38

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 ...


35

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 ...


33

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 ...


30

As you move from left to right across a period, the number of protons in the nucleus increases. The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons). As you move down a column, there are more protons, but there are also more ...


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 ...


28

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 ...


27

The two hydrogens are the same, but some periodic tables show hydrogen in both places to emphasize that hydrogen isn't really a member of the first group or the seventh group. Hydrogen is a diatomic gas in it's elemental state, which is different from the other group one metals (and similar to the group seven elements). At the same time, hydrogen usually ...


24

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 (...


23

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 \...


18

This exception rule is actually orbital filling rule. For two electrons to be in same orbital they need to have different spins (Pauli exclusion principal). This electron pairing requires additional energy and thus it is easier to add electrons if there are free orbitals. When element has a half-filled p sublevel all 3 orbitals have one electron and pairing ...


18

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 ...


17

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 ...


16

Remember, the 'size' of an atom has nothing to do with the size of the nucleus. It has to do with the size of the valence shell (which itself is not well-defined*). So, if we neglect change in electrical attraction, the size should stay the same—a shell is a shell and it need not 'expand' to accomodate electrons. Now, as we add more protons and electrons, ...


16

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

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 ...


14

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.


14

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 ...


13

You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells. As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the ...


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

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

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....


12

You are correct on atomic size being due to number of electrons and their shells and the reason why atomic size decreases from left to right is due to the number of the protons being greater on the right than on the left of the periodic table with same number of shells due to shell theory. (Reference). The concept you are forgetting is Valence Electrons. ...


11

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 ...


10

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.


10

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. ...


10

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

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 ...


10

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. ...


9

I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun. (Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.) From the physicists' "electron ...


9

To make an argument based on electron configuration, one could see that the orbital occupation does affect the melting point: Element Z Electron Config. Melting Point / K ----------------------------------------------------- Vanadium 23 [Ar] 4s2 3d3 2183 Chromium 24 [Ar] 4s1 3d5 2180 Manganese 25 [Ar] 4s2 3d5 1519 ...


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