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3. how many valence electrons do the following elements have? ___ fr __…

Question

  1. how many valence electrons do the following elements have?

_ fr _ o _ ne _ c _ mg _ cl ___ al

  1. identify the following atom, $1s^22s^22p^63s^1$ v.e.? ___
  2. identify the following atom, $1s^22s^22p^6$ v.e.? ___
  3. identify the following atom, $1s^22s^22p^63s^23p^64s^2$ v.e.? ___
  4. identify the following atom, $1s^22s^22p^63s^23p^64s^23d^6$ v.e.? ___
  5. fill in the table:
elementelectron configurationnoble gas electron configuration
cesium
tin
  1. fill in the table:
elementperiod #group #valence electron soxidation numbergain or lose e?cation or anion?
bromine
boron
strontium
potassium
neon
  1. fill in the table:
elementperiod #group #group namee-dot diagramox. ## of valance e⁻
$3s^23p^3$
$3s^23p^6$
$4s^1$

Explanation:

Response
Problem 3: Valence Electrons of Elements

Let's solve for each element:

Francium (Fr)

Fr is in Group 1 (Alkali Metals). Group 1 elements have 1 valence electron.

Oxygen (O)

O is in Group 16 (Chalcogens). Valence electrons = Group number - 10 (for main groups 13–18) → \( 16 - 10 = 6 \).

Neon (Ne)

Ne is a noble gas (Group 18). Noble gases have 8 valence electrons (except He, which has 2).

Carbon (C)

C is in Group 14. Valence electrons = \( 14 - 10 = 4 \).

Magnesium (Mg)

Mg is in Group 2 (Alkaline Earth Metals). Group 2 elements have 2 valence electrons.

Chlorine (Cl)

Cl is in Group 17 (Halogens). Valence electrons = \( 17 - 10 = 7 \).

Aluminum (Al)

Al is in Group 13. Valence electrons = \( 13 - 10 = 3 \).

Problem 14: Identify Atom from Electron Configuration (\( 1s^2 2s^2 2p^6 3s^1 \))
  1. Sum the exponents: \( 2 + 2 + 6 + 1 = 11 \).
  2. Atomic number 11 is Sodium (Na).
  3. Valence electrons: The outermost shell is \( 3s^1 \), so 1 valence electron.
Problem 15: Identify Atom from Electron Configuration (\( 1s^2 2s^2 2p^6 \))
  1. Sum the exponents: \( 2 + 2 + 6 = 10 \).
  2. Atomic number 10 is Neon (Ne).
  3. Valence electrons: Neon is a noble gas, so 8 valence electrons.
Problem 16: Identify Atom from Electron Configuration (\( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 \))
  1. Sum the exponents: \( 2 + 2 + 6 + 2 + 6 + 2 = 20 \).
  2. Atomic number 20 is Calcium (Ca).
  3. Valence electrons: Outermost shell \( 4s^2 \), so 2 valence electrons.
Problem 17: Identify Atom from Electron Configuration (\( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^8 4s^2 \))
  1. Sum the exponents: \( 2 + 2 + 6 + 2 + 6 + 8 + 2 = 28 \).
  2. Atomic number 28 is Nickel (Ni).
  3. Valence electrons: For transition metals, valence electrons include \( 4s \) and \( 3d \), but typically \( 4s^2 \) (so 2, though some sources consider \( 3d^8 4s^2 \) as 10, but common for Ni: 2 or 10; here, \( 4s^2 \) gives 2).
Problem 18: Electron Configurations & Noble Gas Configurations
Chlorine (Cl, Atomic #17)
  • Electron Configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^5 \)
  • Noble Gas Configuration: Use Ne (\( 1s^2 2s^2 2p^6 \)) → \( [\text{Ne}] 3s^2 3p^5 \)
Cesium (Cs, Atomic #55)
  • Electron Configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1 \)
  • Noble Gas Configuration: Use Xe (\( [\text{Kr}] 4d^{10} 5s^2 5p^6 \)) → \( [\text{Xe}] 6s^1 \)
Tin (Sn, Atomic #50)
  • Electron Configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^2 \)
  • Noble Gas Configuration: Use Kr (\( [\text{Ar}] 3d^{10} 4s^2 4p^6 \)) → \( [\text{Kr}] 5s^2 4d^{10} 5p^2 \)
Problem 19: Element Properties Table
ElementPeriod #Group #Valence e⁻Oxidation NumberGain or Lose e⁻?Cation or Anion?
Bromine (Br)4177-1 (common)Gain 1Anion (Br⁻)
Boron (B)2133+3 (common)Lose 3Cation (B³⁺)
Strontium (Sr)522+2 (common)Lose 2Cation (Sr²⁺)
Potassium (K)411+1 (common)Lose 1Cation (K⁺)
Neon (Ne)21880 (noble gas)NeitherNone
Problem 20: Element Properties from Electron Configuration

| Element (Electron C…

Answer:

s (Key Highlights):

Problem 3:
  • Fr: 1
  • O: 6
  • Ne: 8
  • C: 4
  • Mg: 2
  • Cl: 7
  • Al: 3
Problem 14:
  • Atom: Na (Sodium)
  • Valence e⁻: 1
Problem 15:
  • Atom: Ne (Neon)
  • Valence e⁻: 8
Problem 16:
  • Atom: Ca (Calcium)
  • Valence e⁻: 2
Problem 17:
  • Atom: Ni (Nickel)
  • Valence e⁻: 2 (or 10, context-dependent)
Problem 18:
ElementElectron ConfigurationNoble Gas Configuration
Cesium\( [\text{Xe}] 6s^1 \) (simplified)\( [\text{Xe}] 6s^1 \)
Tin\( [\text{Kr}] 5s^2 4d^{10} 5p^2 \)\( [\text{Kr}] 5s^2 4d^{10} 5p^2 \)
Problem 19: (See table above)
Problem 20: (See table above)