Hund's Rule Important Problems By-N.K.D. Sir Please -Rate review and share this lesson

Hund's Rule of maximum Multiplicity According to this rule: "Electrons never pair until no available empty degenerate orbitals are left to him This means an electron always occupies a vacant orbital in the same sub-shell (degenerate orbital) and pairing starts only when all of the degenerate orbitals are filled up. This means that the pairing starts with 2nd electron in a sub-shell, 4th electron in p-sub-shell, 6th electron in d-sub-shell and 8 electron in f-sub-shell By doing this, the electrons stay as far away from each other as possible. This is highly reasonable if we consider the electron-electron repulsion. Hence electrons obey Hund's rule as it results in lower energy state and hence more stability

Extra Stability of Half and fully Filled Orbitals: A particularly stable system is obtained when a set of equivalent orbitals (degenerate orbitals) is either fully filled or half filled, i.e. each containing one or a pair of electrons. This effect is more dominant in d and f sub-shells This means three or six electrons in p-sub-shell, five or ten electrons in d-sub-shell, and seven or fourteen electrons in f-sub-shell forms a stable arrangement. Note this effect when filling of electrons takes place in d sub-shells (for atomic number Z 24, 25, and 29, 30) In the following table you should analyse how to employ the above rules to write electronic configuration of various elements. Electronic configuration of an element is represented by the notation x: number of electrons present in an orbital 1: denotes the sub-shell n: principal quantum number.

How many elements would be in the second period of the periodic table if the spin quantum number m, could have the value-10,? Solution: For second period n 1 m 00 2, hence, , +-,0,-2 1-1 +1.0,- Hence, total number of electrons = 12 (= total values of spin quantum number)

To what effective potential a proton beam be subjected to give its protons a wavelength of 25.2 nm? Solution Energy in Joules charge on the electron in coulomb x potential difference in volts. E eV KE eV (mv)-- 2meV; mv - V2meV From Debroghie equation we have - 2me :. we have 25.2x 10% V-1.299x106Volts 6.626 x 10-34 V2x1.66 x 102 x1.602 x 10-19 x V v 2-1.66 x 10 27 x 1.602 x10 Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles.

If a stationary proton and -particle are accelerated through same potential, then calculate the ratio of their de-Broglie's wavelengths. Solution: 2 (mv)2 2meV for proton pp 2mpepVp for o-particle p- /2m,e V (from de-Broglie's equation) A, pp Given V. V pmae m- 4 units 2m e, V me mp 1 unit 4222 (1) x(1)

The accelerating potential to be imparted to a beam of electrons to give an effective wavelength of 2.42 nm is 2.56 V (B) 0.256 V(D) 3.42 V 0.342 V Solution: -- For electron ,1150 x 10-8 2.42 x 10-7150 (24.2)2- 150 eV 108 cm V-0.256 Volt The quantum number not obtained from the Schrodinger's wave equation is n (B)I m (D) s

Hund's Rule Important Problems By-N.K.D. Sir Please -Rate review and share this lesson

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Niranjay Kumar Dwivedi

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Complete Chemistry faculty for IIT~JEE, and NEET examination.
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Manoj Gurjar

2 years ago

bahut hi sandar

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Kapil Charan

a year ago

thank u sir