Answer
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Hint: To answer this, use the concept of electronegativity. The more electronegative among the two will have a stronger bond and thus will be a weaker acid. Also, higher is the electronegativity, higher will be the bond polarity.
Complete step by step answer:
We know that an acid is a species which will donate its proton that is the ${{H}^{+}}$ ion. Now the stronger or the more acidic compound will be the one which will donate its hydrogen cation that is the proton readily.
Here, we have hydrogen sulphide and water. Therefore among hydrogen sulphide and water, the more acidic is the one which will release its proton readily.
We know that oxygen is more electronegative than sulphur thus the H-O bond is stronger than the S-H bond. The energy required to break the S-H bond is lower than that of O-H bond, this energy is the bond dissociation energy which is lower for the S-H bond. Thus, oxygen atoms will not donate the proton readily and this will make it a weaker acid.
Therefore, the assertion that ${{H}_{2}}S$ is more acidic than ${{H}_{2}}O$ is correct.
Now, the reason that is given to us is that the H-S bond is more polar than H-O.
We know that polarity is dependent on electronegativity. Here, the electronegativity difference in O-H bind is greater than that of S-H bond. Therefore, the H-O bond is more polar than H-S so the reason given to us is incorrect.
From the above discussion we can understand that the assertion is correct but the reason is incorrect.
So, the correct answer is “Option C”.
Note: There are many other theories that we use for describing acids like here we used the concept of proton donation. This concept is given by the Arrhenius theory according to which ${{H}^{+}}$ ion donors are acids and $O{{H}^{-}}$ donors are bases. Similarly, we have Lewis theory, Bronsted-Lowry theory and many others.
Complete step by step answer:
We know that an acid is a species which will donate its proton that is the ${{H}^{+}}$ ion. Now the stronger or the more acidic compound will be the one which will donate its hydrogen cation that is the proton readily.
Here, we have hydrogen sulphide and water. Therefore among hydrogen sulphide and water, the more acidic is the one which will release its proton readily.
We know that oxygen is more electronegative than sulphur thus the H-O bond is stronger than the S-H bond. The energy required to break the S-H bond is lower than that of O-H bond, this energy is the bond dissociation energy which is lower for the S-H bond. Thus, oxygen atoms will not donate the proton readily and this will make it a weaker acid.
Therefore, the assertion that ${{H}_{2}}S$ is more acidic than ${{H}_{2}}O$ is correct.
Now, the reason that is given to us is that the H-S bond is more polar than H-O.
We know that polarity is dependent on electronegativity. Here, the electronegativity difference in O-H bind is greater than that of S-H bond. Therefore, the H-O bond is more polar than H-S so the reason given to us is incorrect.
From the above discussion we can understand that the assertion is correct but the reason is incorrect.
So, the correct answer is “Option C”.
Note: There are many other theories that we use for describing acids like here we used the concept of proton donation. This concept is given by the Arrhenius theory according to which ${{H}^{+}}$ ion donors are acids and $O{{H}^{-}}$ donors are bases. Similarly, we have Lewis theory, Bronsted-Lowry theory and many others.
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