E E E = − = − = − ⇒ = − II Key to brief answer question: . n = l = , , , four sub-shells ⇒ s, p, d, f l = m l = ; one 4s orbital. l = m l = – , , + ; three 4p orbitals. l = m l = – , – , , + , + ; five 4d orbitals.
l = m l = - , – , – , , + , + , + ; seven 4f orbitals Over all Sixteen orbitals. 11th Std Chemistry 11th Std Chemistry - - - - . Orbital l Radial node n – l – Angular node l 2s 4p 5d 4f . i) ground state ↿ ↿ ↿ ↿ ↿ ii) maximum exchange energy ↿ ↿ ↿ ↿ ↿ .
Orbital l 3p x d x y . ∆ ∆≥ ∆ ∆≥ ∆ ∆ ≥ x p h x p Kgm s x m v Kgm s . ( ) π Given ∆ v = . % v = .
× ms – m = . × – Kg ∆= v ms ms ∴∆≥ ∆≥ x Kgm s Kg ms x − m . Electronic configuration of oxygen = 1s 2s 2p ∴ th electron present in 2p x orbital and the quantum numbers are n = , l = , m l = either + or - and s = – / Electronic configuration of chlorine = 1s ⇌ 2s 2p 3s 3p 3p x 3p y 3p z ⇌ ⇌ ⇌ ⇌ ⇌ ⇌ ⇌ ↿ th electron present in 3P Z orbital and the quantum numbers are n = , l = , m l =either + or