Which electron configuration represents the electrons of an atom of neon in an excited state?
2 answers:
Answer: Option (C) is the correct answer.
Explanation:
Atomic number of neon is 10 and its electronic configuration is .
As neon has a vacant 3s orbital also. Therefore, in the excited state an electron will jump into the excited energy level. So, in excited state of neon one electron will jump into the 3s orbital.
Hence, its electronic distribution in the valence shell will become 2-7-1.
Thus, we can conclude that 2-7-1 electron configuration represents the electrons of an atom of neon in an excited state.
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Answer : The value of and K is, -180 kJ/mol and
Explanation :
The balanced cell reaction will be,
The half-cell reactions are:
Oxidation reaction (anode) :
Reduction reaction (cathode) :
Relationship between standard Gibbs free energy and standard electrode potential follows:
where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:
Now we have to calculate the value of 'K'.
where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant =
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:
Therefore, the value of and K is, -180 kJ/mol and