Maximum number of covalent bonds that an oxygen atom can make with hydrogen is 2.
- the ground state electronic configuration of oxygen is 2s² 2p⁴ that means it has 6 electrons in its valence shell and require two electrons are required to complete its octate.
- Two bonds are created when an electron donor atom shares the two needed electrons with oxygen. The ability of two oxygen atoms to share valence electrons results in the creation of a double bond between the two atoms.
- There are no longer any empty orbitals in the octet of oxygen after it is complete. As a result, it is unable to accept more electrons or create more bonds.
Therefore, Oxygen can only generate two bonds because it needs two additional electrons to complete its octet, after which it will run out of empty orbitals in which to receive additional electrons and create additional bonds.
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Answer:
n = 3 to n = 5
Explanation:
According to the Bohr's model of the atom, electrons in an atom absorb energy to move from a lower to higher energy level.
We must note that as we progress away from the nucleus, the energy levels of electrons become closer together. The energy difference between successive levels decreases and the wavelength of light associated with such transitions become longer.
Hence,the absorption of light of the longest wavelength corresponds to n = 3 to n = 5
.
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
Average atomic mass = .374 ( 184.953 amu ) + .626 ( 186.958 amu ) =
186.207 amu
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