Periodic Classification of Elements - Class 10th Science

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

[∵ the last electron enters in d-orbital in the case of Sc (Scandium), thus Sc (Scandium) is a d-block element. For the d-block elements valence electrons are counted 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

[∵ the 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 are counted 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

[∵ the 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 are counted 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 the atom stable, one electron from 4s goes to 3d, and the number of electrons in 3d becomes 5 instead of 4. In 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

[∵ the 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 are counted along with electrons in d-orbital]

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