Easy
Which one of the following configurations represents a noble gas?
- A$1s^2, 2s^2 2p^6, 3s^2$
- B$1s^2, 2s^2 2p^6, 3s^1$
- C$1s^2, 2s^2 2p^6$
- D$1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^2$
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Under normal conditions,which of the following electronic configurations is able to form a dipositive ion?
Easy
View Solution"No two electrons in an atom can have the same set of all four quantum numbers." This principle is called:
$A$ $3p$ orbital has
Values of the four quantum numbers for the last electron in the atom are $n = 4, l = 1, m = +1$ and $s = -1/2$. Atomic number of the atom will be
Medium
View SolutionWhich one is a wrong statement?
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