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1. There is no order for filling atomic orbitals of the same azimuthal quantum number (l) such as px, py, and pz since they are degenerate ie. have the same energy (however they lose their degeneracy in a magnetic field, see Zeeman effect). Therefore, one electron in a p sub-cell can be in either 3 of them and the order of filling doenst matter.
There is a subtle effect when you compare transition metal cations with neutral atoms: the ions have greater effective nuclear charge. This overwhelms electron-electron repulsion and tends to force orbitals into the same shell order they would have with only a single electron (meaning $\color{blue}{3}d$ drops below $\color{blue}{4}s$ in
Well, you'll always be given a periodic table, whether during a test or simply by looking it up, so you can use the periodic table to remember most of it. Just sit down and get to know the periodic table. Break down the patterns into chunks and remember each chunk. In general, for the first three rows: The principal quantum number n corresponds to the row number. The particular block ...
27. I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons. For NX2 the orbitals in increasing energy are: σ1s <σ ∗ 1s <σ2s <σ ∗ 2s <π2px, π2py <σ2pz <π ∗ 2px, π ∗ 2py <σ ∗ 2pz. because it has 14 electrons.
There are five d orbitals starting with the fourth energy level. One electron enters each orbital, having the same spin. Then a second electron enters each orbital, having opposite spin, for a total of 10 electrons. This is similar to the way in which p orbitals fill. Starting with the fourth period, electrons start filling the d orbitals. There are five d orbitals, and like the p orbitals ...
Some thing to take note of in this picture is the order of 3d and 4s. My belief is: 3d has a higher energy level than 4s, and but 4s is still the outermost shell. The reason that transition metal would form colorful ions is that there are empty or half filled d orbitals, so when excited electrons drop back to the d orbitals, they emit energy in ...
This means that it is easier for the electron in the 5"s" orbital to leave. So, the 5"s" electron get ionized first. After the 5"s" electron leave, the next two electrons to be ionized comes from the 4"d" orbital. Therefore, the electronic configuration of "Mo"^ {3+} is " [Kr]" 4"d"^3.
Where the finding of an electron is maximum is known as orbital.The first shell contain s orbital in which two electron can be filled .The second shell contain s & p orbital ,the p orbital can have maximum 6 electrons.The third shell contain S , p & d orbital ,d orbital can have maximum 10 electrons .The forth shell contain s , p , d & f ...
Or, just for one electron, take a neutral atom, e.g. neutral Scandium, remove an electron, so we are looking at the 1+ cation, (maybe assuming that the electronic configuration is now that of the previous element, or maybe checking to find out what is the electronic configuration of that 1+ cation), then add the electron and consider where it's ...
Once you have drawn and labelled molecular orbitals, you fill them in the same order as you fill atomic orbitals. The rules you use for filling atomic orbitals are: 1. Aufbau Principle You place electrons in the lowest energy orbitals available. 2. Pauli Exclusion Principle No orbital may hold more than two electrons, and they must have opposite spin. 3. Hund's Rule Every orbital in a subshell ...