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The number of electrons with $(n+\ell)$ values equal to 3,4 and 5 in an element with atomic number (z) 24 are respectively
(n $=$ principal quantum number and $\ell=$ azimuthal quantum number)
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(n $=$ principal quantum number and $\ell=$ azimuthal quantum number)
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Verified Answer
The correct answer is:
$8,7,5$
$\mathrm{Z}=24$ means the element is $\mathrm{Cr}$.
$$
\mathrm{Cr}=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1
$$
For $(n+1)=3$, we have $2 p$ and 3 s so number of such electrons $=6+2=8$.
For $(n+1)=4$, we have $3 p$ and $4 s$ so number of such electrons $=6+1=7$
For $(n+1)=5$, we have $3 d$ so number of such electrons $=5$.
$$
\mathrm{Cr}=1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^6 3 \mathrm{~d}^5 4 \mathrm{~s}^1
$$
For $(n+1)=3$, we have $2 p$ and 3 s so number of such electrons $=6+2=8$.
For $(n+1)=4$, we have $3 p$ and $4 s$ so number of such electrons $=6+1=7$
For $(n+1)=5$, we have $3 d$ so number of such electrons $=5$.
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