built from corner-sharing SiO4\mathrm{SiO_4} tetrahedra, each drawn as a triangle with the oxygen vertices as dots. Which diagram represents this anion (the oxygen at a shared vertex is common to the two triangles)?
Not attempted
A
Diagram 1
B
Diagram 2
C
Diagram 3
D
Diagram 4
28
Zeolite 4A is an aluminosilicate whose anhydrous framework has the composition (AlO2)12(SiO2)12(\mathrm{AlO_2})_{12}(\mathrm{SiO_2})_{12} with all charge balanced by Na+\mathrm{Na^+}. As a water softener, Ca2+\mathrm{Ca^{2+}} replaces Na+\mathrm{Na^+} ion for ion on a charge basis. Atomic masses: Na=22.99\mathrm{Na}=22.99, Al=26.98\mathrm{Al}=26.98, Si=28.09\mathrm{Si}=28.09, O=16.00\mathrm{O}=16.00, Ca=40.08\mathrm{Ca}=40.08. How many grams of Ca2+\mathrm{Ca^{2+}} can 100 g100\,\mathrm{g} of the anhydrous sodium zeolite remove?
Not attempted
A
14.1 g14.1\,\mathrm{g}
B
28.2 g28.2\,\mathrm{g}
C
7.05 g7.05\,\mathrm{g}
D
10.9 g10.9\,\mathrm{g}
29
List-II gives the ratio of oxygen atoms to silicon atoms. Match List-I with List-II and select the correct combination. List-IList-IIP)Si2O76− (pyrosilicate)1)2.5Q)(SiO3)n2n− (cyclic or single chain)2)2.75R)Si4O116− (double chain)3)3.0S)(Si2O5)n2n− (sheet)4)3.55)4.0\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{Si_2O_7^{6-}}\ \text{(pyrosilicate)} & \text{1)} 2.5 \\ \text{Q)} (\mathrm{SiO_3})_n^{2n-}\ \text{(cyclic or single chain)} & \text{2)} 2.75 \\ \text{R)} \mathrm{Si_4O_{11}^{6-}}\ \text{(double chain)} & \text{3)} 3.0 \\ \text{S)} (\mathrm{Si_2O_5})_n^{2n-}\ \text{(sheet)} & \text{4)} 3.5 \\ & \text{5)} 4.0 \end{array}
Not attempted
A
P-3, Q-4, R-2, S-1
B
P-4, Q-3, R-1, S-2
C
P-4, Q-3, R-2, S-1
D
P-5, Q-4, R-3, S-2
30
A solution contains 0.050 mol0.050\,\mathrm{mol} of Ca(OH)2\mathrm{Ca(OH)_2} and 0.020 mol0.020\,\mathrm{mol} of NaOH. Carbon dioxide (0.085 mol0.085\,\mathrm{mol}) is passed through it until absorbed completely, and the system reaches equilibrium (all sodium and any dissolved calcium remain in solution as carbonate or bicarbonate species; M(CaCO3)=100M(\mathrm{CaCO_3})=100). What mass of CaCO3\mathrm{CaCO_3} is present as a precipitate at the end?
Not attempted
A
3.50 g3.50\,\mathrm{g}
B
1.50 g1.50\,\mathrm{g}
C
2.50 g2.50\,\mathrm{g}
D
5.00 g5.00\,\mathrm{g}
31
0.600 mol0.600\,\mathrm{mol} of copper and 0.400 mol0.400\,\mathrm{mol} of zinc are dissolved together in excess dilute nitric acid. Copper is oxidised to Cu2+\mathrm{Cu^{2+}} with NO as the only gaseous reduction product of nitrate; zinc is oxidised to Zn2+\mathrm{Zn^{2+}} with N2O\mathrm{N_2O} as the only gaseous reduction product (as in the usual equations for dilute acid). What volume of gas, measured at STP, is evolved altogether?
Not attempted
A
14.9 L14.9\,\mathrm{L}
B
13.4 L13.4\,\mathrm{L}
C
8.96 L8.96\,\mathrm{L}
D
11.2 L11.2\,\mathrm{L}
32
Which of the following statements about the oxides and oxo-ions of nitrogen is/are correct?
Not attempted
A
The N-O bond length order is NO+<NO<NO−\mathrm{NO^+ < NO < NO^-}: the bond orders are 33, 2.52.5 and 22.
B
NO2\mathrm{NO_2} dimerises to diamagnetic N2O4\mathrm{N_2O_4} by pairing the unpaired electron located largely on nitrogen, whereas the unpaired electron of NO sits in a π∗\pi^* orbital shared by both atoms and gives little gain on dimerisation.
C
Solid N2O5\mathrm{N_2O_5} is ionic, built from NO2+\mathrm{NO_2^+} and NO3−\mathrm{NO_3^-}.
D
N2O\mathrm{N_2O} has the symmetrical structure N−O−N\mathrm{N-O-N} and is isoelectronic with CO2\mathrm{CO_2} with the oxygen as the central atom.
33
Which of the following statements about bond angles in nitrogen compounds is/are correct?
Not attempted
A
The F-N-F angle of NF3\mathrm{NF_3} (≈102∘\approx102^\circ) is smaller than the H-N-H angle of NH3\mathrm{NH_3}, because the bonding pairs of N-F are drawn towards F, away from nitrogen, reducing bond pair-bond pair repulsion.
B
The H-N-H angle of NH3\mathrm{NH_3} (107.8∘107.8^\circ) is larger than the H-P-H angle of PH3\mathrm{PH_3} (≈93.6∘\approx93.6^\circ), because the bonding pairs are closer to the small nitrogen and repel each other more.
C
The O-N-O angle falls in the order NO2+>NO2>NO2−\mathrm{NO_2^+ > NO_2 > NO_2^-}.
D
The angle in NO2\mathrm{NO_2} (≈134∘\approx134^\circ) is larger than in NO2−\mathrm{NO_2^-} (≈115∘\approx115^\circ) because a single unpaired electron repels the bonding pairs less than a lone pair does.
34
In a Haber reactor a stoichiometric mixture N2:H2=1:3\mathrm{N_2}:\mathrm{H_2}=1:3 (no inerts) is brought to equilibrium at 700 K700\,\mathrm{K} and a constant total pressure of 200 bar200\,\mathrm{bar}. For N2+3H2⇌2NH3\mathrm{N_2+3H_2\rightleftharpoons2NH_3}, Kp=1.0×10−4 bar−2K_p=1.0\times10^{-4}\,\mathrm{bar^{-2}} at 700 K700\,\mathrm{K}. Assuming ideal gases, what is the mole fraction of NH3\mathrm{NH_3} at equilibrium?
Not attempted
Your answer: —
Correct answer: 0.31
35
Which plot correctly shows the boiling points of the group 15 hydrides EH3\mathrm{EH_3} (E == N, P, As, Sb, Bi along the horizontal axis; vertical axis in kelvin, linear scale, only the shape matters)?
Not attempted
A
Plot 1
NPAsSbBi
B
Plot 2
NPAsSbBi
C
Plot 3
NPAsSbBi
D
Plot 4
NPAsSbBi
36
Which of the following statements about the halides of group 15 is/are correct?
Not attempted
A
The hydrolysis of SbCl3\mathrm{SbCl_3} and BiCl3\mathrm{BiCl_3} is only partial and gives the oxychlorides SbOCl and BiOCl, which are insoluble.
B
NF3\mathrm{NF_3} is readily hydrolysed to NH3\mathrm{NH_3} and HOF\mathrm{HOF}, and is a stronger Lewis base than NH3\mathrm{NH_3} because of the electronegative F atoms.
C
NCl3\mathrm{NCl_3} is hydrolysed by water to NH3\mathrm{NH_3} and HOCl\mathrm{HOCl}, in contrast to PCl3\mathrm{PCl_3} which gives H3PO3\mathrm{H_3PO_3} and HCl.
D
The complete hydrolysis of PCl3\mathrm{PCl_3} gives H3PO4\mathrm{H_3PO_4} and HCl, since phosphorus is oxidised by water.
37
Match List-I with List-II and select the correct combination. List-IList-IIP)NO2+1)planar, N-N bonded, diamagneticQ)NO22)linearR)NO2−3)bent (≈134∘), paramagneticS)N2O44)bent (≈115∘), diamagnetic, lone pair on N5)tetrahedral\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{NO_2^+} & \text{1)} \text{planar, N-N bonded, diamagnetic} \\ \text{Q)} \mathrm{NO_2} & \text{2)} \text{linear} \\ \text{R)} \mathrm{NO_2^-} & \text{3)} \text{bent }(\approx134^\circ)\text{, paramagnetic} \\ \text{S)} \mathrm{N_2O_4} & \text{4)} \text{bent }(\approx115^\circ)\text{, diamagnetic, lone pair on N} \\ & \text{5)} \text{tetrahedral} \end{array}
Not attempted
A
P-3, Q-2, R-4, S-1
B
P-2, Q-3, R-4, S-1
C
P-2, Q-4, R-3, S-1
D
P-1, Q-3, R-2, S-4
38
In the brown ring test the coloured species is [Fe(H2O)5(NO)]2+\mathrm{[Fe(H_2O)_5(NO)]^{2+}}. The ligand NO is taken as the nitrosonium ion NO+\mathrm{NO^+}, which makes the oxidation state of iron +1+1. Iron is in a weak-field environment (high spin). Taking the spin-only formula μ=n(n+2)\mu=\sqrt{n(n+2)} in BM, what is the value of μ2\mu^2 for the complex, in BM2\mathrm{BM^2}?
Not attempted
FeOH₂OH₂OH₂OH₂OH₂NO[Fe(H₂O)₅(NO)]²⁺, octahedral
Your answer: —
Correct answer: 15
39
Which of the following statements about ammonia and its reactions is/are correct?
Not attempted
A
Ammonia is far more soluble in water than phosphine mainly because ammonia is a much stronger acid than phosphine.
B
The Haber process is run at 700 K700\,\mathrm{K}, a compromise: lower temperature raises the equilibrium yield but slows the reaction, while high pressure favours the yield since Δng<0\Delta n_g<0.
C
With an excess of chlorine, ammonia gives NCl3\mathrm{NCl_3}; with an excess of ammonia it gives N2\mathrm{N_2} and NH4Cl\mathrm{NH_4Cl}.
D
AgCl dissolves in dilute aqueous ammonia, AgBr dissolves only in concentrated ammonia, and AgI is practically insoluble, because of the falling solubility product of the silver halides.
40
1.239 g1.239\,\mathrm{g} of white phosphorus (P4\mathrm{P_4}, M=123.9 g mol−1M=123.9\,\mathrm{g\,mol^{-1}}) is boiled under an inert atmosphere with excess aqueous NaOH (P4+3NaOH+3H2O→PH3+3NaH2PO2\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}); the gas escapes completely. The remaining solution is acidified and added to excess aqueous AgNO3\mathrm{AgNO_3} (silver ion is reduced to the metal and the hypophosphorous acid is oxidised to H3PO4\mathrm{H_3PO_4}). What mass of silver, in grams, is deposited? (Ag=107.87\mathrm{Ag}=107.87)
Not attempted
Your answer: —
Correct answer: 12.94
41
Which of the following statements about PCl5\mathrm{PCl_5} is/are correct?
Not attempted
A
The dissociation PCl5⇌PCl3+Cl2\mathrm{PCl_5\rightleftharpoons PCl_3+Cl_2} is favoured by an increase of the total pressure at constant temperature.
B
In the gas phase the molecule is trigonal bipyramidal and the two axial P-Cl bonds are longer than the three equatorial ones.
C
Complete hydrolysis follows PCl5+3H2O→H3PO4+5HCl\mathrm{PCl_5+3H_2O\to H_3PO_4+5HCl}, and partial hydrolysis stops at POCl3\mathrm{POCl_3}.
D
In the solid both the cation and the anion of [PCl4]+[PCl6]−\mathrm{[PCl_4]^+[PCl_6]^-} are tetrahedral, because phosphorus is sp3sp^3 in both.
42
Hydrolysis of PCl3\mathrm{PCl_3} gives HCl and a phosphorus acid that is dibasic (it needs exactly two moles of NaOH per mole) and reduces silver(I) to silver metal. Which structure (all P-H and O-H hydrogens shown; a pair of dots is a lone pair) is the acid?
Not attempted
A
Structure 1
POHOHOH
B
Structure 2
POHOHOH
C
Structure 3
POHHOH
D
Structure 4
POOHOHOH
43
0.040 mol0.040\,\mathrm{mol} of phosphorous acid is heated until it has completely disproportionated according to 4H3PO3→3H3PO4+PH34\mathrm{H_3PO_3}\to3\mathrm{H_3PO_4}+\mathrm{PH_3}, the phosphine escaping. The residue is dissolved and titrated with 0.500 M0.500\,\mathrm{M} NaOH to the stage at which every P-OH hydrogen has been neutralised. What volume of NaOH is required?
Not attempted
A
160 mL160\,\mathrm{mL}
B
180 mL180\,\mathrm{mL}
C
120 mL120\,\mathrm{mL}
D
240 mL240\,\mathrm{mL}
44
The figure shows P4O6\mathrm{P_4O_6}, the anhydride of phosphorous acid: the six edges of a P4\mathrm{P_4} tetrahedron each carry one bridging oxygen atom. P4O10\mathrm{P_4O_{10}}, the anhydride of orthophosphoric acid, is obtained from it by adding one terminal oxygen atom, bonded by a P=O double bond, to each phosphorus. Find the total of: (number of P-O-P linkages in P4O6\mathrm{P_4O_6}) ++ (number of P-O-P linkages in P4O10\mathrm{P_4O_{10}}) ++ (number of P=O bonds in P4O10\mathrm{P_4O_{10}}).
Not attempted
OOOOOOPPPPP₄O₆: an O bridges each P–P edge
Your answer: —
Correct answer: 16
45
Match List-I with List-II and select the correct combination. List-IList-IIP)H3PO21)tribasic, three P-O-P linkages in a ringQ)H3PO32)tetrabasic, one P-O-P linkageR)H4P2O73)dibasic, reduces Ag+S)(HPO3)3 (cyclic)4)monobasic, reduces Ag+5)tribasic, no P-O-P linkage, not reducing\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{H_3PO_2} & \text{1)} \text{tribasic, three P-O-P linkages in a ring} \\ \text{Q)} \mathrm{H_3PO_3} & \text{2)} \text{tetrabasic, one P-O-P linkage} \\ \text{R)} \mathrm{H_4P_2O_7} & \text{3)} \text{dibasic, reduces }\mathrm{Ag^+} \\ \text{S)} (\mathrm{HPO_3})_3\ \text{(cyclic)} & \text{4)} \text{monobasic, reduces }\mathrm{Ag^+} \\ & \text{5)} \text{tribasic, no P-O-P linkage, not reducing} \end{array}
Not attempted
A
P-3, Q-4, R-2, S-1
B
P-4, Q-3, R-1, S-2
C
P-5, Q-4, R-3, S-1
D
P-4, Q-3, R-2, S-1
46
At 523 K523\,\mathrm{K} and 1.00 atm1.00\,\mathrm{atm} the equilibrium vapour of PCl5\mathrm{PCl_5} (partly dissociated into PCl3\mathrm{PCl_3} and Cl2\mathrm{Cl_2}) has a density of 4.20 g L−14.20\,\mathrm{g\,L^{-1}}. Take M(PCl5)=208.24 g mol−1M(\mathrm{PCl_5})=208.24\,\mathrm{g\,mol^{-1}} and R=0.082057 L atm K−1mol−1R=0.082057\,\mathrm{L\,atm\,K^{-1}mol^{-1}}. What is KpK_p for PCl5⇌PCl3+Cl2\mathrm{PCl_5\rightleftharpoons PCl_3+Cl_2}?
Not attempted
A
2.47×10−2 atm2.47\times10^{-2}\,\mathrm{atm}
B
2.85×10−2 atm2.85\times10^{-2}\,\mathrm{atm}
C
1.84×10−2 atm1.84\times10^{-2}\,\mathrm{atm}
D
5.76×10−4 atm5.76\times10^{-4}\,\mathrm{atm}
47
In the solid state PCl5\mathrm{PCl_5} is ionic, [PCl4]+[PCl6]−\mathrm{[PCl_4]^+[PCl_6]^-}. How many Cl-P-Cl bond angles of exactly 90∘90^\circ are there in one formula unit of the solid (ideal geometries)?
Not attempted
A
1515
B
1818
C
1212
D
66
48
1.00 L1.00\,\mathrm{L} of dry oxygen (measured at STP) is passed through a silent electric discharge. After the discharge the gas, again measured at STP, occupies 0.96 L0.96\,\mathrm{L}. All of this gas is bubbled through excess neutral potassium iodide solution (O3+2I−+H2O→O2+I2+2OH−\mathrm{O_3+2I^-+H_2O\to O_2+I_2+2OH^-}), and the iodine liberated is titrated with 0.100 M0.100\,\mathrm{M} sodium thiosulfate (I2+2S2O32−→2I−+S4O62−\mathrm{I_2+2S_2O_3^{2-}\to2I^-+S_4O_6^{2-}}). Take the molar volume at STP as 22.4 L mol−122.4\,\mathrm{L\,mol^{-1}}. What volume of thiosulfate solution, in mL, is required?
Not attempted
Your answer: —
Correct answer: 71.43
49
Which of the following statements about sulfur dioxide is/are correct?
Not attempted
A
The bleaching by SO2\mathrm{SO_2} is permanent, because it is caused by oxidation of the coloured matter by nascent oxygen.
B
SO2\mathrm{SO_2} acts as an oxidising agent towards H2S\mathrm{H_2S}, giving sulfur: 2H2S+SO2→3S+2H2O\mathrm{2H_2S+SO_2\to3S+2H_2O}.
C
In water SO2\mathrm{SO_2} reduces Fe3+\mathrm{Fe^{3+}} to Fe2+\mathrm{Fe^{2+}} and turns acidified dichromate from orange to green.
D
The molecule is bent (≈119.5∘\approx119.5^\circ) with both S-O bonds equal in length, the angle being slightly below 120∘120^\circ because of the lone pair on sulfur.
50
A gas containing 10.010.0 mol% SO2\mathrm{SO_2}, 11.011.0 mol% O2\mathrm{O_2} and 79.079.0 mol% N2\mathrm{N_2} is passed over a vanadium(V) oxide catalyst at a constant total pressure of 2.00 bar2.00\,\mathrm{bar}. At the operating temperature, Kp=60.0 bar−1K_p=60.0\,\mathrm{bar^{-1}} for 2SO2+O2⇌2SO32\mathrm{SO_2}+\mathrm{O_2}\rightleftharpoons2\mathrm{SO_3}. Assuming ideal gases and equilibrium, what is the percentage conversion of SO2\mathrm{SO_2} to SO3\mathrm{SO_3}?
Not attempted
Your answer: —
Correct answer: 75.04
51
Consider the oxidation states of sulfur. Which of the following statements is/are correct?
Not attempted
A
The average oxidation state of sulfur in sodium tetrathionate, Na2S4O6\mathrm{Na_2S_4O_6} (a chain −O3S−S−S−SO3−\mathrm{^-O_3S-S-S-SO_3^-}), is +2.5+2.5.
B
In sodium dithionate, Na2S2O6\mathrm{Na_2S_2O_6} (S-S bonded), each sulfur is +5+5.
C
In peroxodisulfuric acid, H2S2O8\mathrm{H_2S_2O_8}, sulfur is in the +7+7 oxidation state.
D
In sodium thiosulfate, Na2S2O3\mathrm{Na_2S_2O_3}, the two sulfur atoms have oxidation states +6+6 (central) and −2-2 (the sulfane sulfur), an average of +2+2.
52
How many of the following species are paramagnetic: O2\mathrm{O_2}, O2−\mathrm{O_2^-}, O22−\mathrm{O_2^{2-}}, O3\mathrm{O_3}, S2\mathrm{S_2} (vapour), SO2\mathrm{SO_2}, NO\mathrm{NO}, ClO2\mathrm{ClO_2}?
Not attempted
A
44
B
66
C
33
D
55
53
Which plot of the first acid dissociation constant, expressed as pKa1\mathrm{p}K_{a1} (vertical axis, linear scale), for the hydrides H2E\mathrm{H_2E} with E == O, S, Se, Te along the horizontal axis, correctly shows the trend?
Not attempted
A
Plot 1
OSSeTe
B
Plot 2
OSSeTe
C
Plot 3
OSSeTe
D
Plot 4
OSSeTe
54
Match List-I with List-II and select the correct combination. List-IList-IIP)SO2+2H2S1)S changes from +6 to +4Q)SO2 into acidified K2Cr2O72)no change in the oxidation state of SR)Cu with hot concentrated H2SO43)S changes from +4 to +6S)SO3 dissolved in concentrated H2SO44)S changes from +4 to 05)S changes from +6 to −2\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{SO_2+2H_2S} & \text{1)} \text{S changes from }+6\text{ to }+4 \\ \text{Q)} \mathrm{SO_2}\text{ into acidified }\mathrm{K_2Cr_2O_7} & \text{2)} \text{no change in the oxidation state of S} \\ \text{R)} \mathrm{Cu}\text{ with hot concentrated }\mathrm{H_2SO_4} & \text{3)} \text{S changes from }+4\text{ to }+6 \\ \text{S)} \mathrm{SO_3}\text{ dissolved in concentrated }\mathrm{H_2SO_4} & \text{4)} \text{S changes from }+4\text{ to }0 \\ & \text{5)} \text{S changes from }+6\text{ to }-2 \end{array}
Not attempted
A
P-3, Q-4, R-2, S-1
B
P-4, Q-3, R-1, S-2
C
P-4, Q-3, R-2, S-1
D
P-1, Q-2, R-3, S-4
55
Which of the following statements about sulfuric acid and its manufacture is/are correct?
Not attempted
A
The conversion of SO2\mathrm{SO_2} to SO3\mathrm{SO_3} in the contact process is carried out at about 100100 atm because the reaction has Δng<0\Delta n_g<0 and the equilibrium conversion is otherwise low.
B
Oleum, H2S2O7\mathrm{H_2S_2O_7}, is obtained by absorbing SO3\mathrm{SO_3} in concentrated H2SO4\mathrm{H_2SO_4} rather than in water, to avoid an acid mist.
C
Concentrated H2SO4\mathrm{H_2SO_4} with NaCl gives HCl, but with NaBr it gives also Br2\mathrm{Br_2} and SO2\mathrm{SO_2}, and with NaI it gives I2\mathrm{I_2} and H2S\mathrm{H_2S}, because HBr and HI are reducing enough to reduce the acid.
D
In peroxodisulfuric acid, H2S2O8\mathrm{H_2S_2O_8}, two sulfur atoms are joined through an O-O peroxide bridge, while sulfur remains in the +6+6 state.
56
Chlorine is passed into warm concentrated NaOH under conditions such that the only products are NaCl, NaClO, NaClO3_3 and water, with n(NaClO):n(NaClO3)=3:1n(\mathrm{NaClO}):n(\mathrm{NaClO_3})=3:1. How many moles of NaOH are consumed for each mole of NaClO3_3 formed?
Not attempted
Your answer: —
Correct answer: 12
57
Chlorine reacts with hot concentrated NaOH to give chloride and an anion ClOx−\mathrm{ClO}_x^- in which chlorine is in its +5+5 oxidation state. In the diagrams a line is a bond and a pair of dots is a lone pair on chlorine (the Cl-O π\pi bonding is not drawn). Which diagram correctly shows the shape of this anion?
Not attempted
A
Diagram 1
ClOOO
B
Diagram 2
ClOOO
C
Diagram 3
ClOO
D
Diagram 4
ClOOOO
58
Which of the following statements about the interhalogen compounds and related ions is/are correct?
Not attempted
A
Interhalogens such as ClF3\mathrm{ClF_3} are less reactive than the parent halogens Cl2\mathrm{Cl_2} and F2\mathrm{F_2} because the X-X' bond is stronger than the X-X bond.
B
BrF3\mathrm{BrF_3} undergoes self-ionisation, 2BrF3⇌BrF2++BrF4−2\mathrm{BrF_3}\rightleftharpoons\mathrm{BrF_2^+}+\mathrm{BrF_4^-}, the cation being bent and the anion square planar.
C
Solid ICl3\mathrm{ICl_3} exists as the planar dimer I2Cl6\mathrm{I_2Cl_6}, in which each iodine is square planar with two bridging chlorine atoms.
D
All seven I-F bonds in IF7\mathrm{IF_7} (pentagonal bipyramid) are of the same length, since the five equatorial and two axial fluorines are related by symmetry.
59
Consider the formation of the aqueous halide ion by 12X2(g)→X−(aq)\tfrac12\mathrm{X_2(g)}\to\mathrm{X^-(aq)} built from the three steps shown in the figure. Use: D(F−F)=158.8D(\mathrm{F{-}F})=158.8, D(Cl−Cl)=242.6D(\mathrm{Cl{-}Cl})=242.6 (both in kJ mol−1\mathrm{kJ\,mol^{-1}}); electron gain enthalpies −333-333 (F\mathrm{F}) and −349-349 (Cl\mathrm{Cl}); hydration enthalpies of the gaseous ions −515-515 (F−\mathrm{F^-}) and −381-381 (Cl−\mathrm{Cl^-}), all in kJ mol−1\mathrm{kJ\,mol^{-1}}. By how many kJ mol−1\mathrm{kJ\,mol^{-1}} is the overall process more exothermic for fluorine than for chlorine?
Not attempted
½X₂(g) → X(g) → X⁻(g) → X⁻(aq)step 1: ½ × bond dissociation enthalpystep 2: electron gain enthalpystep 3: hydration enthalpy of X⁻
Your answer: —
Correct answer: 159.9
60
Which of the following statements about hydrogen fluoride and fluorides of silicon is/are correct?
Not attempted
A
Glass is etched by HF: SiO2+4HF→SiF4+2H2O\mathrm{SiO_2+4HF\to SiF_4+2H_2O}, and with excess HF the product is the hexafluorosilicic acid H2SiF6\mathrm{H_2SiF_6}.
B
Silicon tetrafluoride is hydrolysed by water: 3SiF4+2H2O→SiO2+2H2SiF6\mathrm{3SiF_4+2H_2O\to SiO_2+2H_2SiF_6}, so a gelatinous silica deposit appears.
C
Dilute hydrochloric acid etches glass in the same way as HF because chloride, like fluoride, is a good ligand for silicon.
D
HF boils higher than HCl because of strong hydrogen bonding, but it is the weakest of the hydrohalic acids because of the high H-F bond enthalpy.
61
Which of the following statements about the oxoacids and oxoanions of chlorine is/are correct?
Not attempted
A
In ClO4−\mathrm{ClO_4^-} the chlorine is sp3sp^3 hybridised and all four Cl-O bonds have the same length, which is why perchlorate is a stronger oxidising agent (kinetically and thermodynamically) than hypochlorite in water.
B
The acid strength follows HOCl<HClO2<HClO3<HClO4\mathrm{HOCl < HClO_2 < HClO_3 < HClO_4}, while the oxidising power of the corresponding acids is greatest for HOCl.
C
Hypochlorous acid is a stronger acid than chloric acid because chlorine bears the lower oxidation number in HOCl.
D
ClO2\mathrm{ClO_2} contains chlorine in the +4+4 state, is paramagnetic (odd electron) and is bent.
62
How many of the following species have a zero dipole moment: ClF3\mathrm{ClF_3}, BrF5\mathrm{BrF_5}, IF7\mathrm{IF_7}, ICl4−\mathrm{ICl_4^-}, BrF4−\mathrm{BrF_4^-}, ICl2−\mathrm{ICl_2^-}, XeF4\mathrm{XeF_4}, XeF2\mathrm{XeF_2}, I3−\mathrm{I_3^-}, IO4−\mathrm{IO_4^-}, ClO3−\mathrm{ClO_3^-}, XeOF4\mathrm{XeOF_4}? (Use the ideal VSEPR geometry of each.)
Not attempted
A
66
B
77
C
88
D
99
63
Match List-I with List-II and select the correct combination. List-IList-IIP)ClF31)sp3d2, two lone pairs on the central atomQ)BrF52)sp3d3, no lone pairR)IF73)sp3d, two lone pairsS)ICl4−4)sp3d2, one lone pair5)sp3d, no lone pair\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{ClF_3} & \text{1)} sp^3d^2\text{, two lone pairs on the central atom} \\ \text{Q)} \mathrm{BrF_5} & \text{2)} sp^3d^3\text{, no lone pair} \\ \text{R)} \mathrm{IF_7} & \text{3)} sp^3d\text{, two lone pairs} \\ \text{S)} \mathrm{ICl_4^-} & \text{4)} sp^3d^2\text{, one lone pair} \\ & \text{5)} sp^3d\text{, no lone pair} \end{array}
Not attempted
A
P-3, Q-4, R-2, S-1
B
P-5, Q-4, R-2, S-1
C
P-3, Q-1, R-2, S-4
D
P-4, Q-3, R-1, S-2
64
0.600 mol0.600\,\mathrm{mol} of XeF4\mathrm{XeF_4} is hydrolysed completely by water according to 6XeF4+12H2O→4Xe+2XeO3+24HF+3O26\mathrm{XeF_4}+12\mathrm{H_2O}\to4\mathrm{Xe}+2\mathrm{XeO_3}+24\mathrm{HF}+3\mathrm{O_2}. The solution is acidified and added to excess potassium iodide (XeO3+6I−+6H+→Xe+3I2+3H2O\mathrm{XeO_3+6I^-+6H^+\to Xe+3I_2+3H_2O}; dissolved O2\mathrm{O_2} does not react), and the iodine liberated is titrated with sodium thiosulfate (I2+2S2O32−→2I−+S4O62−\mathrm{I_2+2S_2O_3^{2-}\to2I^-+S_4O_6^{2-}}). How many moles of thiosulfate are consumed?
Not attempted
Your answer: —
Correct answer: 1.2
65
Partial hydrolysis of XeF6\mathrm{XeF_6} with two moles of water gives XeO2F2\mathrm{XeO_2F_2}. It has five electron domains on Xe (four bond domains, one lone pair). Use these VSEPR rules for the trigonal bipyramidal arrangement: the lone pair and the double-bonded oxygen atoms occupy equatorial positions in preference to axial ones, and the more electronegative fluorine atoms occupy axial positions. Which diagram (a pair of dots is the lone pair) shows XeO2F2\mathrm{XeO_2F_2}?
Not attempted
A
Diagram 1
XeOOFF
B
Diagram 2
XeOFOF
C
Diagram 3
XeOOFF
D
Diagram 4
XeFFOO
66
Which of the following statements about xenon compounds is/are correct?
Not attempted
A
In the hydrolysis of XeF4\mathrm{XeF_4} only one-third of the xenon ends up in XeO3\mathrm{XeO_3} (the rest is released as Xe).
B
The numbers of lone pairs on xenon in XeF2\mathrm{XeF_2}, XeF4\mathrm{XeF_4}, XeF6\mathrm{XeF_6} and XeO3\mathrm{XeO_3} are 33, 22, 11 and 11 respectively.
C
The hydrolysis of XeF6\mathrm{XeF_6} to XeO3\mathrm{XeO_3} is not a redox reaction, whereas hydrolysis of XeF2\mathrm{XeF_2} and XeF4\mathrm{XeF_4} is.
D
XeF2\mathrm{XeF_2} acts as a fluoride donor (with PF5\mathrm{PF_5} it gives [XeF]+[PF6]−\mathrm{[XeF]^+[PF_6]^-}), whereas XeF6\mathrm{XeF_6} acts as a fluoride acceptor (with CsF it gives Cs+[XeF7]−\mathrm{Cs^+[XeF_7]^-}).
67
For each of XeO3\mathrm{XeO_3}, XeOF4\mathrm{XeOF_4} and XeO2F2\mathrm{XeO_2F_2} draw the Lewis structure with Xe expanded, every Xe=O\mathrm{Xe{=}O} bond drawn as a double bond, all atoms having complete octets (Xe: lone pairs as needed) and no formal charges (the figure shows XeO2F2\mathrm{XeO_2F_2} as an example). What is the total number of lone pairs on all atoms of the three molecules together?
Not attempted
XeOOFFXeO₂F₂ (Xe=O drawn double)
A
3030
B
3939
C
3333
D
2727
68
0.0200 mol0.0200\,\mathrm{mol} of xenon and 0.0350 mol0.0350\,\mathrm{mol} of fluorine are heated in a sealed nickel vessel. The only products are XeF2\mathrm{XeF_2} and XeF4\mathrm{XeF_4}, and all the xenon reacts (the fluorine is also completely consumed). What mass of XeF4\mathrm{XeF_4}, in grams, is formed? (Xe=131.29\mathrm{Xe}=131.29, F=19.00\mathrm{F}=19.00)
Not attempted
Your answer: —
Correct answer: 3.11
69
Match List-I with List-II and select the correct combination. List-IList-IIP)XeF2+H2O1)XeO3 and HF onlyQ)XeF4+H2O2)Xe, O2 and HF onlyR)XeF6 (excess H2O)3)Xe, XeO3, O2 and HFS)XeF6 + 1 H2O4)XeOF4 and HF5)XeO2F2 and HF\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\ \text{P)} \mathrm{XeF_2+H_2O} & \text{1)} \mathrm{XeO_3}\text{ and HF only} \\ \text{Q)} \mathrm{XeF_4+H_2O} & \text{2)} \mathrm{Xe},\ \mathrm{O_2}\text{ and HF only} \\ \text{R)} \mathrm{XeF_6\ (excess\ H_2O)} & \text{3)} \mathrm{Xe},\ \mathrm{XeO_3},\ \mathrm{O_2}\text{ and HF} \\ \text{S)} \mathrm{XeF_6\ +\ 1\ H_2O} & \text{4)} \mathrm{XeOF_4}\text{ and HF} \\ & \text{5)} \mathrm{XeO_2F_2}\text{ and HF} \end{array}
Not attempted
A
P-2, Q-3, R-1, S-4
B
P-3, Q-2, R-4, S-1
C
P-2, Q-3, R-4, S-1
D
P-5, Q-3, R-1, S-2
70
A diver breathes air (79% N279\%\ \mathrm{N_2} by volume) at a total pressure of 5.00 atm5.00\,\mathrm{atm} long enough for the dissolved nitrogen to equilibrate; the blood volume is 5.00 L5.00\,\mathrm{L} and the solubility of N2\mathrm{N_2} in blood at 310 K310\,\mathrm{K} is 6.1×10−4 mol L−1 atm−16.1\times10^{-4}\,\mathrm{mol\,L^{-1}\,atm^{-1}} (Henry's law). He then ascends so quickly that the extra dissolved gas, relative to equilibrium with air at 1.00 atm1.00\,\mathrm{atm}, comes out of solution as bubbles. What volume of N2\mathrm{N_2} gas (in mL, at 310 K310\,\mathrm{K} and 1.00 atm1.00\,\mathrm{atm}, R=0.082057 L atm K−1mol−1R=0.082057\,\mathrm{L\,atm\,K^{-1}mol^{-1}}) is released? (Replacing nitrogen by helium, which is much less soluble, avoids this.)
Not attempted
Your answer: —
Correct answer: 245.17
Henry's law uses the partial pressure of nitrogen, not the total pressure.
At depth: p(N2)=0.79×5.00=3.95 atmp(\mathrm{N_2})=0.79\times5.00=3.95\,\mathrm{atm}. At the surface: p(N2)=0.79 atmp(\mathrm{N_2})=0.79\,\mathrm{atm}. The excess is 3.16 atm3.16\,\mathrm{atm}. Dissolved excess =3.16×6.1×10−4×5.00=9.64×10−3 mol=3.16\times6.1\times10^{-4}\times5.00=9.64\times10^{-3}\,\mathrm{mol}. V=nRT/P=9.64×10−3×0.082057×310/1.00=0.2452 L=245.2 mLV=nRT/P=9.64\times10^{-3}\times0.082057\times310/1.00=0.2452\,\mathrm{L}=245.2\,\mathrm{mL}. Using the total pressure instead of the partial pressure of nitrogen would give 310 mL310\,\mathrm{mL} (wrong).