Consider the reaction below.
2 answers:
Answer : The correct option is
Explanation :
According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.
Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.
The given equilibrium reaction will be,
In this reaction, and
are acid and base and
and
are conjugate acid and base respectively.
From this we conclude that, is a base-conjugate acid pair.
Hence, the correct option is
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The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
Answer:
Hence the order of the transition will be : d < a < c < b
Explanation:
where,
= energy of
orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy of n = 3 in an hydrogen atom:
Energy of n = 4 in an hydrogen atom:
Energy of n = 5 in an hydrogen atom:
a) n = 2 to n = 4 (absorption)
b) n = 2 to n = 1 (emission)
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
d) n = 4 to n = 3 (emission)
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
h = Planck's constant
frequency of the wave
So, the increasing order of magnitude of the energy difference :
And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :
: d < a < c < b