Inside the atom

The problem asks for the maximum electron count for a single $d$-sublevel. In electron configuration, each sublevel type ($s,p,d,f$) has a specific electron capacity.

In studying Atomic Structure and Electronic Configuration, it's important to recall the maximum electron count for each sublevel. The typical capacities are: $s$-sublevel: $2$, $p$-sublevel: $6$, $d$-sublevel: $10$, $f$-sublevel: $14$.

Option A lists $2$ electrons, which is the capacity for an $s$-sublevel, not a $d$-sublevel.

Option B lists $6$ electrons, which matches the capacity for a $p$-sublevel.

Option D lists $14$ electrons, which is the capacity of an $f$-sublevel.

Based on memorized capacities, a $d$-sublevel holds up to $10$ electrons, which aligns with Option C.

The task is to find the maximum number of electrons that can fit in a $d$-sublevel. To solve this, we need to know the number of orbitals in that sublevel.

In the structure of atoms, sublevels contain a specific, odd number of orbitals: $s$-sublevel: $1$ orbital, $p$-sublevel: $3$ orbitals, $d$-sublevel: $5$ orbitals, $f$-sublevel: $7$ orbitals.

The Pauli Exclusion Principle states that each orbital can hold up to $2$ electrons with opposite spins. This is true for any orbital, regardless of the sublevel it belongs to.

Multiply the number of orbitals by the number of electrons per orbital: $extTotalelectrons=5extorbitalsimes2extelectrons/orbital$ $extTotalelectrons=10$

Review the options to confirm: * Option A is wrong: $2$ is for an $s$-sublevel. * Option B is wrong: $6$ is for a $p$-sublevel. * Option D is wrong: $14$ is for an $f$-sublevel. Our calculation confirms Option C is correct.

The question is about the maximum electrons in a $d$-sublevel, which can be determined from the $d$-block of the Periodic Table.

On the Periodic Table, elements are grouped into blocks by the sublevel being filled with electrons. The $d$-block, containing transition metals, is centrally located (Groups 3-12).

Per the Aufbau principle, moving across a period adds one proton to the nucleus and one electron to the valence shell. Therefore, the number of elements in a block matches the maximum electrons the sublevel can hold.

For a row in the $d$-block (e.g., Period 4), count: Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), Zinc (Zn). There are 10 elements across.

Using other block widths: * Option A: 2 electrons fit an $s$-block. * Option B: 6 electrons fit a $p$-block. * Option D: 14 electrons fit an $f$-block.

The $d$-block spans 10 elements, so a $d$-sublevel holds 10 electrons.

To find the electron capacity of a $d$-sublevel, identify its angular momentum quantum number ($l$). The $l$ values are: $s$-sublevel: $l=0$, $p$-sublevel: $l=1$, $d$-sublevel: $l=2$, $f$-sublevel: $l=3$. Use $l=2$ for the $d$-sublevel.

The formula to calculate maximum electrons is: $extMaxelectrons=2(2l+1)$. Here, $(2l+1)$ indicates the number of orbitals in the sublevel, and $2$ shows the max electrons each orbital can have (Pauli Exclusion Principle).

Insert $l=2$ for the $d$-sublevel into the formula: $\text{Max electrons}=2(2(2)+1)=2(4+1)=2(5)=10$. This calculation shows a $d$-sublevel has 5 orbitals, each holding 2 electrons.

Verify other options: * Option A (2): Matches an $s$-sublevel ($l=0\to 2$). * Option B (6): Matches a $p$-sublevel ($l=1\to 6$). * Option D (14): Matches an $f$-sublevel ($l=3\to 14$).

The quantum number formula confirms a $d$-sublevel can hold 10 electrons, matching Option C.

In the periodic table, Group 1 elements like Lithium, Sodium, and Potassium are known for their high reactivity due to a single electron in their outer shell. When they come into contact with water, they react very energetically.

When an alkali metal ($M$) reacts with water, the standard equation is: $2M(s)+2H_{2}O(l)\to 2MOH(aq)+H_2(g)$ This equation indicates that the reaction produces a metal hydroxide and hydrogen gas.

While alkali metals do form oxides when reacting with oxygen, their interaction with water results in hydroxides, not oxides, and doesn't release oxygen gas.

Group 1 metals are famously highly reactive with water. For instance, Potassium reacts so intensely that the hydrogen it produces often catches fire, burning with a lilac flame.

The phrase "metal acid" is a misconception. The resulting $MOH$ is a strong base, which is why these are termed "alkali metals."

According to the general equation, the reaction yields a metal hydroxide and hydrogen gas, corresponding to the description in Option D.

To tackle this, recall that Group 1 alkali metals (like Lithium, Sodium, and Potassium) are extremely reactive due to having one electron in their outer shell, which is easily lost.

In lab settings, metals like sodium or potassium react vigorously on contact with water. Due to their high reactivity, they are stored under oil to prevent accidental reactions with air moisture. Thus, any option suggesting no reaction can be excluded.

The name "Alkali metals" hints at their ability to create a basic solution upon reacting with water. These metals don't form acids, so options suggesting acid formation can be disregarded.

During the reaction, gas bubbles form, and the gas can be identified by the "squeaky pop" test with a lit splint, confirming it as hydrogen gas ($H_2$). Oxygen gas would reignite a glowing splint, and the metals form hydroxides instead of oxides in water.

By eliminating the options involving no reaction, acid formation, and oxygen gas, the remaining chemical equation for this reaction (where $M$ is the alkali metal) is: $2M(s)+2H_{2}O(l)ightarrow2MOH(aq)+H_2(g)$ This verifies that the products are a metal hydroxide and hydrogen gas.

Elements are arranged in the periodic table into blocks (s, p, d, and f) based on the subshell being filled by their outermost electrons. The s-block includes the first two columns: Group 1 and Group 2, along with Helium.

The s-subshell features a single orbital capable of accommodating two electrons. Elements in Group 1, like Lithium and Sodium, hold one electron in their outer s-subshell, resulting in a typical configuration of: $n{s}^1$

For elements in Group 2, such as Beryllium and Magnesium, an additional electron fills the s-subshell to its maximum. Consequently, their electron configuration terminates with: $n{s}^2$

Consider why these don't suit the s-block: - Option A ($n{f}^{14}$) is linked to the f-block (lanthanides and actinides). - Option C ($n{d}^{10}$) corresponds to the d-block (transition metals). - Option D ($n{p}^1$) is associated with the start of the p-block (Group 13).

With the s-block only comprising two groups, the electron configuration of an s-block element can only conclude with $n{s}^1$ or $n{s}^2$.

For elements in the s-block, their outermost electrons reside in an s-subshell, meaning any valid configuration must end with '$s$'.

We're looking solely for s-subshell endings, so we dismiss any configurations ending in $f$, $d$, or $p$ subshells, as these belong to different blocks: - Option A with an $f$-subshell ($n{f}^{14}$) is of the f-block. - Option C with a $d$-subshell ($n{d}^{10}$) is of the d-block. - Option D with a $p$-subshell ($n{p}^1$) is of the p-block.

After eliminating $f$, $d$, and $p$ subshell options, the remaining choice with an s-subshell is correct. The s-subshell can hold up to 2 electrons, with $n{s}^1$ for Group 1 and $n{s}^2$ for Group 2.

The focus is on the s-block of the periodic table, which includes elements in Group 1 and Group 2, plus Helium, where the outer electrons fill an s-subshell.

To determine configurations, consider the s-subshell capacity: 1. An s-subshell has one orbital. 2. By the Pauli Exclusion Principle, one orbital can contain two electrons. Therefore, possible configurations are: - $n{s}^1$ (one electron) - $n{s}^2$ (two electrons)

The question is about the s-block, so any configurations with $p$, $d$, or $f$ subshells for outer electrons must be incorrect.

With $n{s}^1$ typical for alkali metals, the only other s-block configuration occurs when the orbital is full, resulting in $n{s}^2$, as seen in alkaline earth metals and helium.

Using the Period Number Method, start by identifying bromine's ($\text{Br}$) spot on the periodic table. As mentioned, bromine is situated in the 4th period.

When exploring the periodic table's structure, it's important to note that the period number shows the highest principal energy level ($n$) that contains electrons in an atom's ground state. As bromine is in the 4th period, its electrons extend up to the 4th principal energy level ($n=4$).

Since bromine's maximum energy level is $n=4$, options can be dismissed as follows: * Option 1 and 2: Levels 1 and 2 must fill up before reaching level 4. * Option 3: Though level 3 accommodates electrons (like the $3d$ subshell), it isn't the highest level in period 4.

With bromine being in period 4, the top principal energy level with electrons is $n=4$. This is validated by its electron configuration: $[\text{Ar}]3d^{10}4s^{2}4p^5$ The number "4" in $4s$ and $4p$ signifies the principal energy level.

To find the electron setup, first, locate bromine's atomic number ($Z$) on the periodic table. It's in the 4th period and 17th group, with atomic number 35, meaning it has 35 electrons in a neutral ground-state atom.

Following the Aufbau principle, orbitals are filled in order of rising energy: $1s,2s,2p,3s,3p,4s,3d,4p$. Bromine's ($Z=35$) full electron configuration is: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^5$ Alternatively, the condensed form using Argon is: $[Ar]3d^{10}4s^{2}4p^5$

In electron configurations, the leading numbers before subshells (like the 4 in $4s^2$ and $4p^5$) denote the principal quantum number ($n$), showing the principal energy level. Reviewing the configuration: - Electrons are in levels $n=1,n=2,n=3$, and $n=4$. - Despite $3d$ filling after $4s$, the largest coefficient is 4.

Based on the highest $n$ value found, assess the options: * Option 1 (1): This is the lowest energy level, not the top. * Option 2 (2): Electrons are here, but it's not the highest level. * Option 3 (3): Although the $3d$ subshell has many electrons, the $4s$ and $4p$ subshells are in a higher principal level ($n=4$).

Bromine's placement in period 4 indicates its valence electrons are in $n=4$ level. Thus, the highest principal energy level with electrons is 4.