Easy
The electronic configuration of $Sc$ $(Z = 21)$ is:
- A$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1$
- B$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^2$
- C$1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^3$
- D$1s^2 2s^2 2p^6 3s^2 3p^2 4s^2 3d^2$
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Similar Questions
The orbital angular momentum of an electron in a $d$ orbital is equal to:
If the outermost electrons of an element are represented as $(a)$ $3s^1$,$(b)$ $2p^3$,and $(c)$ $3p^5$,what will be the atomic numbers of these elements?
Easy
View SolutionArrange the following orbitals in decreasing order of energy:
$A$. $n = 3, l = 0, m = 0$
$B$. $n = 4, l = 0, m = 0$
$C$. $n = 3, l = 1, m = 0$
$D$. $n = 3, l = 2, m = 1$
The correct option for the order is:
$A$. $n = 3, l = 0, m = 0$
$B$. $n = 4, l = 0, m = 0$
$C$. $n = 3, l = 1, m = 0$
$D$. $n = 3, l = 2, m = 1$
The correct option for the order is:
Two statements are given below:
Statement $I$: In $H$ atom,the energy of $2s$ and $2p$ orbitals is same.
Statement $II$: In $He$ atom,the energy of $2s$ and $2p$ orbitals is same.
The correct answer is:
Statement $I$: In $H$ atom,the energy of $2s$ and $2p$ orbitals is same.
Statement $II$: In $He$ atom,the energy of $2s$ and $2p$ orbitals is same.
The correct answer is:
Assertion :- In $H$-atom,the energy of $3d$-level is smaller than $4s$-level.
Reason :- In multi-electron species,an orbital with lower value of $(n+\ell)$ has energy smaller than the orbital with larger value of $(n+\ell)$.
Reason :- In multi-electron species,an orbital with lower value of $(n+\ell)$ has energy smaller than the orbital with larger value of $(n+\ell)$.
Easy
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