If n = 6, the correct sequence for filling of electrons will be :

Question

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