# Tag Info

## Hot answers tagged periodic-trends

41

Here are the $\ce{H-X-H}$ bond angles and the $\ce{H-X}$ bond lengths: \begin{array}{lcc} \text{molecule} & \text{bond angle}/^\circ & \text{bond length}/\pu{pm}\\ \hline \ce{H2O} & 104.5 & 96 \\ \ce{H2S} & 92.3 & 134 \\ \ce{H2Se}& 91.0 & 146 \\ \hline \end{array} The traditional textbook explanation would argue that the ...

39

Starting point: 2s orbitals are lower in energy than 2p orbitals. The $\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally ...

36

The inert pair effect describes the preference of late p-block elements (elements of the 3rd to 6th main group, starting from the 4th period but getting really important for elements from the 6th period onward) to form ions whose oxidation state is 2 less than the group valency. So much for the phenomenological part. But what's the reason for this ...

32

Each of these molecules has a pair of electrons in an orbital - this is termed a "lone pair" of electrons. It is the lone pair of electrons that makes these molecules nucleophilic or basic. As you move down the column from nitrogen to bismuth, you are placing your outermost shell of electrons, including the lone pair, in a larger and more diffuse orbital (...

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Introduction The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size. In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other. Understanding the ...

28

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated. $$\begin{array}{cc} \hline \ce{M} & E^\circ(\ce{M+}/\ce{M}) \\ \hline \ce{Li} & -3.045 \\ \ce{Na} & -2.714 \... 28 The short answer is that you can find a power-law fit (1.61Z^{1.1}) with low average error. I'd never really thought about it much, but after downloading the IUPAC Atomic Weights, I decided to do some curve fitting. Here's a linear fit between atomic number and atomic mass: As you say, the fit isn't very good for small Z, but the overall fit isn't bad ... 25 It's a good question, and I agree that it can be quite confusing, but Chemguide is (at least partly*) correct about this. I'd phrase it this way, which I hope is clearer. First, think about what would happen if the "paired electron" factor was not a factor at all. We would then expect that, going from B to Ne, there would be a roughly constant ... 24 This is due to the transition metal contraction. Bromine has the electron configuration \ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation ... 21 Though it does go against your intuition, you've actually mentioned the answer in your question. Stibane has a higher boiling point than ammonia/azane on account of van der Waals interactions (owing to the larger size of the antimony atom). Our teacher had actually posed this question to us during my first year of high-school. All of us were incredulous ... 18 The question asks why water has a larger angle than other hydrides of the form \ce{XH2} in particular \ce{H2S} and \ce{H2Se}. There have been other similar questions, so an attempt at a general answer is given below. There are, of course, many other triatomic hydrides, \ce{LiH2}, \ce{BeH2}, \ce{BeH2}, \ce{NH2}, etc.. It turns out that some are ... 18 Down the group, atoms of the alkali metals increase in both atomic and ionic radii, due to the addition of electron shells. This results in the charge density of their corresponding cations decreasing down the group. Thus, as we go down the group, the cations become "softer". Now, note that the hydride ion is "hard", having high negative charge density. ... 17 I feel very ashamed to answer my own question, but having found a possible answer, I don’t see why I shouldn’t share it with the community. I would like to start by bringing in this table of orbital energies (from the appendix I linked to in the question) as a reference point for further explanations:$$\begin{array}{ccc} \hline \text{Element} & \text{...

15

Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine. See this.

15

TL;DR $\ce{BCl3}$ does not dimerize to $\ce{B2Cl6}$ due to a conflict between the short $\ce{B-Cl}$ bond length in $\ce{BCl3}$ $(1.74~\mathring{\mathrm{A}}$ in my calculations$)$ and the long $\ce{B-Cl}$ bond lengths $(\sim 2~\mathring{\mathrm{A}})$ that would be required for the $\ce{B2Cl2}$ core of a $\ce{B2Cl6}$ dimer. $\ce{B2Cl6}$ does appear to have a ...

14

The density of an element is related to how many atoms can be placed in a given volume and the weight of the nuclei. Therefore, the smaller the atomic radius of an atom and the higher the atomic number of the nucleus, the greater the density of the element. The very small atomic radius of osmium results in a small metal-metal separation. This small atomic ...

13

I would say the most important reason why no development for effective nuclear charges is that the concept of effective nuclear charges has fallen to disuse in the recent years. I can think of two important reasons: Effective nuclear charges is intended to be used to explain electronic structure of isolated atoms. To this day some atoms can still pose a ...

13

First note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better. Second note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves ...

13

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals. This is because of something called penetration . The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals. ...

13

There isn't really much sense in memorizing the periodic table. The elements you often use you will know them by heart after a while. And you can always use a table when you need it for the others. That being said if you really want to do it, mnemonics are probably the best solution to memorizing the whole table. You can find some here : https://www....

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The simplest way to look at this trend is through VSEPR theory, which produces a very good qualitative understanding of molecular geometry even if it is not compatible with modern molecular orbital theory. However, VSEPR is not the whole story for the heavier Group V and Group VI hydrides. According to VSEPR, electron domains (i.e. bonds and lone pairs) are ...

12

Francium is radioactive. All isotopes of francium are radioactive. The most stable isotope of francium has a half life of only 22 minutes. From the Wikipedia article I linked: Bulk francium has never been viewed. ... However preparing such a sample is impossible, since the extreme heat of decay (the half-life of its longest-lived isotope is only 22 ...

12

Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry). Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram: Crystal studies of both gold and platinum has revealed ...

11

The electron being gained by fluorine would be taken in to a much smaller 2p orbital and requires more electron coupling energy than that of much larger 3p orbital of chlorine. Therefore, energy released during the electron gaining process of fluorine is less than that of chlorine.

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You can memorize the periodic table in one night, simply by emulating best-practice memorization techniques and doing what memory experts do. Common sense, right? Memory experts and world champion memory ‘athletes’ activate the enormous natural power of their visual memory by using visualization and association mnemonic techniques. That’s a fancy way of ...

10

Two types of metallic character In fact, there are two type of metallic character if you look at the metal from the chemical point of view or if you look at the metal from the physical point of view: So it really depends on how do you define metallic character Chemical metallic character Since metallic character in chemistry is defined as: the tendency of ...

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The pattern is better expressed this way: Row 1: 2 elements Row 2: 2+6 elements Row 3: 2+6 elements Row 4: 2+6+10 elements Row 5: 2+6+10 elements Row 6: 2+6+10+14 elements Row 7: 2+6+10+14 elements The reason comes down to how the electrons fill the available energy levels. The thing that differentiates one element from another is the proton number, and ...

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The letters are related to the electron orbitals, which were originally observed through spectroscopy. The lines shown in the spectroscope were named sharp, principal, diffuse and fine (or fundamental). With a strong magnetic or electrostatic field, these separate into one, three, five or seven lines, or energy levels. There can be up to two electrons (with ...

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I will be using an approach which has been enlisted in the following book for answering this question: Arrow Pushing in Inorganic Chemistry ;A Logical Approach to the Chemistry of the Main-Group Elements To start off, please go through my answer to this question: Why is chromate stable in basic medium and dichromate stable in acidic medium? to read about the ...

9

As you go down any group, there are extra electron shells (for example Lithium has 2 shells, Francium has 7) which cause electron shielding. Because the force of attraction must extend further and through these shells to the outermost shell, the force becomes less so there is less strong bonding occuring. Thus, it is easier to break these less strong bonds, ...

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