Arrange the following atomic subshells in order of increasing energy according to the Aufb… — JAMB UTME

Arrange the following atomic subshells in order of increasing energy according to the Aufbau principle and the $(n+l)$ rule, starting with the subshell of lowest energy:

1 $5s$
2 $4d$
3 $5p$
4 $4f$

Answer

The correct order of subshells from lowest to highest energy is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

According to the Aufbau principle and the

(n+l)

rule, subshells fill in order of increasing

(n+l)

value.

5s

has

(n+l) = 5+0 = 5

, making it lowest in energy. Both

4d

(

(n+l) = 4+2 = 6

) and

5p

(

(n+l) = 5+1 = 6

) have a sum of

6

, but

4d

has lower energy than

5p

because its principal quantum number

n=4

is smaller.

4f

has

(n+l) = 4+3 = 7

, placing it highest in energy. Thus, the correct sequence is

5s \rightarrow 4d \rightarrow 5p \rightarrow 4f

Step-by-Step Solution

Calculate the

(n+l)

value for each atomic subshell

For

5s

n=5, l=0 \implies n+l = 5

.
For

4d

n=4, l=2 \implies n+l = 6

.
For

5p

n=5, l=1 \implies n+l = 6

.
For

4f

n=4, l=3 \implies n+l = 7

According to Madelung's rule, orbitals fill in order of increasing

(n+l)

values.

Order subshells by increasing

(n+l)

sum

5s

(

(n+l)=5

) is lowest in energy, while

4f

(

(n+l)=7

) is highest.

Subshells with smaller

(n+l)

values are filled before those with larger

(n+l)

values.

Break ties for subshells with identical

(n+l)

values (

4d

and

5p

)

4d

(

n=4

) has lower energy than

5p

(

n=5

When two subshells share the same

(n+l)

value, the subshell with the smaller principal quantum number

n

is lower in energy.

Assemble the complete sequence from lowest to highest energy

5s < 4d < 5p < 4f

Combines the

(n+l)

rule and the tie-breaking principal quantum number rule.

Key Concept

Aufbau Principle and the (n+l) Rule

Question

Answer

Step-by-Step Solution

Key Concept

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