Electrons and periodic patterns

What you know: A p drawing shows two lobes on opposite sides of the nucleus. What you want: How many orbitals that one drawing shows.

1. Recall what an orbital is: one region where an electron is likely to be found.
2. The two lobes are two parts of that same probability region. They belong together as one shape.
3. So one two-lobed drawing is One orbital. Like any orbital, it holds at most two electrons.

Why it matters: A p subshell has three orbitals, shown as three separate two-lobed drawings, not six lobes counted one by one.

Watch out: Counting lobes instead of orbitals gives 2, which is wrong.

That’s right: One orbital

What you know: A p subshell is drawn as three two-lobed shapes pointing in different directions (p_x, p_y, p_z). What you want: How many orbitals are in one p subshell.

1. Each differently oriented drawing is one separate orbital.
2. Count the orientations: x, y, and z make Three orbitals.

Check: Each row of the p block is six columns wide. Six electrons at two per orbital needs exactly 3 orbitals.

Watch out: Don't count lobes (6) or electrons (6). The question asks for orbitals.

That’s right: Three

What you know: The arrows ↑↑ show two electrons with the same spin. What you want: Whether both can sit in one orbital.

1. Use the orbital rule: one orbital holds at most two electrons, and they must have opposite spins.
2. ↑↑ has two parallel spins, not opposite ones.
3. So the answer is No. A paired orbital must be drawn ↑↓.

Why it matters: Parallel arrows are fine in two different orbitals, as when a p subshell fills singly (↑ | ↑ | ↑) before pairing.

That’s right: No

What you know: The label 3p⁵. What you want: What each part means.

1. The leading number 3 is the shell (n = 3).
2. The letter p names the subshell type in that shell.
3. The superscript 5 counts electrons in that subshell.
4. Together: Shell 3; p subshell; five electrons.

Why it matters: 5 is less than the p maximum of 6, so 3p⁵ is possible. In boxes it is ↑↓ | ↑↓ | ↑.

Watch out: The superscript counts electrons, not orbitals. A p subshell always has three orbitals.

That’s right: Shell 3; p subshell; five electrons

What you know: Hydrogen's electron moves from n = 3 to n = 2 in the Bohr model. What you want: What happens to the energy.

1. In Bohr's model, a smaller n is a lower energy level. So n = 2 is lower than n = 3.
2. Moving to a lower level means the electron loses energy.
3. That lost energy leaves the atom as light. So It gives off light carrying the energy difference.

Why it matters: The reverse move, n = 2 to n = 3, would need the atom to absorb exactly that same amount of energy.

Watch out: Moving closer to the nucleus releases energy; it does not take energy in.

That’s right: It gives off light carrying the energy difference.

What you know: Bohr pictured electrons on fixed circular orbits. What you want: What the quantum model uses instead.

1. The quantum model keeps Bohr's energy levels (n).
2. It drops the idea of an exact path, because electrons do not travel in fixed tracks.
3. Instead it uses Orbitals: regions where an electron is likely to be found.

Why it matters: This is why orbital drawings are probability regions, not racetracks the electron follows.

Watch out: "Orbit" and "orbital" sound alike but are different ideas.

That’s right: Orbitals: regions where an electron is likely to be found.

What you know: ν = 1.00 × 10¹⁴ Hz (s⁻¹) and c = 3.00 × 10⁸ m/s. What you want: Wavelength λ in nm.

1. Rearrange c = λν to λ = c ÷ ν.
2. λ = (3.00 × 10⁸ m/s) ÷ (1.00 × 10¹⁴ s⁻¹) = 3.00 × 10⁻⁶ m. The s⁻¹ and /s cancel, leaving meters.
3. Convert to nm: 3.00 × 10⁻⁶ m × (1 nm / 10⁻⁹ m) = 3.00 × 10³ nm. Three significant figures, matching the data.

Check: Reverse it: 3000 × 10⁻⁹ m × 1.00 × 10¹⁴ s⁻¹ should give 3.00 × 10⁸ m/s.

Watch out: Divide c by ν; don't multiply. And convert m to nm by dividing by 10⁻⁹ (multiplying by 10⁹).

That’s right: 3.00 × 10³ nm

What you know: ν = 2.00 × 10¹⁴ Hz (s⁻¹) and c = 3.00 × 10⁸ m/s. What you want: Wavelength λ in nm.

1. Rearrange c = λν to λ = c ÷ ν.
2. λ = (3.00 × 10⁸ m/s) ÷ (2.00 × 10¹⁴ s⁻¹) = 1.50 × 10⁻⁶ m. The s⁻¹ and /s cancel, leaving meters.
3. Convert to nm: 1.50 × 10⁻⁶ m × (1 nm / 10⁻⁹ m) = 1.50 × 10³ nm. Three significant figures, matching the data.

Check: Reverse it: 1500 × 10⁻⁹ m × 2.00 × 10¹⁴ s⁻¹ should give 3.00 × 10⁸ m/s.

Watch out: Divide c by ν; don't multiply. And convert m to nm by dividing by 10⁻⁹ (multiplying by 10⁹).

That’s right: 1.50 × 10³ nm

What you know: ν = 3.00 × 10¹⁴ Hz (s⁻¹) and c = 3.00 × 10⁸ m/s. What you want: Wavelength λ in nm.

1. Rearrange c = λν to λ = c ÷ ν.
2. λ = (3.00 × 10⁸ m/s) ÷ (3.00 × 10¹⁴ s⁻¹) = 1.00 × 10⁻⁶ m. The s⁻¹ and /s cancel, leaving meters.
3. Convert to nm: 1.00 × 10⁻⁶ m × (1 nm / 10⁻⁹ m) = 1.00 × 10³ nm. Three significant figures, matching the data.

Check: Reverse it: 1000 × 10⁻⁹ m × 3.00 × 10¹⁴ s⁻¹ should give 3.00 × 10⁸ m/s.

Watch out: Divide c by ν; don't multiply. And convert m to nm by dividing by 10⁻⁹ (multiplying by 10⁹).

That’s right: 1.00 × 10³ nm

What you know: ν = 4.00 × 10¹⁴ Hz (s⁻¹) and c = 3.00 × 10⁸ m/s. What you want: Wavelength λ in nm.

1. Rearrange c = λν to λ = c ÷ ν.
2. λ = (3.00 × 10⁸ m/s) ÷ (4.00 × 10¹⁴ s⁻¹) = 7.50 × 10⁻⁷ m. The s⁻¹ and /s cancel, leaving meters.
3. Convert to nm: 7.50 × 10⁻⁷ m × (1 nm / 10⁻⁹ m) = 750. nm. Three significant figures, matching the data. The decimal point in 750. shows the zero counts.

Check: Reverse it: 750 × 10⁻⁹ m × 4.00 × 10¹⁴ s⁻¹ should give 3.00 × 10⁸ m/s.

Watch out: Divide c by ν; don't multiply. And convert m to nm by dividing by 10⁻⁹ (multiplying by 10⁹).

That’s right: 750. nm

What you know: ν = 5.00 × 10¹⁴ Hz (s⁻¹) and c = 3.00 × 10⁸ m/s. What you want: Wavelength λ in nm.

1. Rearrange c = λν to λ = c ÷ ν.
2. λ = (3.00 × 10⁸ m/s) ÷ (5.00 × 10¹⁴ s⁻¹) = 6.00 × 10⁻⁷ m. The s⁻¹ and /s cancel, leaving meters.
3. Convert to nm: 6.00 × 10⁻⁷ m × (1 nm / 10⁻⁹ m) = 600. nm. Three significant figures, matching the data. The decimal point in 600. shows both zeros count.

Check: Reverse it: 600 × 10⁻⁹ m × 5.00 × 10¹⁴ s⁻¹ should give 3.00 × 10⁸ m/s.

Watch out: Divide c by ν; don't multiply. And convert m to nm by dividing by 10⁻⁹ (multiplying by 10⁹).

That’s right: 600. nm

What you know: Two photons: 400 nm and 700 nm. What you want: Which carries more energy.

1. Use E = hc/λ. h and c are constants, so energy depends only on λ.
2. λ is in the denominator, so a smaller wavelength gives a larger energy.
3. 400 nm is shorter than 700 nm, so 400 nm has more energy.

Check: 700/400 = 1.75, so the 400 nm photon has 1.75 times the energy. Violet light (400 nm) is more energetic than red (700 nm).

Watch out: Bigger wavelength does not mean bigger energy. It is the opposite.

That’s right: 400 nm

What you know: In emission, an atom gives off a photon. What you want: Whether the electron moves up or down in energy.

1. The photon carries energy away from the atom.
2. Energy is conserved, so the electron must end with less energy than it started with.
3. So emission moves the electron Down to a lower-energy state.

Why it matters: The photon's energy equals the size of the gap between the two levels.

That’s right: Down

What you know: In absorption, an atom takes in a photon. What you want: Whether the electron moves up or down.

1. The photon's energy is added to the atom.
2. It is absorbed only if it matches an allowed gap between levels.
3. The electron gains that energy, so it moves Up to a higher-energy state.

Why it matters: Absorption and emission are opposites: absorption goes up, emission goes down.

That’s right: Up

What you know: c = λν with c in m/s and ν in s⁻¹. What you want: The unit λ must use.

1. Write the units: m/s = λ × s⁻¹.
2. For the units to match on both sides, λ must be in meters.
3. So λ must use Meters. Convert nm to m first (1 nm = 10⁻⁹ m).

Check: (m) × (s⁻¹) = m/s, which matches c's units.

Watch out: Plugging 500 nm straight in as "500" makes the answer off by 10⁹.

That’s right: Meters

What you know: E = hν, and h is a constant. What you want: What happens to E when ν doubles.

1. E is directly proportional to ν: E = h × ν.
2. Replace ν with 2ν: E_new = h × 2ν = 2 × (hν).
3. The answer: It doubles.

Check: Try numbers. ν = 1.00 × 10¹⁴ s⁻¹ gives E = 6.63 × 10⁻²⁰ J; ν = 2.00 × 10¹⁴ s⁻¹ gives E = 1.33 × 10⁻¹⁹ J, twice as much.

Watch out: Don't confuse this with wavelength. Doubling λ would halve the energy.

That’s right: It doubles

What you know: Successive ionization energies: 590, 1145, and 4912 kJ/mol. What you want: The number of valence electrons.

1. Compare each step, not the raw numbers. 590 → 1145 is about 1.9 times larger.
2. 1145 → 4912 is about 4.3 times larger. That is the unusually large jump.
3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
4. So the first two removed were valence electrons: 2.

Check: These are the measured values for calcium (Ca), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

That’s right: 2

What you know: Successive ionization energies: 550, 1064, and 4138 kJ/mol. What you want: The number of valence electrons.

1. Compare each step, not the raw numbers. 550 → 1064 is about 1.9 times larger.
2. 1064 → 4138 is about 3.9 times larger. That is the unusually large jump.
3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
4. So the first two removed were valence electrons: 2.

Check: These are the measured values for strontium (Sr), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

That’s right: 2

What you know: Successive ionization energies: 503, 965, and 3458 kJ/mol. What you want: The number of valence electrons.

1. Compare each step, not the raw numbers. 503 → 965 is about 1.9 times larger.
2. 965 → 3458 is about 3.6 times larger. That is the unusually large jump.
3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
4. So the first two removed were valence electrons: 2.

Check: These are the measured values for barium (Ba), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

That’s right: 2

What you know: Successive ionization energies: 900, 1757, and 14,849 kJ/mol. What you want: The number of valence electrons.

1. Compare each step, not the raw numbers. 900 → 1757 is about 2.0 times larger.
2. 1757 → 14,849 is about 8.5 times larger. That is the unusually large jump.
3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
4. So the first two removed were valence electrons: 2.

Check: These are the measured values for beryllium (Be), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

That’s right: 2

What you know: Successive ionization energies: 738, 1451, and 7733 kJ/mol. What you want: The number of valence electrons.

1. Compare each step, not the raw numbers. 738 → 1451 is about 2.0 times larger.
2. 1451 → 7733 is about 5.3 times larger. That is the unusually large jump.
3. The jump comes right before the third electron. That third electron must be a core electron, much closer to the nucleus.
4. So the first two removed were valence electrons: 2.

Check: These are the measured values for magnesium (Mg), a Group 2 atom, which shows exactly this pattern: a big jump between the second and third ionizations.

Watch out: Each ionization energy is always higher than the last, so a rise alone means nothing. Look for the rise that is much bigger than the others.

That’s right: 2

What you know: Na and K are neutral atoms in the same group. What you want: Which is generally larger.

1. Same group, so use the down-a-group trend.
2. Going down a group adds an occupied shell. The outer electrons sit farther from the nucleus, and inner electrons shield them.
3. K is one row below Na, so K is larger.

Why it matters: Neutral atomic size grows down and to the left. K is lower, so it should be bigger.

Watch out: K has more protons, but more protons does not make it smaller here. The extra shell wins down a group.

On the periodic table: Sodium (Na) is in Period 3 and potassium (K) is in Period 4, both in Group 1, so K has one more occupied shell.

That’s right: K

What you know: Na and Cl are neutral atoms in the same period. What you want: Which is generally smaller.

1. Same period, so use the across-a-row trend.
2. Across a period, protons are added while electrons go into the same main shell. Shielding barely changes.
3. The stronger pull draws the electrons in. Cl is farther right, so Cl is smaller.

Why it matters: Size grows down and to the left; Na is at the left edge, so it should be the larger one.

Watch out: More electrons does not mean bigger across a row. Cl has more electrons but is smaller.

On the periodic table: Sodium (Na) and chlorine (Cl) are both in Period 3, and Cl is farther right, so Cl is generally smaller.

That’s right: Cl

What you know: Core electrons are the inner electrons, below the valence shell. What you want: Why removing one takes much more energy.

1. Core electrons sit in a lower shell, closer to the nucleus.
2. They are shielded by fewer electrons, so they feel more of the nuclear pull.
3. By then the ion is already positive, which pulls even harder. In short: It is more tightly bound.

Why it matters: This is why ionization-energy data show a huge jump right after the last valence electron is removed.

That’s right: It is more tightly bound

What you know: Two quantities: electron affinity and ionization energy. What you want: Whether they are the same.

1. Ionization energy is the energy needed to remove an electron from a gaseous atom.
2. Electron affinity concerns adding an electron to a gaseous atom.
3. Removing and adding are different processes, so the answer is No.

Why it matters: The energy changes differ too. Removing an electron always takes energy in, while adding an electron to many atoms, such as Cl, releases energy.

Watch out: The two trends look similar on the table, but that does not make them the same thing.

That’s right: No

What you know: First ionization energy generally increases across a row. What you want: Whether it always does.

1. "Generally" is the clue: trends are patterns, not exception-free laws.
2. Period 2 has two known exceptions: B is lower than Be, and O is lower than N.
3. These come from subshell energy (B's electron is in 2p) and paired-electron repulsion (O's first paired 2p electron). So the answer is No.

Why it matters: One exception is enough to make "always" false.

On the periodic table: Beryllium (Be) and boron (B) sit side by side in Period 2, as do nitrogen (N) and oxygen (O). Each pair is one step to the right across the same row.

That’s right: No