which of the following electronic transitions represents an increase in energy in a multi - electron atom?
answer:
a 4s→3d
b 5p→4d
c 4d→5s
d 4f→6s
e 5d→6s

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Accepted Answer

# Brief Explanations:
In a multi - electron atom, the energy of an orbital is determined by the $n + l$ rule (where $n$ is the principal quantum number and $l$ is the azimuthal quantum number).
- For option A:
    - For $4s$ orbital, $n = 4$ and $l=0$, so $n + l=4+0 = 4$.
    - For $3d$ orbital, $n = 3$ and $l = 2$, so $n + l=3 + 2=5$.
    - Since $4<5$, the $3d$ orbital has higher energy than the $4s$ orbital. So, the transition $4s
ightarrow3d$ represents an increase in energy.
- For option B:
    - For $5p$ orbital, $n = 5$ and $l = 1$, so $n + l=5+1=6$.
    - For $4d$ orbital, $n = 4$ and $l = 2$, so $n + l=4 + 2=6$. When $n + l$ values are equal, the orbital with higher $n$ has higher energy. So, $5p$ has higher energy than $4d$, and the transition $5p
ightarrow4d$ is a decrease in energy.
- For option C:
    - For $4d$ orbital, $n = 4$ and $l = 2$, so $n + l=4+2 = 6$.
    - For $5s$ orbital, $n = 5$ and $l=0$, so $n + l=5+0 = 5$. Since $6>5$, $4d$ has higher energy than $5s$, and the transition $4d
ightarrow5s$ is a decrease in energy.
- For option D:
    - For $4f$ orbital, $n = 4$ and $l = 3$, so $n + l=4+3 = 7$.
    - For $6s$ orbital, $n = 6$ and $l=0$, so $n + l=6+0 = 6$. Since $7>6$, $4f$ has higher energy than $6s$, and the transition $4f
ightarrow6s$ is a decrease in energy.
- For option E:
    - For $5d$ orbital, $n = 5$ and $l = 2$, so $n + l=5+2 = 7$.
    - For $6s$ orbital, $n = 6$ and $l=0$, so $n + l=6+0 = 6$. Since $7>6$, $5d$ has higher energy than $6s$, and the transition $5d
ightarrow6s$ is a decrease in energy.

# Answer:
A. $4s
ightarrow3d$

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