Chemistry: Atoms First Review

CHM 111 Module 3 Practice Notes

Lesson 1: Light Waves and Energy

This lesson focuses on fundamental relationships between the properties of light waves: wavelength, frequency, and energy. Key constants and formulas are used to solve various problems.

Fundamental Formulas for Light

  • Speed of Light: The speed of light (c) is constant in a vacuum and relates wavelength (\lambda) and frequency (\nu):
    c = \lambda \nu
    Where: \text{c} = 3.00 \times 10^8 \text{ m/s}

  • Energy of a Photon: The energy (E) of a quantum or photon of light is directly proportional to its frequency (\nu) and inversely proportional to its wavelength (\lambda):
    E = h \nu
    E = \frac{hc}{\lambda}
    Where: \text{h} = 6.626 \times 10^{-34} \text{ J\cdot s} (Planck's constant)

Practice Problems and Solutions

  1. Frequency of a Light Wave given Wavelength:

    • Question: What is the frequency of a light wave with a wavelength of 410.0 \text{ nm}?

    • Solution Steps:

      1. Convert wavelength from nanometers to meters: \lambda = 410.0 \text{ nm} = 410.0 \times 10^{-9} \text{ m}

      2. Use the formula \nu = \frac{c}{\lambda} .

      3. Calculate: \nu = \frac{3.00 \times 10^8 \text{ m/s}}{410.0 \times 10^{-9} \text{ m}} = 7.317 \times 10^{14} \text{ s}^{-1} \text{ (or Hz)}

  2. Energy of a Quantum of Light given Frequency:

    • Question: What is the energy of a quantum or wave of light of frequency 4.31 \times 10^{14} \text{ Hz}?

    • Solution Steps:

      1. Use the formula E = h \nu.

      2. Calculate: E = (6.626 \times 10^{-34} \text{ J\cdot s}) \times (4.31 \times 10^{14} \text{ Hz}) = 2.857 \times 10^{-19} \text{ J}

  3. Wavelength of Light given Frequency (in nm):

    • Question: What is the wavelength of light with frequency 6.62 \times 10^{14} \text{ Hz} in nm?

    • Solution Steps:

      1. Use the formula \lambda = \frac{c}{\nu}.

      2. Calculate: \lambda = \frac{3.00 \times 10^8 \text{ m/s}}{6.62 \times 10^{14} \text{ Hz}} = 4.532 \times 10^{-7} \text{ m}

      3. Convert to nm: 4.532 \times 10^{-7} \text{ m} = 453.2 \text{ nm}

  4. Energy of a Light Wave given Wavelength:

    • Question: What is the energy of a light wave with wavelength 662 \text{ nm}?

    • Solution Steps:

      1. Convert wavelength to meters: \lambda = 662 \text{ nm} = 662 \times 10^{-9} \text{ m}

      2. Use the formula E = \frac{hc}{\lambda}.

      3. Calculate: E = \frac{(6.626 \times 10^{-34} \text{ J\cdot s}) \times (3.00 \times 10^8 \text{ m/s})}{662 \times 10^{-9} \text{ m}} = 3.001 \times 10^{-19} \text{ J}

  5. Frequency of Violet Light given Wavelength:

    • Question: A certain violet light has a wavelength of 413 \text{ nm}. What is its frequency?

    • Solution Steps:

      1. Convert wavelength to meters: \lambda = 413 \text{ nm} = 413 \times 10^{-9} \text{ m}

      2. Use the formula \nu = \frac{c}{\lambda} .

      3. Calculate: \nu = \frac{3.00 \times 10^8 \text{ m/s}}{413 \times 10^{-9} \text{ m}} = 7.264 \times 10^{14} \text{ Hz}

Lesson 2: Electron Transitions in Hydrogen Atoms

This lesson explores the energy changes and emitted light when an electron moves between energy levels in a hydrogen atom. The Rydberg equation is central to these calculations.

Rydberg Equation and Energy Transitions

  • Rydberg Equation (for wavelength): This equation predicts the wavelength of light emitted or absorbed during electron transitions in hydrogen-like atoms.
    \frac{1}{\lambda} = RH \left(\frac{1}{n1^2} - \frac{1}{n_2^2}\right)
    Where:

    • R_H = 1.097 \times 10^7 \text{ m}^{-1} (Rydberg constant)

    • n_1 = lower energy level (final state for emission, initial for absorption)

    • n_2 = higher energy level (initial state for emission, final for absorption)

  • Energy Change for Transitions: The energy change (\Delta E) associated with an electron transition can also be calculated.
    \Delta E = K \left(\frac{1}{ni^2} - \frac{1}{nf^2}\right)
    Where:

    • K = 2.179 \times 10^{-18} \text{ J} (ionization energy of H from ground state, a positive value representing the magnitude)

    • n_i = initial energy level

    • n_f = final energy level

  • Energy and Wavelength Relationship: Once \Delta E is known, the wavelength can be found using \lambda = \frac{hc}{\Delta E}. For emission, \Delta E is negative, but the magnitude is used to calculate wavelength.

Practice Problems and Solutions

  1. Wavelength of Light Emitted (n=3 to n=2):

    • Question: An electron in a hydrogen atom transitions from the n=3 energy level to the n=2 energy level. Calculate the wavelength of light emitted during this transition.

    • Solution Steps:

      1. Identify n1 = 2 and n2 = 3 (for emission, n1 < n2).

      2. Apply the Rydberg equation: \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \left(\frac{1}{2^2} - \frac{1}{3^2}\right)

      3. Calculate the term in parentheses: \left(\frac{1}{4} - \frac{1}{9}\right) = \left(\frac{9-4}{36}\right) = \frac{5}{36}

      4. \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \times \frac{5}{36} = 1.5236 \times 10^6 \text{ m}^{-1}

      5. Solve for \lambda: \lambda = \frac{1}{1.5236 \times 10^6 \text{ m}^{-1}} = 6.563 \times 10^{-7} \text{ m}

      6. Convert to nm: 6.563 \times 10^{-7} \text{ m} = 656.3 \text{ nm}

  2. Energy and Wavelength of Photon Emitted (n=4 to n=1):

    • Question: An electron in a hydrogen atom transitions from the n=4 energy level to the n=1 energy level. Calculate the energy of the photon emitted and its wavelength.

    • Solution Steps:

      1. Identify ni = 4 and nf = 1.

      2. Calculate energy emitted: \Delta E = K \left(\frac{1}{nf^2} - \frac{1}{ni^2}\right) = (2.179 \times 10^{-18} \text{ J}) \left(\frac{1}{1^2} - \frac{1}{4^2}\right).
        (Note: The solution in the transcript uses \Delta E = K (\frac{1}{ni^2} - \frac{1}{nf^2}) implicitly to get a positive energy value for emitted photon. Let's use it for consistency if it represents the magnitude of emitted energy).
        \Delta E = (2.179 \times 10^{-18} \text{ J}) \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = (2.179 \times 10^{-18} \text{ J}) \left(1 - \frac{1}{16}\right) = (2.179 \times 10^{-18} \text{ J}) \times \frac{15}{16} = 2.043 \times 10^{-18} \text{ J}

      3. Calculate wavelength: \lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \text{ J\cdot s}) \times (3.00 \times 10^8 \text{ m/s})}{2.043 \times 10^{-18} \text{ J}} = 9.723 \times 10^{-8} \text{ m}

      4. Convert to nm: 9.723 \times 10^{-8} \text{ m} = 97.23 \text{ nm}

  3. Wavelength and Frequency of Light Emitted (n=4 to n=2):

    • Question: For a hydrogen atom, calculate the wavelength and frequency of the light emitted when an electron transitions from the n=4 level to the n=2 level.

    • Solution Steps:

      1. Identify n1 = 2 and n2 = 4.

      2. Apply Rydberg equation: \frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = (1.097 \times 10^7 \text{ m}^{-1}) \left(\frac{1}{4} - \frac{1}{16}\right)

      3. Calculate the term in parentheses: \left(\frac{4-1}{16}\right) = \frac{3}{16}

      4. \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \times \frac{3}{16} = 2.056875 \times 10^6 \text{ m}^{-1}

      5. Solve for \lambda: \lambda = \frac{1}{2.056875 \times 10^6 \text{ m}^{-1}} = 4.862 \times 10^{-7} \text{ m} = 486.2 \text{ nm}

      6. Calculate frequency: \nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \text{ m/s}}{4.862 \times 10^{-7} \text{ m}} = 6.17 \times 10^{14} \text{ Hz}

  4. Energy of Emitted Photon in eV (n=7 to n=4):

    • Question: In a hydrogen atom, an electron moves from the n=7 level to the n=4 level. What is the energy of the emitted photon? Provide the result in electron volts (eV). (1 \text{ Joule} = 6.242 \times 10^{18} \text{ eV}).

    • Solution Steps:

      1. Identify ni = 7 and nf = 4.

      2. Calculate energy emitted in Joules: \Delta E = K \left(\frac{1}{nf^2} - \frac{1}{ni^2}\right) = (2.179 \times 10^{-18} \text{ J}) \left(\frac{1}{4^2} - \frac{1}{7^2}\right)

      3. Calculate term in parentheses: \left(\frac{1}{16} - \frac{1}{49}\right) = \left(\frac{49-16}{16 \times 49}\right) = \frac{33}{784}

      4. \Delta E = (2.179 \times 10^{-18} \text{ J}) \times \frac{33}{784} = 9.172 \times 10^{-20} \text{ J}

      5. Convert to eV: E_{eV} = (9.172 \times 10^{-20} \text{ J}) \times (6.242 \times 10^{18} \text{ eV/J}) = 0.5725 \text{ eV}

  5. Wavelength of Light Emitted (n=8 to n=6) using Rydberg Equation:

    • Question: In a hydrogen atom, an electron transitions from n=8 to n=6. Use the Rydberg equation to calculate the wavelength of the light emitted in nanometers (nm).

    • Solution Steps:

      1. Identify n1 = 6 and n2 = 8.

      2. Apply Rydberg equation: \frac{1}{\lambda} = R_H \left(\frac{1}{6^2} - \frac{1}{8^2}\right) = (1.097 \times 10^7 \text{ m}^{-1}) \left(\frac{1}{36} - \frac{1}{64}\right)

      3. Calculate term in parentheses: \left(\frac{64-36}{36 \times 64}\right) = \frac{28}{2304} = \frac{7}{576}

      4. \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \times \frac{7}{576} = 1.334 \times 10^5 \text{ m}^{-1}

      5. Solve for \lambda: \lambda = \frac{1}{1.334 \times 10^5 \text{ m}^{-1}} = 7.501 \times 10^{-6} \text{ m}

      6. Convert to nm: 7.501 \times 10^{-6} \text{ m} = 7501 \text{ nm}

Lesson 3: Matter Waves and Quantum Numbers

This lesson introduces the wave-particle duality of matter, specifically de Broglie's hypothesis, and quantum numbers that describe electron states.

De Broglie Wavelength

  • Formula: Louis de Broglie proposed that particles, like electrons, can exhibit wave-like properties. The de Broglie wavelength (\lambda) is given by: \lambda = \frac{h}{mv} Where:

    • h = 6.626 \times 10^{-34} \text{ J\cdot s} (Planck's constant)

    • m = mass of the particle in kg

    • v = velocity of the particle in m/s

Practice Problems and Solutions (De Broglie Wavelength)

  1. Wavelength of a Duck:

    • Question: Calculate the wavelength of a 5.00 \text{ Kg} duck walking around a pond at 20.0 \text{ cm} per second.

    • Solution Steps:

      1. Convert velocity to m/s: v = 20.0 \text{ cm/s} = 0.200 \text{ m/s}

      2. Use de Broglie wavelength formula: \lambda = \frac{6.626 \times 10^{-34} \text{ J\cdot s}}{(5.00 \text{ Kg}) \times (0.200 \text{ m/s})} = 6.63 \times 10^{-34} \text{ m}

  2. Speed of a Plane:

    • Question: Calculate the speed of a plane if it has a wavelength of 1.671 \times 10^{-41} \text{ m} and a mass of 6.60 \times 10^7 \text{ grams}.

    • Solution Steps:

      1. Convert mass to kg: m = 6.60 \times 10^7 \text{ g} = 6.60 \times 10^4 \text{ Kg}

      2. Rearrange de Broglie formula to solve for velocity: v = \frac{h}{m \lambda}

      3. Calculate: v = \frac{6.626 \times 10^{-34} \text{ J\cdot s}}{(6.60 \times 10^4 \text{ Kg}) \times (1.671 \times 10^{-41} \text{ m})} = 6.0 \text{ m/s}

Quantum Numbers

Quantum numbers describe the state of an electron in an atom. There are four types:

  • Principal Quantum Number (n): Describes the main energy level or shell. n = 1, 2, 3, …

  • Angular Momentum (Azimuthal) Quantum Number (l): Describes the shape of the orbital and subshell. l = 0, 1, …, (n-1).

    • l=0 corresponds to s orbital

    • l=1 corresponds to p orbital

    • l=2 corresponds to d orbital

    • l=3 corresponds to f orbital

  • Magnetic Quantum Number (ml): Describes the orientation of the orbital in space. ml = -l, …, 0, …, +l

  • Spin Quantum Number (ms): Describes the intrinsic angular momentum (spin) of the electron. ms = +\frac{1}{2} or \text{-}\frac{1}{2}

Practice Problems and Solutions (Quantum Numbers for Last Electron)

  1. Chromium (Cr):

    • Electron configuration: [Ar] 4s^1 3d^5 (exception, 4s^2 3d^4 is not observed due to half-filled d-subshell stability)

    • Last electron in 3d^5 (the 5th d-electron when filling from 4s^1):

      • n = 3

      • l = 2 (for d orbital)

      • ml = +2 (or any of -2, -1, 0, +1, +2 for the d-electrons, assuming the last one is filled into a new orbital with ml=+2 based on Hund's rule, all spins up)

      • m_s = +\frac{1}{2}

      • *Correction/Clarification based on reference solution: The reference gives 4,0,0, \pm \frac{1}{2} for Cr, indicating consideration of the 4s^1 electron. If it's the *last* electron filled in the standard Aufbau principle before the exception for Cr, it might be the 4s electron for 4s^23d^4. But since Cr is an exception, the last valence electron is usually considered. The provided answer is ambiguous; for a 4s^1 3d^5 configuration, if we consider filling in a 4s^2 3d^4 pattern and then promoting, or just the highest energy electron, it's typically the 3d electrons. If it's strictly the last one added to reach 4s^1 3d^5, it would be one of the 3d electrons. Let's reinterpret the reference solution's 4,0,0, \pm \frac{1}{2} to refer to the 4s^1 electron, as it's often the last one unique in the Aufbau process.*

      • Thus, if we consider the unique last electron as 4s^1:

        • n=4

        • l=0

        • m_l=0

        • m_s = \pm \frac{1}{2}

  2. Gold (Au):

    • Electron configuration: [Xe] 6s^1 4f^{14} 5d^{10} (exception: 6s^2 5d^9 is not observed)

    • Last electron (for 6s^1 from [Xe]…6s^1): (similar to Cr, considering the s^1 electron as the 'last' in terms of valence shell outermost electron)

      • n=6

      • l=0

      • m_l=0

      • m_s = \pm \frac{1}{2}

      • Alternatively, if considering the one that makes it 5d^{10} from 5d^9, it would be n=5, l=2, ml=\text{any}, ms=\text{-}1/2.

  3. Carbon (C):

    • Electron configuration: 1s^2 2s^2 2p^2

    • Last electron (the second 2p electron):

      • n = 2

      • l = 1 (for p orbital)

      • ml = 0 (or +1) (based on Hund's rule, filling orbitals with parallel spins, it would be the second p orbital, so ml=0 if first was -1)

      • m_s = +\frac{1}{2}

  4. Cl$^-$ ion:

    • Chlorine (Cl) configuration: [Ne] 3s^2 3p^5

    • Cl$^-$ (gains one electron): [Ne] 3s^2 3p^6

    • Last electron (3p^6):

      • n = 3

      • l = 1 (for p orbital)

      • ml = +1 (the last one to complete the p subshell, pairing up previous electrons in ml=-1, 0, +1)

      • m_s = -\frac{1}{2} (paired with an existing electron)

  5. Ca$^{2+}$ ion:

    • Calcium (Ca) configuration: [Ar] 4s^2

    • Ca$^{2+}$ (loses two electrons from 4s): [Ar]

    • Last electron (this refers to the last electron in Argon configuration, which is the 3p^6 electron):

      • n = 3

      • l = 1 (for p orbital)

      • m_l = +1

      • m_s = -\frac{1}{2}

Electron Configurations and Element Identification

Rewrite the configurations using noble gas abbreviations.

  1. 1s^22s^22p^3: Nitrogen (N) -> [He] 2s^2 2p^3

  2. 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3: Antimony (Sb) -> [Kr] 5s^2 4d^{10} 5p^3

  3. 1s^22s^22p^63s^23p^64s^13d^{10}: Copper (Cu) -> [Ar] 4s^1 3d^{10} (exception)

  4. 1s^22s^22p^63s^1: Sodium (Na) -> [Ne] 3s^1

  5. 1s^22s^22p^63s^23p^64s^23d^5: Manganese (Mn) -> [Ar] 4s^2 3d^5

  6. 1s^22s^22p^2: Carbon (C) -> [He] 2s^2 2p^2

  7. 1s^22s^22p^63s^23p^64s^23d^{10}4p^4: Selenium (Se) -> [Ar] 4s^2 3d^{10} 4p^4

  8. 1s^22s^22p^63s^23p^6: Argon (Ar) -> [Ne] 3s^2 3p^6

  9. 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2: Strontium (Sr) -> [Kr] 5s^2

  10. 1s^22s^22p^63s^23p^64s^1: Potassium (K) -> [Ar] 4s^1

Ground State Full Electron Configurations

Write the full electron configuration for each element or ion.

  1. Na: 1s^2 2s^2 2p^6 3s^1

  2. Pb: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^2

  3. Sr: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2

  4. Ag: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^1 4d^{10} (exception)

  5. Ti: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^2

  6. Cl: 1s^2 2s^2 2p^6 3s^2 3p^5

  7. Hg: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10}

  8. O$^{2-}$: 1s^2 2s^2 2p^6 (isoelectronic with Neon)

  9. Mn$^{2+}$: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 (Mn is [Ar] 4s^2 3d^5, loses 4s electrons first)

  10. Ca$^{2+}$: 1s^2 2s^2 2p^6 3s^2 3p^6 (isoelectronic with Argon)

  11. Cu: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10} (exception)

Lessons 4-5: Periodic Trends and Element Classification

This section covers key periodic trends that describe how atomic properties vary across the periodic table, along with categorizing elements.

Electronegativity

  • Definition: Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.

  • Periodic Trend: Electronegativity generally increases from left to right across a period and decreases from top to bottom down a group.

  1. More Electronegative Atom in Each Pair:
    a. Cl (Chlorine is to the right of Bromine in the same group, and higher up) - Note: Electronegativity increases up a group, so Cl is more electronegative than Br.
    b. O (Oxygen is to the right of Nitrogen in the same period)
    c. O (Oxygen is above Sulfur in the same group)
    d. S (Sulfur is to the right of Phosphorus in the same period)
    e. N (Nitrogen is higher and to the right of Silicon)

  2. Increasing Electronegativity Order:
    a. H, C, N, O, F (Generally increases from left to right; Hydrogen is an outlier but less electronegative than C)
    b. I, Br, Cl, F (Increases up a group)
    c. H, P, S, O, F (Based on general trends, with H being the least and F the most. P is lower and further left than S, O, F.)

Atomic Radius

  • Definition: Atomic radius is a measure to describe the size of atoms. It is typically defined as half the distance between the nuclei of two identical atoms that are bonded together.

  • Periodic Trend: Atomic radius generally decreases from left to right across a period and increases from top to bottom down a group.

  1. Larger Atomic Radius in Each Pair:
    a. K (Potassium is below Lithium in the same group)
    b. Ca (Calcium is to the left of Nickel in the same period)
    c. Ga (Gallium is below Boron, even though B is in the same column, Ga is further down in the periodic table compared to B's relative position compared to period)
    d. C (Carbon is to the left of Oxygen in the same period)
    e. Br (Bromine is below Chlorine in the same group)
    f. Ba (Barium is below Beryllium in the same group)
    g. Si (Silicon is to the left of Sulfur in the same period)
    h. Au (Gold is below Iron in the same group)

  2. Why atoms get smaller left to right in a period:

    • As you move from left to right across a period, the atomic number increases, meaning the number of protons in the nucleus increases. This leads to a stronger Coulombic attraction between the positively charged nucleus and the negatively charged electron cloud. While electrons are added in the same principal energy level, the increasing nuclear charge pulls the electron cloud in more tightly, resulting in a smaller atomic radius.

  3. Larger Radius in Atom/Ion Pair:
    a. Na (Neutral atom has a larger radius than its cation; loosing an electron reduces electron-electron repulsion and increases effective nuclear charge on remaining electrons, pulling them closer)
    b. S$^{2-}$ (Anion has a larger radius than its neutral atom; gaining electrons increases electron-electron repulsion and adds to the electron cloud size)
    c. I$^{-}$ (Anion larger than neutral atom)
    d. Al (Neutral atom larger than its cation)

Ionization Energy

  • Definition: Ionization energy is the energy required to remove an electron from a gaseous atom or ion in its ground state.

  • First Ionization Energy: Specifically refers to the energy required to remove the first electron from a neutral gaseous atom.

  • Periodic Trend: Ionization energy generally increases from left to right across a period and increases from bottom to top within a group.

  1. Increasing Ionization Energy Order:
    a. Sr, Mg, Be (Increases up a group)
    b. Cs, Ba, Bi (General trend: Cs is an alkali metal, Ba is an alkaline earth metal, Bi is a post-transition metal; Ionization energy generally increases across a period. Cs is furthest left and lowest. Ba is next. Bi is furthest right and highest compared to the other two.)
    c. Na, Al, S (Increases across a period)

  2. What is Ionization Energy? What is First Ionization Energy?

    • Ionization energy is the energy required to remove an electron from a gaseous atom (or ion) in its ground state.

    • First ionization energy is the energy required to remove the first electron from a neutral gaseous atom.

  3. Periodic Trend for First Ionization Energy:

    • First ionization energy increases from left to right across a period.

    • First ionization energy increases from the bottom of a group to the top of a group.

Element Classification and Group Names

  1. Identify as Metals, Nonmetals, or Metalloids:
    a. Calcium: Metal
    b. Chlorine: Nonmetal
    c. Zinc: Metal
    d. Silicon: Metalloid
    e. Phosphorus: Nonmetal
    f. Arsenic: Metalloid
    g. Sodium: Metal
    h. Hydrogen: Nonmetal
    i. Xenon: Nonmetal
    j. Uranium: Metal
    k. Promethium: Metal

  2. Identify Group (Family) Name:
    a. Calcium: Alkaline earth metal
    b. Chlorine: Halogen
    c. Zinc: Transition metal (no specific group name for Zn column of main group)
    d. Silicon: No specific group name (in Group 14, Carbon group)
    e. Phosphorus: Pnictogen (Group 15)
    f. Arsenic: Pnictogen (Group 15)
    g. Sodium: Alkali metal
    h. Hydrogen: No group name (unique element)
    i. Xenon: Noble gas
    j. Uranium: Actinide series
    k. Promethium: Lanthanide series

  3. Identify as Main Group, Transition, or Inner Transition Elements:
    a. Calcium: Main group element
    b. Chlorine: Main group element
    c. Zinc: Transition element (d-block)
    d. Silicon: Main group element (p-block)
    e. Phosphorus: Main group element (p-block)
    f. Arsenic: Main group element (p-block)
    g. Sodium: Main group element
    h. Hydrogen: Main group element (often placed in Group 1, but unique)
    i. Xenon: Main group element (noble gas, p-block)
    j. Uranium: Inner transition element (actinide)
    k. Promethium: Inner transition element (lanthanide)

  4. Why elements in the same group have similar properties:

    • Elements in the same group (or family) have similar physical and chemical properties because they have the same valence electron configurations. The arrangement and number of valence electrons are the primary determinants of an atom's chemical behavior and many of its physical properties, especially how it will interact and bond with other atoms.