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1800-102-2727If you have 30 elements and task to arrange them according to their similar chemical and physical properties, on the basis of which property you try to arrange them?
Electronic configurations tell us about their outermost electronic configuration (valence electrons), we can easily arrange them to their similar chemical and physical property.
Table of contents
Electrons are filled in orbitals of an element under a set of rules based on the parameters of energy, indistinguishability and orientation.
Rules for Filling Electrons in Orbitals
Electrons are filled in various orbitals in order of their increasing energies. Orbitals having the lowest energy are filled first. The order in which orbitals increase in energy is called the orbital sequence.
Note: Energy of single-electron species depends only on the principal quantum number (n).
Order of Energy, 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < …
Note: Energy of single-electron species depends on the principal quantum number (n) & azimuthal quantum number (l).
The energy of orbitals depends on (n+l) rule.
Letter code (subshell) |
Value of l |
s |
0 |
p |
1 |
d |
2 |
f |
3 |
Subshell |
Value of l |
Value of n |
(n+l) |
1s |
0 |
1 |
1 |
2s |
0 |
2 |
2 |
2p |
1 |
2 |
3 |
3s |
0 |
3 |
3 |
Note: if two subshells with same (n+l) value, subshell with lower ‘n’ value has lower energy
Order of Energy, 1s < 2s < 2p < 3s < 3p < 3d < 4s < 4p < 4d < 4f < …
No two electrons in an atom can have the same set of all four quantum numbers.
No electron pairing takes place in the orbitals in a sub-shell until each orbital is occupied by one electron with parallel spin. Exactly half-filled and fully filled orbitals make the atoms more stable, i.e., p^{3}, p^{6}, d^{5},d^{10},f^{7}&f^{14} the configuration is most stable.
Maximum spin multiplicity = 2|S|+1
|S| = Modulus of the maximum spin of an atom
Name of element |
Symbol |
Atomic number |
Electronic configuration |
Hydrogen |
H |
1 |
H_{1}- 1s^{1} |
Helium |
He |
2 |
He_{2}- 1s^{2} |
Lithium |
Li |
3 |
Li_{3}- 1s^{2},2s^{1} |
Berillium |
Be |
4 |
Be_{4}- 1s^{2},2s^{2} |
Boron |
B |
5 |
B_{5}- 1s^{2},2s^{2}2p^{1} |
Carbon |
C |
6 |
C_{6}- 1s^{2},2s^{2}2p^{2} |
Nitrogen |
N |
7 |
N_{7}- 1s^{2},2s^{2}2p^{3} |
Oxygen |
O |
8 |
O_{8}- 1s^{2},2s^{2}2p^{4} |
Flourine |
F |
9 |
F_{9}- 1s^{2},2s^{2}2p^{5} |
Neon |
Ne |
10 |
Ne_{10}- 1s^{2},2s^{2}2p^{6} |
Sodium |
Na |
11 |
Na_{11}- 1s^{2},2s^{2}2p^{6},3s^{1} |
magnesium |
Mg |
12 |
Mg_{12}- 1s^{2},2s^{2}2p^{6},3s^{2} |
Aluminium |
Al |
13 |
Al_{13}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{1} |
Silicon |
Si |
14 |
Si_{14}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{2} |
Phosphorous |
P |
15 |
P_{15}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{3} |
Sulphur |
S |
16 |
S_{16}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{4} |
Chlorine |
Cl |
17 |
Cl_{17}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{5} |
Argon |
Ar |
18 |
Ar_{18}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6} |
Potassium |
K |
19 |
K_{19}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{1} |
Calcium |
Ca |
20 |
Ca_{20}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2} |
Scandium |
Sc |
21 |
Sc_{21}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{1},4s^{2 or} 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{1} |
Titanium |
Ti |
22 |
Ti_{22}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{2},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{2} |
Vanadium |
V |
23 |
V_{23}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{3},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{3} |
Chromium |
Cr |
24 |
Cr_{24}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{5},4s^{1} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{1},3d^{5} |
Manganese |
Mn |
25 |
Mn_{25}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{5},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{5} |
Iron |
Fe |
26 |
Fe_{26}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{6},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{6} |
Cobalt |
Co |
27 |
Co_{27}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{7},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{7} |
Nickel |
Ni |
28 |
Ni_{28}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{8},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{8} |
Copper |
Cu |
29 |
C_{u29}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{10},4s^{1} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{1},3d^{10} |
Zinc |
Zn |
30 |
Zn_{30}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{10},4s^{2} or 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{2},3d^{10} |
Rules to write condensed electronic configuration or noble gas configuration
(n-2)f^{0-14}(n-1)d^{0-10}ns^{0-2}np^{0-6};
where, n outermost (valence) shell
(n-1) penultimate shell
(n-2) anti-penultimate shell
Example: Write the condensed electronic configuration of flerovium (Fl_{114}).
Answer: We know, the last element of period 6 is Rn_{86}.
114-86 = 28, only we have arranged 28 electrons in their valence, penultimate & anti-penultimate shells according to increasing (n+l) rule
This element belongs to period 7 because period 6 is completely filled and the last element of period 7 is
Og_{118}. So, n = 7, (n-1) =6 and (n-2) =5
Subshell |
n |
l |
(n+l) |
Preference Rank |
Maximum capacity of electron |
Actual number of electrons |
7s |
7 |
0 |
7 |
1 |
2 |
2 |
7p |
7 |
1 |
8 |
4 |
6 |
2 |
6d |
6 |
2 |
8 |
3 |
10 |
10 |
5f |
5 |
3 |
8 |
2 |
14 |
14 |
Note: subshells having same (n+l) value, subshell with the lower value of n is preferred
Condensed electronic configurations of elements
H_{1}- 1s^{1}
He_{2}- 1s^{2}
Li_{3}- [He] 2s^{1}
Be_{4}- [He] 2s^{2}
B_{5}- [He] 2s^{2}2p^{1}
C_{6}- [He] 2s^{2}2p^{2}
N_{7}- [He] 2s^{2}2p^{3}
O_{8}- [He] 2s^{2}2p^{4}
F_{9}- [He] 2s^{2}2p^{5}
Ne_{10}- [He] 2s^{2}2p^{6}
Na_{11}- [Ne] 3s^{1}
Mg_{12}- [Ne] 3s^{2}
Al_{13}- [Ne] 3s^{2}3p^{1}
Si_{14}- [Ne] 3s^{2}3p^{2}
P_{15}- [Ne] 3s^{2}3p^{3}
S_{16}- [Ne] 3s^{2}3p^{4}
Cl_{17}- [Ne] 3s^{2}3p^{5}
Ar_{18}- [Ne] 3s^{2}3p^{6}
K_{19}- [Ar] 4s^{1}
Ca_{20}- [Ar] 4s^{2}
Sc_{21}- [Ar] 3d^{1},4s^{2} or [Ar] 4s^{2},3d^{1}
Ti_{22}- [Ar] 3d^{2},4s^{2} or [Ar] 4s^{2},3d^{2}
V_{23}- [Ar] 3d^{3},4s^{2 }or [Ar] 4s^{2},3d^{3}
Cr_{24}- [Ar] 3d^{5},4s^{1} or [Ar] 4s^{1},3d^{5}
Mn_{25}- [Ar] 3d^{5},4s^{2} or [Ar] 4s^{2},3d^{5}
Fe_{26}- [Ar] 3d^{6},4s^{2} or [Ar] 4s^{2},3d^{6}
Co_{27}- [Ar] 3d^{7},4s^{2} or [Ar] 4s^{2},3d^{7}
Ni_{28}- [Ar] 3d^{8},4s^{2} or [Ar] 4s^{2},3d^{8}
Cu_{29}- [Ar] 3d^{10},4s^{1} or [Ar] 4s^{1},3d^{10}
Zn_{30}- [Ar] 3d^{10},4s^{2} or [Ar] 4s^{2},3d^{10}
A. Identification of color of metallic ions compounds
Compounds generally exhibits colours due to excitation and deexcitation of electrons. In compounds of d block elements colors are produced by mainly 2 reasons
The energy of excitation relates to the frequency of light absorbed when an electron from a lower energy d orbital is pushed to a higher energy d orbital. This frequency is usually in the visible range. The colour seen matches the light absorbed's complementary colour.
Case 1: if unpaired electrons are present then, the ion exhibits paramagnetic behaviour.
Case 2: if all electrons are fully paired then, the ion exhibits paramagnetic behaviour.
Case 3: if half-filled subshells are present, the ion exhibits ferromagnetic behaviour (d^{5}- extreme case of paramagnetism).
With the help of electronic configuration we can easily find its spin magnetic moment,
n=number of unpaired electron
Example 1: Find the spin magnetic moment of d^{6} configuration?
Answer: d^{6 }configuration :
number of unpaired electron = 4
Maximum spin multiplicity = 2|S|+1
S = total spin
E.g- for d^{6} configuration
Electrons having the same spin and energy present in degenerate orbitals can exchange their positions and in this exchange process, the energy is released and the released energy is termed exchange energy.
The higher the number of exchanges in a particular configuration, the stability of the configuration becomes higher.
The exchange energy is the basis for Hund's rule, which allows maximum multiplicity, that is electron pairing is possible only when all the degenerate orbitals contain one electron each.
more the number of exchange ∝ more the stability of configuration
=maximum number of possible exchange
Where n is the total number of electrons having same energy and spin
r = 2 (minimum 2 electrons required for exchange)
E.g- for d^{8} configuration
d^{8} configuration:
Case 1:
Number of electrons having same energy and spin (n) = 5
Minimum number of electrons required for exchange = 2
Case 2:
Number of electrons having same energy and spin (n) = 3
Minimum number of electrons required for exchange = 2
Total number of possible exchange = 10 + 3 = 13
We can categorize elements having same chemical properties on the basis of their same outer electronic configuration.
E.g: Li_{3}- 1s^{2},2s^{1}
Na_{11}- 1s^{2},2s^{2}2p^{6},3s^{1}
K_{19}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{6},4s^{1}
Li, Na, K have same outermost electronic conjugation ns^{1}
F_{9}- 1s^{2},2s^{2}2p^{5 }and Cl_{17}- 1s^{2},2s^{2}2p^{6},3s^{2}3p^{5 }have same chemical behaviour both have same outer electronic configuration ns^{2}np^{5}
Rules to identify group number, period number & block name of elements
Eg. F_{9} [He],2s^{2}2p^{5}; n=2 - belongs to 2^{nd} the period of periodic table
Na_{11} [Ne]3s^{1}; n=3 - belongs to 3^{rd} the period of periodic table
Eg.
F_{9} [He],2s^{2}2p^{5}; the last electron enters in p a subshell. So, Fluorine is a p block element
Na_{11} [Ne]3s^{1}; the last electron enters in s a subshell. So, Sodium is a s block element
E.g
F_{9} [He],2s^{2}2p^{5}; p block elements. So, Group no = 7 + 10 = 17
Na_{11} [Ne]3s^{1}; s block elements. So, Group no = 1
Fe_{26}1s^{2},2s^{2}2p^{6},3s^{2}3p^{6}3d^{6},4s^{2}; d block elements. So, Group no = 6 + 2 = 8
Video link:
Q1. Find the spin magnetic moment of manganese (Mn_{25})
Answer: (A)
Solution:
Electronic configuration of Mn = Mn_{25}- [Ar] 3d^{5},4s^{2} or [Ar] 4s^{2},3d^{5}
^{}
number of unpaired electron = 5
Q2. Which ion in their aqueous solution exhibit colour
A. Ti^{3+}
B. Sc^{3+}
C. Cu^{+}
D. Zn^{2+}
Answer: (A)
Solution:
Due to d-d transition generally ions having electronic configuration d^{1} to d^{9} exhibit colour and d^{0} to d^{10} are colourless.
Ti-[Ar] 3d^{2}4s^{2} and Ti^{3+}-[Ar] 3d^{1}, number of unpaired electron = 1 & it exhibit colour.
Sc-[Ar] 3d^{1}4s^{2} and Sc^{3+}-[Ar] 3d^{0}, number of unpaired electrons = 0 & it doesn’t exhibit colour.
Cu-[Ar] 3d^{10}4s^{1} and Cu^{+}-[Ar] 3d^{10}4s^{0}, number of unpaired electrons = 0 & it doesn’t exhibit colour.
Zn-[Ar] 3d^{10}4s^{2} and Zn^{2+}-[Ar] 3d^{10}4s^{0}, number of unpaired electrons = 0 & it doesn’t exhibit colour.
Q3. Condensed electronic configuration of F is
A. 1s^{2},2s^{2}2p^{5}
B. [He] 2s^{2}2p^{5}
C. K - 2, L - 7
D. All of these
Answer: (B)
Condensed electronic configuration of F is [He] 2s^{2}2p^{5}
Q 4. Carbon in its ground state is
A. Monovalent
B. Divalent
C. Trivalent
D. tetravalent
Answer: (B)
Solution: electronic configuration of C is 1s^{2}, 2s^{2}2p^{2}
^{}
Question 1. Can we write exactly the correct configurations of all elements by strictly following (n + l) rule, Pauli exclusion principle, and Hund's rule?
Answer: No, we can see many configurations which don’t obey these rules (specially Aufbau). Eg. Cr, Cu, Pd, Pt, etc.
Question 2. Is the outermost electronic configuration of elements along with a group of the periodic tables always the same?
Answer: Not always, we can observe in many cases,
Eg- He - 1s^{2} but Ne - 1s^{2},2s^{2}2p^{6}
We can observe the same in many cases in transition metal series.
Question 3. Is it necessary to start filling any orbitals with an upward arrow?
Answer: No, you can start filling orbitals with upward or downward spin but make sure, no electron pairing takes place in the orbitals in a sub-shell until each orbital is occupied by one electron with parallel spin.
Question 4. What are exchange energy and pairing energy?
Answer: Exchange energy is the energy released when two or more electrons with the same spin exchange their positions in the degenerate orbitals of a subshell. The more the options for exchange, the more the electron’s stability. The number of exchange pairs is maximum in half-filled orbitals, hence it is more stable compared to partially filled orbitals.
Pairing energy refers to the energy released with paired electrons sharing one orbital. The more the paired electrons the more the atom’s stability. The number of pairs is maximum in fully filled orbitals, hence it is more stable compared to partially filled orbitals.
Related topics:
Isotopes Isobars, Isotones & Isodiaphers |
Discovery of neutron |
Rutherford atomic model |
Thompson’s Atomic Model |
Atomic number and Mass number |
Quantum numbers |