(B) According to the $(n+l)$ rule,electrons fill orbitals in order of increasing $(n+l)$ values.For $n=6$:$6s: n+l = 6+0 = 6$$4f: n+l = 4+3 = 7$$5d: n

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$ns \to (n-2)f \to (n-1)d \to np$

(B) According to the $(n+l)$ rule,electrons fill orbitals in order of increasing $(n+l)$ values.For $n=6$:$6s: n+l = 6+0 = 6$$4f: n+l = 4+3 = 7$$5d: n+l = 5+2 = 7$$6p: n+l = 6+1 = 7$Comparing the orbitals with the same $(n+l)$ value,the one with the lower $n$ value fills first.Thus,the order is $6s \to 4f \to 5d \to 6p$,which corresponds to $ns \to (n-2)f \to (n-1)d \to np$.

Class 11 Chemistry Structure of Atom Quantum number, Electronic configuration and Shape of orbitals

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If the principal quantum number $n=6,$ the correct sequence of filling of electrons will be

JEE MAIN 2015 All JEE MAIN

A
$ns \to (n-2)f \to np \to (n-1)d$
B
$ns \to (n-2)f \to (n-1)d \to np$
C
$ns \to np \to (n-1)d \to (n-2)f$
D
$ns \to (n-1)d \to (n-2)f \to np$

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Structure of Atom

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Quantum number, Electronic configuration and Shape of orbitals

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Describe the orbitals with the following quantum numbers using $s, p, d, f$ notation:
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The electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$ corresponds to:

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If the principal quantum number $n=6,$ the correct sequence of filling of electrons will be

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Describe the orbitals with the following quantum numbers using $s, p, d, f$ notation:
$(a) n = 1, l = 0$
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