JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 5)
Correct statements for an element with atomic number 9 are:
A. There can be 5 electrons for which $m_s = +\frac{1}{2}$ and 4 electrons for which $m_s = -\frac{1}{2}$.
B. There is only one electron in $p_z$ orbital.
C. The last electron goes to orbital with $n = 2$ and $l = 1$.
D. The sum of angular nodes of all the atomic orbitals is 1.
Choose the correct answer from the options given below:
Explanation
$$\begin{aligned} &\text { Element with atomic number } 9 \text { is Fluorine }\\ &F(9)=1 s^2 2 s^2 2 p^5 \end{aligned}$$
_8th_April_Evening_Shift_en_5_1.png)
(A) 5 electrons can be up-spin $\left[\mathrm{m}_{\mathrm{s}}=+\frac{1}{2}\right]$ and 4 electrons can be down spin $\left[\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\right]$
(B) Unpaired electron can be in anyone of $p_x, p_y$ or $p_z$ orbital
(C) Last electron is in 2 p subshell with $\mathrm{n}=2, \ell=1$
(D) Angular node for s-orbital $=0$ while of each p -orbital $=1$
Sum of all angular node $=3$
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