What is the electronic configuration of an element with Z=9?

The element is Fluorine (F), and its configuration is $1s^2 2s^2 2p^5$.

How many electrons have spin +1/2 in Fluorine?

There are 5 electrons with $m_s = +1/2$ and 4 with $m_s = -1/2$ (or vice versa).

What are the quantum numbers for the last electron of Fluorine?

The last electron enters a 2p orbital, so $n=2$ and $l=1$.

How many angular nodes does an s-orbital have?

An s-orbital ($l=0$) has zero angular nodes.

How many angular nodes does a p-orbital have?

A p-orbital ($l=1$) has exactly one angular node.

What is the total number of angular nodes in Fluorine?

Total angular nodes = $\sum l$. For $1s^2 2s^2 2p^5$, it is $0+0+(5 \times 1) = 5$.

Is statement B correct for Fluorine?

No, in $2p^5$, two p-orbitals are fully filled and one has a single electron. Any of $p_x, p_y, p_z$ could be the one with a single electron, but it is not a 'must' for $p_z$ specifically in a general state.

What does l=1 signify?

It signifies the Azimuthal quantum number for a p-subshell.

Can an element with Z=9 have 5 electrons with same spin?

Yes, following Hund's rule and Pauli's principle, the distribution allows for a 5/4 split in spin states.

Why is statement D incorrect?

Statement D says the sum of angular nodes is 1, but there are 5 electrons in p-orbitals, each contributing 1 angular node.

Correct statements for an element with atomic number 9 are: A. There can be 5 electrons for which ms = +1/2...

Correct statements for an element with atomic number 9 are: A. There can be 5 electrons for which ms = +1/2… | JEE Main Chemistry

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ChemistryAtomic Structure

QMCQ

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.

✅ Correct Answer

A and C Only

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Full Solution

Identify the Element

The element with atomic number $Z = 9$ is Fluorine (F). Its ground state electronic configuration is:

$$F (Z=9): 1s^2 2s^2 2p^5$$

Evaluate Statement A (Spin States)

In $1s^2$, there are 1 ($+1/2$) and 1 ($-1/2$). In $2s^2$, same. In $2p^5$, following Hund’s rule, we have 3 electrons with same spin (say $+1/2$) and 2 with opposite ($-1/2$).

Total $+1/2 = 1(1s) + 1(2s) + 3(2p) = 5$. Total $-1/2 = 1(1s) + 1(2s) + 2(2p) = 4$. Statement A is Correct.

Evaluate Statement B (Orbital Occupancy)

The $2p$ subshell has 3 orbitals: $p_x, p_y, p_z$. For 5 electrons, two orbitals must be full (2e each) and one must be half-filled (1e). Since these orbitals are degenerate, the single electron could be in any of them. It’s not a definitive rule for $p_z$. Statement B is Incorrect.

Evaluate Statement C (Quantum Numbers)

The “last” electron (9th electron) enters the $2p$ subshell.

For $2p$, Principal quantum number $n = 2$ and Azimuthal quantum number $l = 1$. Statement C is Correct.

Evaluate Statement D (Angular Nodes)

Angular nodes for any orbital is equal to $l$. Total angular nodes is the sum of $l$ values for every electron.

$1s^2 (l=0) \rightarrow 0$ nodes. $2s^2 (l=0) \rightarrow 0$ nodes. $2p^5 (l=1) \rightarrow 1$ node per electron. Total = $5 \times 1 = 5$. Statement D says sum is 1, so it is Incorrect.

Concluding Correct Statements

Based on the analysis, only statements A and C are scientifically accurate for Fluorine.

Final Selection

Match with options provided in the image: Option (C) states “A and C Only”.

Final Answer: A and C Only

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Theory

1. Electronic Configuration & Hund’s Rule

The electronic configuration of an atom describes the distribution of electrons in its atomic orbitals. For Fluorine (Z=9), the configuration is $1s^2 2s^2 2p^5$. According to Hund’s Rule of Maximum Multiplicity, for a given electron configuration, the lowest energy term is the one with the greatest value of spin multiplicity. This means in the $2p$ subshell, electrons first occupy all three orbitals ($p_x, p_y, p_z$) singly with parallel spins before pairing starts. For 5 electrons, this results in three $+1/2$ spins and two $-1/2$ spins within that subshell, contributing to the overall spin imbalance in the atom.

2. Quantum Numbers (n and l)

Quantum numbers are the “coordinates” of an electron. The principal quantum number ($n$) defines the main energy level or shell ($n=1, 2, 3…$). The azimuthal quantum number ($l$) defines the shape of the orbital and the subshell ($l=0$ for s, $l=1$ for p, $l=2$ for d). For the valence electrons of Fluorine, they reside in the $L$-shell ($n=2$) and specifically in the p-subshell ($l=1$). Understanding these values is critical for determining the energy, size, and shape of the region where the electron is most likely to be found.

3. Orbital Nodes (Radial and Angular)

Nodes are regions in an atom where the probability of finding an electron is zero. There are two types: Radial nodes and Angular nodes. Radial nodes depend on both $n$ and $l$ (Formula: $n – l – 1$). Angular nodes depend solely on the azimuthal quantum number $l$. An s-orbital ($l=0$) has no angular nodes (it is spherically symmetrical). A p-orbital ($l=1$) has one angular node (a nodal plane). In an atom like Fluorine, every electron in a p-orbital contributes exactly one angular node to the total wave function description of that specific state.

4. Spin Quantum Number (ms)

The spin quantum number $m_s$ describes the intrinsic angular momentum of an electron. It can only have two values: $+1/2$ (spin up) or $-1/2$ (spin down). Pauli’s Exclusion Principle states that no two electrons in an atom can have the same set of four quantum numbers. Therefore, an orbital can hold a maximum of two electrons, and they must have opposite spins. This governs why the $1s$ and $2s$ subshells in Fluorine have an equal number of up and down spins, while the partially filled $2p$ subshell creates the net spin difference.

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FAQs

Why is statement B specifically considered incorrect?

Because $p_x, p_y,$ and $p_z$ are degenerate (same energy). There is no physical reason why the single electron must stay in $p_z$ specifically; it could be in $p_x$ or $p_y$.

What is the formula for total nodes?

Total nodes in a single orbital is $(n – 1)$. This is the sum of radial nodes $(n-l-1)$ and angular nodes $(l)$.

How many radial nodes does the 2p orbital have?

Using $n-l-1$: $2 – 1 – 1 = 0$. So, $2p$ orbitals have zero radial nodes.

Can $m_s$ be 5 for -1/2 and 4 for +1/2?

Yes. The choice of which spin is “up” or “down” is arbitrary. The ratio 5:4 is what matters.

What is the shape of the angular node in a p-orbital?

It is a flat plane passing through the nucleus, perpendicular to the axis of the orbital (e.g., the xy-plane for a $p_z$ orbital).

Is Fluorine paramagnetic or diamagnetic?

Fluorine is paramagnetic because it has one unpaired electron in its $2p$ subshell.

What if Z was 10 (Neon)?

Then there would be 5 electrons of each spin, it would be diamagnetic, and it would have 6 angular nodes total.

Does $l=1$ always mean a p-orbital?

Yes, the azimuthal quantum number $l=1$ is uniquely assigned to p-subshells across all energy levels.

Is the “last electron” always the highest energy one?

Usually, yes. In Z=9, the 9th electron is in the highest energy occupied subshell ($2p$).

Why does $2s$ have 0 angular nodes?

Because for any s-orbital, $l=0$. Since angular nodes $= l$, s-orbitals never have angular nodes.

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