The quantum number of four electrons are given below : n = 4, l = 2, ml =–2, ms = –1/2 n = 3, l = 2, ml = 1, ms = +1/2 n = 4, l = 1, ml = 0, ms = +1/2 n = 3, l = 1, ml = 1, ms = –1/2 The correct order of their increasing enegies will be :

Structure of Atom PYQ — JEE Main 2019 · 8 Apr · Shift 1 · Q12 with Solution

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Correct answer: A

Identify the orbitals from the quantum numbers

Given electrons:

I: $n=4,\ l=2$
This is a $4d$ electron.
II: $n=3,\ l=2$
This is a $3d$ electron.
III: $n=4,\ l=1$
This is a $4p$ electron.
IV: $n=3,\ l=1$
This is a $3p$ electron.

The values of $m_l$ and $m_s$ do not affect the energy in a multi-electron atom for such ordering questions. Energy depends mainly on $n$ and $l$ .

Use the $(n+l)$ rule

For multi-electron atoms, orbital energy increases with increasing value of $(n+l)$ . If two orbitals have the same $(n+l)$ value, then the orbital with smaller $n$ has lower energy.

Now calculate:

I: $4d \Rightarrow n+l = 4+2 = 6$
II: $3d \Rightarrow n+l = 3+2 = 5$
III: $4p \Rightarrow n+l = 4+1 = 5$
IV: $3p \Rightarrow n+l = 3+1 = 4$

Arrange in increasing energy

Lowest: IV ( $3p$ ), since $n+l=4$
Next compare II ( $3d$ $3 d$ ) and III ( $4p$ $4 p$ ), both have $n+l=5$ $n + l = 5$
- Lower $n$ means lower energy
- So $3d < 4p$ , i.e. II < III
Highest: I ( $4d$ ), since $n+l=6$

Therefore,

\boxed{\text{IV} < \text{II} < \text{III} < \text{I}}

Match with options

This corresponds to Option A.

Comparison with stored correct answer

Stored correct answer: A
Derived answer: A
So, the derived answer agrees with the stored answer.

More from Structure of Atom

View written solutionFree

Correct answer: A

Identify the orbitals from the quantum numbers

Given electrons:

I: $n=4,\ l=2$
This is a $4d$ electron.
II: $n=3,\ l=2$
This is a $3d$ electron.
III: $n=4,\ l=1$
This is a $4p$ electron.
IV: $n=3,\ l=1$
This is a $3p$ electron.

The values of $m_l$ and $m_s$ do not affect the energy in a multi-electron atom for such ordering questions. Energy depends mainly on $n$ and $l$ .

Use the $(n+l)$ rule

For multi-electron atoms, orbital energy increases with increasing value of $(n+l)$ . If two orbitals have the same $(n+l)$ value, then the orbital with smaller $n$ has lower energy.

Now calculate:

I: $4d \Rightarrow n+l = 4+2 = 6$
II: $3d \Rightarrow n+l = 3+2 = 5$
III: $4p \Rightarrow n+l = 4+1 = 5$
IV: $3p \Rightarrow n+l = 3+1 = 4$

Arrange in increasing energy

Lowest: IV ( $3p$ ), since $n+l=4$
Next compare II ( $3d$ $3 d$ ) and III ( $4p$ $4 p$ ), both have $n+l=5$ $n + l = 5$
- Lower $n$ means lower energy
- So $3d < 4p$ , i.e. II < III
Highest: I ( $4d$ ), since $n+l=6$

Therefore,

\boxed{\text{IV} < \text{II} < \text{III} < \text{I}}

Match with options

This corresponds to Option A.

Comparison with stored correct answer

Stored correct answer: A
Derived answer: A
So, the derived answer agrees with the stored answer.

Structure of Atom question

Structure of Atom question

More from Structure of Atom

Structure of Atom question

Structure of Atom question

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