Easy
What are the electronic configurations of elements with atomic numbers $Z = 117$ and $Z = 120$?
- A$[Rn] 5f^{14} 6d^{10} 7s^2 7p^5$ and $[Og] 8s^2$
- B$[Rn] 5f^{14} 6d^{10} 7s^2 7p^6$ and $[Og] 8s^1$
- C$[Rn] 5f^{14} 6d^{10} 7s^2 7p^3$ and $[Og] 8s^2 8p^1$
- D$[Rn] 5f^{14} 6d^{10} 7s^2 7p^4$ and $[Og] 8s^2 8p^2$
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Similar Questions
The correct order of decreasing energy for the electrons whose quantum numbers $n$ and $l$ are given below, is
$A$. $n=5, l=2$
$B$. $n=5, l=0$
$C$. $n=4, l=3$
$D$. $n=4, l=1$
$A$. $n=5, l=2$
$B$. $n=5, l=0$
$C$. $n=4, l=3$
$D$. $n=4, l=1$
In Niobium $(Z = 41)$,the number of electrons with $m = -1$ will be:
Medium
View SolutionThe sum of paired electrons present in $l = 2$ orbitals for the species $Fe^{2+}$,$Co^{2+}$,and $Ni^{2+}$ is .......
Difficult
View SolutionDescribe the orbitals with the following quantum numbers using $s, p, d, f$ notation:
$(a) n = 1, l = 0$
$(b) n = 3, l = 1$
$(c) n = 4, l = 2$
$(d) n = 4, l = 3$
$(a) n = 1, l = 0$
$(b) n = 3, l = 1$
$(c) n = 4, l = 2$
$(d) n = 4, l = 3$
Medium
View SolutionThe orbital angular momentum of an electron in an $s$ orbital is:
DifficultIIT 1996
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