Periodic Classification of Elements

Electronic Configuration of elements

(1) Electronic Configuration of H (Hydrogen)

Atomic number = 1

Electronic configuration: 1s¹

Or, 1

Valence electrons = 1

(2) Electronic Configuration of He (Helium)

Atomic number = 2

Electronic configuration: 1s²

Or, 2

Valence electrons = 0

(Because He(helium) is an inert gas, and hence valence electros = 0)

(3) Electronic Configuration of Li (Lithium)

Atomic number = 3

Electronic configuration 1s²2s¹

Or, 2, 1

Or, [He] 1

Or, [He] 2s¹

Valence electrons = Electrons in outermost orbit = 1

(4) Electronic Configuration of Be (Beryllium)

Atomic number: 4

Electronic configuration: 1s²2s²

Or, [He] 2

Or, [He] 2s²

Or, 2, 2

Valence electrons = Electrons in outermost orbit = 2

(5) Electronic Configuration of B (Boron)

Atomic number: 5

Electronic configuration: 1s²2s²2p¹

Or, [He] 3

Or, [He] 2s² 2p¹

Or, 2, 3

Valence electrons = Number of electrons in outermost orbit = 3

(6) Electronic Configuration of C (Carbon)

Atomic number: 6

Electronic configuration of C(carbon) : 1s²2s²2p²

Or, [He] 4

Or, [He] 2s² 2p²

Or, 2, 4

Valence electrons of C (Carbon) = Number of electrons in outermost orbit = 4

(7) Electronic Configuration of N (Nitrogen)

Atomic number: 7

Electronic configuration of C(carbon) : 1s²2s²2p³

Or, [He] 5

Or, [He] 2s² 2p³

Or, 2, 5

Valence electrons of N (Nitrogen) = Number of electrons in outermost orbit = 5

(8) Electronic Configuration of O (Oxygen)

Atomic number: 8

Electronic configuration of O (Oxygen) : 1s²2s²2p⁴

Or, [He] 6

Or, [He] 2s² 2p⁴

Or, 2, 6

Valence electrons of O (Oxygen) = Number of electrons in outermost orbit = 6

(9) Electronic Configuration of F (Fluorine)

Atomic number: 9

Electronic configuration of F (Fluorine) : 1s²2s²2p⁵

Or, [He] 7

Or, [He] 2s² 2p⁵

Or, 2, 7

Valence electrons of F (Fluorine) = Number of electrons in outermost orbit = 7

(10) Electronic Configuration of Ne (Neon)

Atomic number: 10

Electronic configuration of Ne (Neon) : 1s²2s²2p⁶

Or, [He] 8

Or, [He] 2s² 2p⁸

Or, 2, 8

Valence electrons of Ne (Neon) = Number of electrons in outermost orbit = 0

Since, Ne (Neon) is an inert gas, thus valence electrons of Ne (Neon) = 0

(11) Electronic Configuration of Na (Sodium)

Atomic number: 11

Electronic configuration of Na (Sodium) : 1s²2s²2p⁶3s¹

Or, 2, 8, 1

Or, [Ne] 1

Or, [Ne] 3s¹

Valence electrons of Na (Sodium) = Number of electrons in outermost orbit = 1

(12) Electronic Configuration of Mg (Magnesium)

Atomic number of Mg (Magnesium)= 12

Electronic configuration of Mg (Magnesium) : 1s²2s²2p⁶3s²

Or, 2, 8, 2

Or, [Ne] 2

Or, [Ne] 3s²

Valence electrons of Mg (Magnesium) = Number of electrons in outermost orbit = 2

(13) Electronic Configuration of Al (Aluminium)

Atomic number of Al (Aluminium)= 13

Electronic configuration of Al (Aluminium) : 1s²2s²2p⁶ 3s²3p¹

Or, 2, 8, 3

Or, [Ne] 2, 1

Or, [Ne] 3s²3p¹

Valence electrons of Al (Aluminium) = Number of electrons in outermost orbit = 3

(14) Electronic Configuration of Si (Silicon)

Atomic number of Si (Silicon) = 14

Electronic configuration of Si (Silicon) : 1s²2s²2p⁶ 3s²3p²

Or, 2, 8, 4

Or, [Ne] 2, 2

Or, [Ne] 3s²3p²

Valence electrons of Si (Silicon) = Number of electrons in outermost orbit = 4

(15) Electronic Configuration of P (Phosphorous)

Atomic number of P (Phosphorous) = 15

Electronic configuration of P (Phosphorous) : 1s²2s²2p⁶ 3s²3p³

Or, [Ne] 3s²3p³

Or, [Ne] 2, 3

Or, 2, 8, 5

Valence electrons of P (Phosphorous) = Number of electrons in outermost orbit = 5

(16) Electronic Configuration of S (Sulphur)

Atomic number of S (Sulphur) = 16

Electronic configuration of S (Sulphur) : 1s²2s²2p⁶ 3s²3p⁴

Or, [Ne] 3s²3p⁴

Or, [Ne] 2, 4

Or, 2, 8, 6

Valence electrons of S (Sulphur) = Number of electrons in outermost orbit = 6

(17) Electronic Configuration of Cl (Chlorine)

Atomic number of Cl (Chlorine) = 17

Electronic configuration of Cl (Chlorine) : 1s²2s²2p⁶ 3s²3p⁵

Or, [Ne] 3s²3p⁵

Or, [Ne] 2, 5

Or, 2, 8, 7

Valence electrons of Cl (Chlorine) = Number of electrons in outermost orbit = 7

(18) Electronic Configuration of Ar (Argon)

Atomic number of Ar (Argon) = 18

Electronic configuration of Ar (Argon) : 1s²2s²2p⁶ 3s²3p⁶

Or, [Ne] 3s²3p⁶

Or, [Ne] 2, 6

Or, 2, 8, 8

Valence electrons of Ar (Argon) = Number of electrons in outermost orbit = 0

Since, Ar (Argon) is an inert gas, thus its valence electrons = 0

(19) Electronic Configuration of K (Potassium)

Atomic number of K (Potassium) = 19

Electronic configuration of K (Potassium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Or, Electronic configuration of K (Potassium): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s¹

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle)]

Or, [Ar] 4s¹

Or, [Ar] 4s¹

Or, 2, 8, 8, 1

Valence electrons of K (Potassium) = Number of electrons in outermost orbit = 1

(20) Electronic Configuration of Ca (Calcium)

Atomic number of Ca (Calcium) = 20

Electronic configuration of Ca (Calcium) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹

Or, Electronic configuration of Ca (Calcium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁰ 4s²

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle)]

Or, [Ar] 4s²

Or, [Ar] 4s²

Or, 2, 8, 8, 2

Valence electrons of Ca (Calcium) = Number of electrons in outermost orbit = 2

(21) Electronic Configuration of Sc (Scandium)

Atomic number of Sc (Scandium) = 21

Electronic configuration of Sc (Scandium) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹

Or, Electronic configuration of Sc (Scandium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle) And in the case of Sc (Scandium) after filling the 4s orbital electrons started to fill the 3d orbital.]

Or, [Ar] 4s² 3d¹

Or, [Ar] 3d¹ 4s²

Or, 2, 8, 8, 3

Valence electrons of Sc (Scandium) = Number of electrons in outermost orbit = 3

[∵ last electron enters in `d`-orbital in the case of Sc (Scandium), thus Sc (Scandium) is a `d`-block element. For `d`-block elements valence electrons is counting along with electrons in `d`-orbital]

(22) Electronic Configuration of Ti (Titanium)

Atomic number of Ti (Titanium) = 22

Electronic configuration of Ti (Titanium) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

Or, Electronic configuration of Ti (Titanium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle) And in the case of Ti (Titanium) after filling the 4s orbital electrons started to fill the 3d orbital.]

Or, [Ar] 4s² 3d²

Or, [Ar] 3d² 4s²

Or, 2, 8, 8, 4

Valence electrons of Ti (Titanium) = Number of electrons in outermost orbit = 4

[∵ last electron enters in `d`-orbital in the case of Ti (Titanium), thus Ti (Titanium) is a `d`-block element. For `d`-block elements valence electrons is counting along with electrons in `d`-orbital]

(23) Electronic Configuration of V (Vanadium)

Atomic number of V (Vanadium) = 23

Electronic configuration of V (Vanadium) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d³

Or, Electronic configuration of Ti (Titanium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s²

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle) And in the case of V (Vanadium) after filling the 4s orbital electrons started to fill the 3d orbital.]

Or, [Ar] 4s² 3d³

Or, [Ar] 3d³ 4s²

Or, 2, 8, 8, 5

Valence electrons of V (Vanadium) = Number of electrons in outermost orbit = 5

[∵ last electron enters in `d`-orbital in the case of V (Vanadium), thus V (Vanadium) is a `d`-block element. For `d`-block elements valence electrons is counting along with electrons in `d`-orbital]

(24) Electronic Configuration of Cr (Chromium)

Atomic number of Cr (Chromium) = 24

Electronic configuration of Cr (Chromium) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵

Or, Electronic configuration of Cr (Chromium) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹

[∵ 4s has less energy level than 3d, thus electrons will fill 4s before filling the 3d orbital (According to Aufbau 's Principle) And in the case of Cr (Chromium) after filling the 4s orbital electrons started to fill the 3d orbital.]

According to Hund's Rule, Completely half filled orbitals are more stable than less than completely half filled orbitals.

Thus, in the case of Cr (Chromium) in order to make atom stable, one electron from `4s` goes to `3d` and number of electrons in `3d` becomes `5` instead of `4`. By this way, there is only 1 (one) electron left in `4s` instead of `2`.

Thus, electronic configuration of Cr (Chromium) becomes [Ar] 4s¹ 3d⁵

Or, [Ar] 4s¹ 3d⁵

Or, [Ar] 3d⁵ 4s¹

Or, 2, 8, 8, 6

Valence electrons of Cr (Chromium) = Number of electrons in outermost orbit = 6

[∵ last electron enters in `d`-orbital in the case of Cr (Chromium), thus Cr (Chromium) is a `d`-block element. For `d`-block elements valence electrons is counting along with electrons in `d`-orbital]