We are given the following chemical reaction:
aMnO₄⁻ + bSO₃²⁻ + H₂O → xMnO₂ + ySO₄²⁻ + zOH⁻
Let's balance the equation step by step:
The manganese on the left-hand side is in the form of MnO₄⁻ (with oxidation state +7). On the right-hand side, it is in the form of MnO₂ (oxidation state +4). Thus, to balance manganese atoms, we need:
a = 2, x = 2
The sulfur on the left-hand side is in the form of SO₃²⁻, and on the right-hand side, it is in the form of SO₄²⁻. To balance sulfur, we need:
b = 5, y = 5
To balance oxygen, we need to ensure that the number of oxygen atoms on both sides is the same. The left-hand side has 4 from MnO₄⁻ and 3 from SO₃²⁻, giving a total of 7 oxygen atoms, and the right-hand side has 4 from MnO₂ and 4 from SO₄²⁻. We adjust the oxygen atoms by balancing with water (H₂O) molecules.
We determine that:
a = 8, y = 6
The values of a and y are: a = 8, y = 6
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below: