answer the following questions.
1. describe how the quantum theory of atomic structure differs from bohr’s theory. (hint: focus on electrons)
2. differentiate orbit from orbital.
3. what do electron configurations tell us about an atom?
4. how many orbitals are in the fourth energy level?
5. how many orbitals of each type are there? s- ______ p- ______ d- ______ f- ______
6. draw the 2s and 2px orbital. ■ 2s ■ 2pₓ ■
7. write the electron configurations of the following elements using the full and condensed forms.
a. phosphorus ______ ______
b. nickel ______ ______
c. neon ______ ______
d. potassium ______ ______
e. titanium ______ ______

Question

answer the following questions.
1. describe how the quantum theory of atomic structure differs from bohr’s theory. (hint: focus on electrons)
2. differentiate orbit from orbital.
3. what do electron configurations tell us about an atom?
4. how many orbitals are in the fourth energy level?
5. how many orbitals of each type are there? s- ______ p- ______ d- ______ f- ______
6. draw the 2s and 2px orbital. ■ 2s ■ 2pₓ ■
7. write the electron configurations of the following elements using the full and condensed forms.
   a. phosphorus ______ ______
   b. nickel ______ ______
   c. neon ______ ______
   d. potassium ______ ______
   e. titanium ______ ______

Accepted Answer

### Question 1
# Brief Explanations:
Bohr's theory: Electrons move in fixed, circular orbits (defined paths) around the nucleus, with definite energy levels (electrons can jump between levels by absorbing/emitting photons). Quantum theory: Electrons don't have definite paths; we describe their location using probability (orbitals, regions of high probability). Quantum theory uses quantum numbers (n, l, mₗ, mₛ) to describe electron states, and electrons have wave - particle duality. Bohr's model is a simplified, early model; quantum theory is more accurate, based on wave mechanics and probability.
# Answer:
- Bohr's theory: Electrons follow fixed, circular orbits (definite paths) around the nucleus, with discrete energy levels (electrons transition between levels by photon absorption/emission).
- Quantum theory: Electrons do not have definite paths; their location is described by probability (orbitals, regions of high electron - probability density). It uses quantum numbers ($n$, $l$, $m_l$, $m_s$) to define electron states and accounts for wave - particle duality of electrons.

### Question 2
# Brief Explanations:
Orbit (Bohr's model): A definite, circular path where an electron moves around the nucleus (e.g., in Bohr's hydrogen atom model, electrons orbit like planets around the sun, with fixed radii and energies). Orbital (quantum theory): A three - dimensional region in space around the nucleus where there is a high probability of finding an electron. It is described by quantum numbers and has specific shapes (s - spherical, p - dumbbell - shaped, etc.). Orbits are definite paths (classical concept), while orbitals are probability - based (quantum concept).
# Answer:
- **Orbit**: A definite, circular path (in Bohr’s model) where an electron moves around the nucleus (classical, definite - path concept).
- **Orbital**: A 3D region around the nucleus with high probability of finding an electron (quantum, probability - based concept, described by quantum numbers, with shapes like spherical (s), dumbbell - shaped (p), etc.).

### Question 3
# Brief Explanations:
Electron configurations show the distribution of electrons in an atom's orbitals. They tell us: 1. The energy levels (shells, $n$) and sub - levels (sub - shells, $l$) occupied by electrons. 2. The number of valence electrons (electrons in the outermost shell), which determines chemical reactivity (e.g., atoms with 8 valence electrons are stable, like noble gases). 3. The atom's chemical properties (e.g., metals have few valence electrons, non - metals have more; this affects bonding, ionization energy, etc.). 4. The atom's magnetic properties (e.g., unpaired electrons make an atom paramagnetic).
# Answer:
Electron configurations tell us:
- The distribution of electrons in an atom’s energy levels (shells) and sub - levels (sub - shells).
- The number of valence electrons (outermost - shell electrons), which determines chemical reactivity (e.g., stability, bonding behavior).
- Chemical properties (e.g., metallic/non - metallic character, ability to form ions/bonds).
- Magnetic properties (e.g., presence of unpaired electrons indicates paramagnetism).

### Question 4
# Explanation:
## Step 1: Recall the formula for the number of orbitals in an energy level
The number of orbitals in an energy level with principal quantum number $n$ is given by the formula $N = n^{2}$.
## Step 2: Substitute $n = 4$ into the formula
For the fourth energy level, $n=4$. So we calculate $N = 4^{2}=16$.
# Answer:
The number of orbitals in the fourth energy level is $\boldsymbol{16}$.

### Question 5
# Explanation:
## Step 1: Recall the number of orbitals for each sub - shell
- For an $s$ - subshell ($l = 0$), the magnetic quantum number $m_l$ can only take the value $0$. So the number of $s$ - orbitals is $1$.
- For a $p$ - subshell ($l = 1$), the magnetic quantum number $m_l$ can take the values $- 1,0, + 1$. So the number of $p$ - orbitals is $3$.
- For a $d$ - subshell ($l = 2$), the magnetic quantum number $m_l$ can take the values $-2,-1,0, + 1, + 2$. So the number of $d$ - orbitals is $5$.
- For an $f$ - subshell ($l = 3$), the magnetic quantum number $m_l$ can take the values $-3,-2,-1,0, + 1, + 2, + 3$. So the number of $f$ - orbitals is $7$.
# Answer:
s - $\boldsymbol{1}$; p - $\boldsymbol{3}$; d - $\boldsymbol{5}$; f - $\boldsymbol{7}$

### Question 6
# Brief Explanations:
- **2s orbital**: It is a spherical (symmetric around the nucleus) region of space. It is a single orbital (since $s$ - subshell has 1 orbital). The 2s orbital is larger in size than the 1s orbital and has a node (a region with zero electron probability) between the nucleus and the main electron - density region.
- **2pₓ orbital**: The $p$ - orbitals are dumbbell - shaped. The 2pₓ orbital is oriented along the $x$ - axis (with lobes along the $x$ - axis and a nodal plane (the $yz$ - plane) passing through the nucleus, where electron probability is zero).

(Note: For drawing, the 2s can be represented as a sphere (with a smaller inner sphere - like region for the node, though often simplified as a single sphere), and the 2pₓ as a dumbbell with lobes on the $x$ - axis, and a circle (or oval) at the nucleus representing the nodal plane perpendicular to the $x$ - axis.)
# Answer:
- **2s orbital**: Draw a sphere (symmetric around the nucleus) to represent the 2s orbital. It has a spherical shape with a node (a region of zero electron probability) between the nucleus and the main electron - density region (often simplified as a single sphere for basic representation).
- **2pₓ orbital**: Draw a dumbbell - shaped figure with lobes along the $x$ - axis. There is a nodal plane (e.g., the $yz$ - plane) passing through the nucleus (represented as a line or a circle at the nucleus, perpendicular to the $x$ - axis) where electron probability is zero.

### Question 7
#### Part (a): Phosphorus (Atomic number $Z = 15$)
# Explanation:
## Step 1: Determine the electron filling order
The electron filling order follows the Aufbau principle: $1s\to2s\to2p\to3s\to3p$.
## Step 2: Write the full electron configuration
Phosphorus has 15 electrons. Filling the orbitals: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$.
## Step 3: Write the condensed electron configuration
The noble gas before phosphorus is neon ($Ne$, with electron configuration $1s^{2}2s^{2}2p^{6}$). So the condensed configuration is $[Ne]3s^{2}3p^{3}$.
# Answer:
- Full: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$
- Condensed: $[Ne]3s^{2}3p^{3}$

#### Part (b): Nickel (Atomic number $Z = 28$)
# Explanation:
## Step 1: Determine the electron filling order
The electron filling order is $1s\to2s\to2p\to3s\to3p\to4s\to3d$.
## Step 2: Write the full electron configuration
Nickel has 28 electrons. Filling the orbitals: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$ (note: $4s$ fills before $3d$ according to Aufbau, and then $3d$ is filled).
## Step 3: Write the condensed electron configuration
The noble gas before nickel is argon ($Ar$, with electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$). So the condensed configuration is $[Ar]4s^{2}3d^{8}$.
# Answer:
- Full: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{8}$
- Condensed: $[Ar]4s^{2}3d^{8}$

#### Part (c): Neon (Atomic number $Z = 10$)
# Explanation:
## Step 1: Determine the electron filling order
The electron filling order is $1s\to2s\to2p$.
## Step 2: Write the full electron configuration
Neon has 10 electrons. Filling the orbitals: $1s^{2}2s^{2}2p^{6}$.
## Step 3: Write the condensed electron configuration
Neon is a noble gas, so its condensed configuration is the same as its full configuration (or we can consider it as $[Ne]$ for elements after it, but for neon itself, the full and condensed (in terms of noble gas core) is $1s^{2}2s^{2}2p^{6}$ (since there is no noble gas before neon with a smaller atomic number that can be used as a core in the same way; however, if we consider, the noble gas core for neon is itself).
# Answer:
- Full: $1s^{2}2s^{2}2p^{6}$
- Condensed: $1s^{2}2s^{2}2p^{6}$ (or $[Ne]$ when used as a core for other elements)

#### Part (d): Potassium (Atomic number $Z = 19$)
# Explanation:
## Step 1: Determine the electron filling order
The electron filling order is $1s\to2s\to2p\to3s\to3p\to4s$ (note: $4s$ fills before $3d$ for potassium as $3d$ is higher in energy than $4s$ for $n = 4$ and $n=3$ respectively in terms of filling order).
## Step 2: Write the full electron configuration
Potassium has 19 electrons. Filling the orbitals: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$.
## Step 3: Write the condensed electron configuration
The noble gas before potassium is argon ($Ar$, with electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$). So the condensed configuration is $[Ar]4s^{1}$.
# Answer:
- Full: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$
- Condensed: $[Ar]4s^{1}$

#### Part (e): Titanium (Atomic number $Z = 22$)
# Explanation:
## Step 1: Determine the electron filling order
The electron filling order is $1s\to2s\to2p\to3s\to3p\to4s\to3d$.
## Step 2: Write the full electron configuration
Titanium has 22 electrons. Filling the orbitals: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2}$ ( $4s$ fills before $3d$, then $3d$ is filled with 2 electrons).
## Step 3: Write the condensed electron configuration
The noble gas before titanium is argon ($Ar$, with electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$). So the condensed configuration is $[Ar]4s^{2}3d^{2}$.
# Answer:
- Full: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{2}$
- Condensed: $[Ar]4s^{2}3d^{2}$

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