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Experimentally, and surprisingly, there is only circumstantial evidence for most short lived species in any type of reaction, i.e. no one has spectral or structural evidence for a transition state, such as often portrayed in textbooks, other than for example NaI dissociation or atom-diatomic molecule reactions e.g. H+OH etc. done in ultra-low vacuum. For ...

1

This is a standard exam question pertaining to the kinetics material that is typically presented in a first year general chemistry course. Usually the coverage is restricted to discussion of zero, first and second order kinetics and this is what gets tested on exams. From the photo of the exam question, the decay is apparently first order, i.e., $$\mathrm C(... 0 Firstly, you're right that the slopes at all points on this curve will be negative as it's a downward sloping curve. The rate of reaction w.r.t. reactants is the negative of the rate of consumption of reactants provided there's 1:1 stoichiometric ratio between the reactants and products. But in this question, nowhere is it mentioned that this curve depicts ... 0 Don't count bonds. Count electrons. Here all the bonds to boron are polarized away from that atom as boron is less electronegative than both hydrogen and nitrogen. Since the boron also has no valence-shell lone pairs we count zero valence electrons dominated by the boron, versus the neutral atom having three. That drop from three valence electron to zero ... 8 First and most obviously, if something in solid state forms ions by shuffling atoms around, the end result must yield the same ratio of atoms. Take for example \ce{PF5}: this exists as \ce{[PF4+][PF6-]}. If you add the atoms together, you get \ce{P2F10} which is 2\times\ce{PF5}. The same is true for the proposed structure \ce{[XeF5+][F-]} which ... 1 Iron: 1st shell: 2 2nd shell: 8 3rd shell: 14 4th shell: 2 The electronic configuration of elements is generally mentioned in the latter form, with proper description of the electrons and the orbitals into which they fit. You should read this wikipedia article, along with this one and this Chemistry Libretexts article. You could also check out a high ... 0 Pure Tin may be fine as a food container. However, Tin ore is accociated with the ores of Lead and Antimon. Traces of these may find their way into Tinware. I believe there is an metallurgy motive to alloy Tin with Lead as well. Lead tends to react when in contact with acidic food, and it accumulates in the body. This was not taken seriously for a long time.... 2 As per this source, the idea and definition of oxidation state is based on the following principle: The oxidation number of an atom in a molecule is based on a formalism that forces a covalent compound to possess complete ionic character and may be defined as the charge that an atom would have if all bonds to it are broken such that the ligands ... 1 It appears that a clean preparation of stannous (or stannic nitrate), other than by the possible direct action of gaseous NO2 on SnO (or SnO2?), as suggested generally by Wikipedia below, is not readily found. To quote: Conversion to nitrates NO2 is used to generate anhydrous metal nitrates from the oxides:[10] \ce{MO + 3 NO2 → M(NO3)2 + NO} ... 0 The formula [Cr(H_2O)_5Cl]Cl2.H_2O describes the crystal. It contains the cation [Cr(H_2O)_5Cl]^{2+}, twice the anion Cl^- and one water molecule in it's lattice. The IUPAC name you gave Pentaaquachloridochromium(III) chloride only describes the cation, so it does not take the anion and the water molecule. 1 Yes, you should account for the water molecule outside the coordination sphere. In most cases, you do not need to do so, but if you are looking for specificity and/or want to describe the physical or chemical properties of this particular hydrate, it is advised to also mention the number of molecules of water of crystalliation. The correct IUPAC name, in ... 0 The electronic configuration of \ce{Cr} is: [\ce{Ar}]4s^2 3d^4. Because of very little energy difference between 4s- and 3d- orbitals, it can lose 6 of it's valence electrons and thus form 6 covalent bonds. 12 The state (sol) or (solv.) stands for "solvated". Since you are dissolving \ce{HF} in liquid \ce{SbF5}, there is no water in that system (which would cause hydrolysis). So it is not correct to use (aq), since it is a nonaqueous system. Here are some other examples: Autoprotolysis of liquid ammonia:$$\ce{2NH3 (l) <=> NH4+ (sol) + NH2- (sol)}$$... 0 Per this source, Chemical Reviews: Transition Metal-Tin Chemistry, I surmise that a potential safety issue with Tin stems from its ability to create complexes, which can act in distinctly catalytic roles in organic-based reactions. In particular, the author discusses the co-ordination chemistry around stannylenes, \ce{R2Sn}, to explain a considerable ... 0 Per a google search: Alloy 316/316L is molybdenum-bearing austenitic stainless steel. The higher nickel and molybdenum content in this grade allows it to demonstrate better overall corrosion resistant properties than 304, especially with regard to pitting and crevice corrosion in chloride environments. Also, another source on Alloy 316L : Cr Chromium ... 7 \ce{Na2} and \ce{K2} do exist in the sodium and potassium vapors. This can be proved by analyzing the highly diluted flames produced by hot vapors of these metals in contact with diluted \ce{Cl2} vapor, as proved by Polanyi and co-workers. See the following references: M. Polanyi, Atomic Reactions, Williams and Norgate, London (1932); M.G. Evans, M. ... 13 The first thing that should be said is that there's no difference between a coordinate bond (dative bond) and an ordinary covalent bond. Yes, the electrons "come from different places"; but the molecule doesn't actually know this, nor does it care. Once a covalent bond is formed, it is a covalent bond, regardless of where the electrons "come from". Some of ... 5 Your textbook is right. Lots of metals form covalent bonds. In the case of lithium chloride such bonding is one explanation for the solubility of this compound in organic solvents (see this answer). One additional example you might want to know about is Grignard reagents, a class of highly basic compounds in which carbon is covalently bonded to magnesium. ... 1 The comment "My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts" is not precisely accurate. Limited solubility in water does not necessarily imply a lack of solubility in an acidic medium. For example, I have prepared Silver acetate (moderately water-soluble, see Table) from the action of ... -2 In the bonds M^{z+}-O-H, where M can be Mn or Cr, the atom M is positively charged. When this charge z is large, M repells the H atom stronger than if z is small. So the molecule containing these bonds is a stronger acid. For example, Z = 6 in H_2CrO_4 and Z = 7 in HMnO_4. They both contain at least one M-O-H bond. As Z is high ... 0 As to the question does gaseous NH3 directly react with metals, other than complex formations, a more accurate short answer is likely no. A better answer is a reaction may proceed from heat (or other energy sources, like light) induced breakdown products of ammonia interacting with the metal. A review of the literature suggests that if ammonia is heated to ... 1 Wikipedia reports that magnesium nitride can be prepared: By passing dry nitrogen over heated magnesium: {\displaystyle{\begin{matrix}{}\\{\ce {{3Mg} + N2->[{\ce{800^{\circ }C}}]Mg3N2}}\\{}\end{matrix}}} or ammonia: {\displaystyle{\begin{matrix}{}\\{\ce{{3Mg} + 2NH3->[{\ce{700^{\circ }C}}]{Mg3N2} + 3H2}}\\{}\end{matrix}}} No ... 3 There is nothing special about fractional indices in chemistry, it is just a pointer signifying we are dealing with a non-stoichiometric compound. Group 6 metal are known not only for their regular oxides existing in numerous polymorphic modifications, but also for forming homologous series of nonstoichiometric oxides \ce{\ce{M^{VI}_nO_{3n-1}}} and \ce{\... 1 As your textbook notes, the relative energy of atomic orbitals is not fixed as you start ionizing your atom. As you ionize, all orbitals drop in energy. However, higher angular momentum orbitals drop more than lower ones, and specifically in your case, the \mathrm{3d} orbitals drop below the \mathrm{4s} orbitals. So consider vanadium, the atom has a \... 3 There are multiple issues here: You should not assume that there is a simple rate law. Chemistry is complicated, and even simple reactions may proceed through multiple pathways in parallel. More generally, you should not assume that you are given enough information to determine the solution. Just because the stoichiometric coefficients are there, does not ... -1 Nobody can determine the order of a reaction with just the stoichiometric coefficients. For example look at the reaction I_2 + Acetone -> ICH_2COCH_3 + HI. It has a rate corresponding to the law : -d([I_2])/dt = k[H^+][Acetone]. The concentration [I_2] does not appear in the rate law. It may seem incredible, unbelievable. Nobody can guess it. It ... 3 Four-member rings can be formed with simple oxy-anion ligands such as nitrate. An example with iron as the central ion is given here. G. R. A Wyllie, O. Q. Munro, C. E. Schulz, W. R. Scheidt, "Structural and Physical Characterization of (Nitrato)iron(III) Porphyrinates [Fe(por)(NO3)] — Variable Coordination of Nitrate", Polyhedron 2007 Oct 10, 26(16): ... 1 I agree with Buck Thorn's explanation on \ce{SO2} concentration in aqueous phase. Thus, when dissolve in water (or when is added to water), the initial reaction of \ce{SO2} with water is shown in the following reaction (Ref.1):$$\ce{SO2 (g) + H2O (l) -> H2SO3 (aq)}$$Then, formed \ce{H2SO3} would stabilize following equilibrium:$$\ce{H2SO3 + ...

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If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to ...

1

There is band structure formed in the metallic lattice in the crystals of d and f block elements due to partially filled orbitals. But as you move along the periodic table to 'Zn' , you will find that the orbitals are completely filled hence there is no possibility for inter-orbital mixing. Another thing worth noting is that due to the filled d-orbital, zn ...

1

A BCC has 6 octahedral holes and 12 tetrahedral holes. At each face of the bcc, there is one octahedral hole. There is also an octahedral hole on each edge. We divide the holes found on the faces by 2 as they lay between two unit cells, and we divide the holes found on the edges by 4 because they likewise lay between 4 unit cells. There are 6 faces and ...

2

It seems that for BeF2 the hydration enthalpy of Be2+ is really high (due to Be2+ being a small cation and having a high charge density) so the energy required to break the lattice is smaller than that gained from hydration of Be2+ while for something like MgF2 the enthalpy of the lattice is greater than the energy you would get from hydration of Mg2+. In ...

-2

In the Zinc atom, the levels $4s$ and $3d$ are completely occupied. This increases the stability of the atom. And electrons need more energy to be removed.

0

$P_4O_{10}$ is known to be a dehydrating agent. It removes $2H$ and $1O$ from any substance it may encounter. In order to withdraw $H_2O$ from nitric acid, it must find at least 2 $H$ atoms. One molecule$HNO_3$ is not enough. So you have to use at least two molecules of $HNO_3$. What remains from $2 HNO_3$ after removing $1 H_2O$ ? By subtraction you find ...

4

As explained in the other answers, water ice may not actually become cubic at low temperature. In fact, if formed at higher temperature as the usual hexagonal phase, water ice will remain so when cooled. There is no thermodynamic driving force to rearrange the hexagonal structure to a cubic one. We can expand on the earlier answers in two, make that three,...

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Aluminum and gallium are on different periods. When an additional shell is (fictitiously) added to aluminum, the size is seen to decrease. This is, as correctly mentioned, due to the poor shielding offered by the $(n-1)d$ orbitals to the incoming electrons in the ultimate $p$ subshell. However, between copper and zinc, the matter seems to be different–their ...

3

As correctly noted in the question, $\ce{CO}$ and $\ce{N2}$ are iso-electronic, so they should be comparable. But a difference remains: $\ce{CO}$ is hetero-nuclear, $\ce{N2}$ is homo-nuclear. The latter case typically means that there is better overlap between the orbitals (because they are of similar extent/energy). This translates into a greater gap ...

1

Going across from left to right on period 2 and 3, the likelihood of forming stable hydroxides seems to go down. Lithium hydroxide, boron hydroxide, indium hydroxide, they all exhibit pretty good stability. As atomic radius decreases, the electronegative oxygens can be expected to drawn closer. Even nitric acid utilizes resonance to maintain stability. This ...

4

The question presupposes that $\ce{N(OH)3}$ does not exist. However, Ab initio molecular dynamics evidence of a new stable symmetric Cs structure for N(OH)3 Chemical Physics Letters volume 435, pages 34-38 says: We point out that very few studies have addressed N(OH)3 and this study is relevant in the context of its possible detection in the gas phase....

1

why is it that metal groups decrease in melting points going down This isn't quite true for all metals. I think the only metals that behave like like this are the alkali metals. Group 6 metals increase in melting point. Calcium has a higher melting point than Magnesium. Etc. more shells exist thus ions in the lattice will be larger so more IMFs ...

4

It isn't just nitrogen that fails to form an orthoacid with the formula $\ce{X(OH)3}$. The same thing happens with phosphorus. The phosphorous acid $\ce{H3PO3}$ is not $\ce{P(OH)3}$, with three equivalent $\ce{OH}$ groups as you might believe. This phosphorous acid has one H atom that is impossible to neutralize. Its structure is better described by $\ce{... 12 Although the acid and its salts are unknown, orthonitrite esters have been found, and from a natural source at that [1]. The orthonitrite function, which is trivalent rather than monovalent as nitrate would be, is incorporated into a macrocycle that hinders its decomposition. The picture below is from Ref. 1. Reference: 1. Leanne Murray,* Tang K. Lim,* ... 10 Well let's not focus on your text but on the actual question here about$\ce{N(OH)3}$. I can only give you some ideas here but not a definite answer. Whenever we have tests for our first semester students there are some who forgot what a nitrate was, or nitric acid. But as we ask for a nitrate as trivial name an ortho-nitrate is an accepted answer as well. ... 0 Acid strength is determined by the stability of the conjugate base. Here are the factors that influence stability of the conjugate base: Ion size (when going down the rows of the periodic table) Electronegativity (when going in the same row of the periodic table across) Resonance Inductive Effects If you look at the conjugate bases, in acetic acid, the ... 0 The correct answer is, indeed, in my opinion choice (B) as is noted. The action of KI on CuSO4 creates, in situ, the very unstable CuI2 (think of the corresponding salt with iron, Ferric iodide, whose very existence is doubted). Per the web from Wikipedia: The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. This ... 2 This is a relatively advanced analytical chemistry question and it requires that you remember the color of all common transition metal ions and their very common precipitates. You have to remember the colors of halogens in solution as well. Instead of worrying about the color of iron thiosulfate (which is bluish green), think about the first step. KI reacts ... -4 In aqueous solution, the substance$FeS_4O_6$is dissociated into a mixture of the ion$Fe^{2+}$and of$S_4O_6^{2-}$. The ion$Fe^{2+}$is pale green, but so pale that its solutions are considered as colorless. The ion$S_4O_6^{2-}$is colorless. 5 Depending whether you have a molecular, metallic or ionic compound, the independently moving particles in the liquid state are molecules, atoms and ions, respectively. So for a molecular solid, you have to break the intermolecular forces to turn it into a liquid (or make them non-persistent so that the interaction partners can change over time). For a ... 0 Diborane is supposed to be a dimer of BH3, which is an electron-deficient species so it needs to look for a way to gain some respite from the drought of electron density. So it adjusts for this 3c-2e bonded system where it shares the B-H σ-bond electron density and shares it to the Boron atom of another BH3 unit, forming a dimer. The 'bridging' bond has a ... 0 Solubility is one of the present scientific problems that nobody is able to explain thoroughly. As you are speaking of Calcium compounds, look at the series of Calcium compounds made with halogens.$\ce{CaCl2, CaBr2, CaI2}$are so soluble in water that they can be dissolved in less than their weight of water.$\ce{CaF2}\$ should be similar. It is not. On the ...

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