Q13 / 155 · WBJEE 2019 · Chemistry
Which one of the following electronic arrangements is absurd?
- A$n=3, l=1, m=-1$
- B$n=3, l=0, m=0$
- Correct answerC$n=2,l=0, m=-1$
- D$n=2, l=1, m=0$
Solution
The three quantum numbers $n, l$ and $m$ that describe an orbital has integer values of 0,1,2,3 and so on.
The electronic arrangement of option (c) is absurd because
$$
\begin{array}{l}
n=2 \\
l=0 \\
m=0
\end{array}
$$
To sum up, where $n=2, l=0, m=0$ Hence, for $l=0,$ value of $m$ should be zero.
The electronic arrangement of option (c) is absurd because
$$
\begin{array}{l}
n=2 \\
l=0 \\
m=0
\end{array}
$$
To sum up, where $n=2, l=0, m=0$ Hence, for $l=0,$ value of $m$ should be zero.Hence, the correct answer is option: 'C'.