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Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3 d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [ d orbitals], there are 2 l + 1 = 5 values of m l , meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4 p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the ( n – 1) shell next to the n shell to bring that ( n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons ( l = 3, 2 l + 1 = 7 m l values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the ( n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.

Quantum numbers and electron configurations

What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?


The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1 s , 2 s , 2 p , 3 s , 3 p , 4 s , . . . The 15 electrons of the phosphorus atom will fill up to the 3 p orbital, which will contain three electrons:

This figure provides the electron configuration 1 s superscript 2 2 s superscript 2 2 p superscript 6 3 s superscript 2 3 p superscript 3. It includes a diagram with two individual squares followed by 3 connected squares, a single square, and another connected group of 3 squares all in a single row. The first square is labeled below as, “1 s.” The second is similarly labeled, “2 s.” The first group of connected squares is labeled below as, “2 p.” The square that follows is labeled, “3 s,” and the final group of three squares is labeled, “3 p.” All squares except the last group of three squares has a pair of half arrows: one pointing up and the other down. Each of the squares in the last group of 3 contains a single upward pointing arrow.

The last electron added is a 3 p electron. Therefore, n = 3 and, for a p -type orbital, l = 1. The m l value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these m l values is correct. For unpaired electrons, convention assigns the value of + 1 2 for the spin quantum number; thus, m s = + 1 2 .

Check your learning

Identify the atoms from the electron configurations given:

(a) [Ar]4 s 2 3 d 5

(b) [Kr]5 s 2 4 d 10 5 p 6


(a) Mn (b) Xe

Got questions? Get instant answers now!

The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in [link] or [link] . For instance, the electron configurations (shown in [link] ) of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.

In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4 s into the 3 d orbital to gain the extra stability of a half-filled 3 d subshell (in Cr) or a filled 3 d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5 s 2 4 d 3 . Experimentally, we observe that its ground-state electron configuration is actually [Kr]5 s 1 4 d 4 . We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5 s orbital are larger than the gap in energy between the 5 s and 4 d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.

Questions & Answers

what is molecule
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What is the generic name for the compound
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CH4 , it is the simplest alkane
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Getrude Reply
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Because when CO2 dissolves in water forming a weak acid. CO does not dissolve in water as it has strong triple bond.
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which donate H+ or accept lone pair of electron
kinetic theory of matter and gas law
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material containing clay minerals. Clays develop plasticity when wet, due to a molecular film of water surrounding the clay particles, but become hard, brittle and non–plastic upon drying or firing. Most pure clay minerals are white or light-coloured, but natural clays show a variety of colours
due iron oxide. The four types of clay are Earthenware clay, Stoneware clay, Ball clay, and Porcelain. All of them can be used to make pottery, but the end result would differ a lot thanks to their different textures, colors, and flexibilities.
And do you know that god has created human from clay (وَلَقَدۡ خَلَقۡنَا ٱلۡإِنسَـٰنَ مِن صَلۡصَـٰلࣲ مِّنۡ حَمَإࣲ مَّسۡنُونࣲ) [سورة الحجر 26] And We did certainly create man out of clay from an altered black mud. You can install Quran from paly store for free with translations.
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draw a periodic table
You will arrange the elements into row and coloumns according to increasing proton number. You may want to use symbols or their names. Hydrogen, Helium, etc. God has created all these elements from nothing, in Islam we know God is the creator.
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It is the chemistry concerning molecules that have Carbon skeletons and hydrogen atoms. We find organic molecules like in plants, living derivatives, etc.
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Anything that can be to cutting from all dimensions to halve. So you end up with 4 cubes of 5 cm side. Repeat with one of the cubes. 10, 5, 2.5, .., 0 1st 2nd 3rd Nth Un= a(r) ^ n-1
Anything that has mass and can reflect or absorb waves. GOD created everything from nothing only he can destroy it as prooved.
Suppose you have a cube of side 10 cm. Then you start cutting from all dimensions to halve. So you end up with 4 cubes of 5 cm side. Repeat with one of the cubes. 10, 5, 2.5, .., 0 1st 2nd 3rd Nth Un= a(r) ^ n-1 0= 10 (1/2)^n-1 0= (1/2) ^ n-1 Log0= (n-1) Log(1/2) - infinity =( n-1)
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Practice Key Terms 7

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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