Answer to Question #127019 in General Chemistry for Rosé A

Question

The arrangement of orbitals on the basis of energy is based upon their (n+l ) value. Lower the value of (n+l ), lower is the energy. For orbitals having same values of (n+l), the orbital with lower value of n will have lower energy.

I. Based upon the above information, arrange the following orbitals in the increasing order of energy.

(a) 1s, 2s, 3s, 2p

(b) 4s, 3s, 3p, 4d

(c) 5p, 4d, 5d, 4f, 6s

(d) 5f, 6d, 7s, 7p

Answer 1

$In\ (a)\ (n+1)\ values\ are\newline \ \ \ 1s=1+0=1$

$2s=2+0=2\newline 3s=3+0=3\newline 2p=2+1=3$

Therefore, the increasing order in the energy is $1s\ <2s<2p<3s$

$In\ (b)\ (n+1)\ \newline$

$4s=4+0=4\newline 3s+3+0=3\newline 3p=3+1=4\newline 4d=4+2=6$

Increasing order, $3s<3p<4s<4d$

$In\ (c),$

$5p=5+1=6\newline 4d=4+2=6\newline 5d=5+2=7\newline 4f=4+3=7\newline 6s=6+0=6$

Increasing order,$\ 4d<5p<6s<4f<5d$

$In\ (d),$

$5f=5+3=8\newline 6d=6+2=8\newline 7s=7+0=7\newline 7p=7+1=8$

Therefore, increasing order is, $7s<5f<6d<7p$