Question 8 of 32

1 answer:

3 0

Sodium, atomic radius increases as we go down and right in the periodic table, which means that cesium is the element with the biggest atomic radius as it is located as the farthest bottom right of the periodic table, anyways, the atomic number of aluminum is 13 which means that the distribution of electrons here is K:2, L:8, M:3,, number of levels = the atomic period which the atom is located in, last number in the last electronic level (or the valency) = the atomic group which the atomic is located in, so in this case aluminum is located in group 3 period 3 at the periodic table, we have to find the atom which is located the nearest to the bottom right using the same method, sodium is located in group 1 period 3, boron is located in group 3 period 2, silicon is located in group 4 period 3, nitrogen is located in group 5 period 2,,, so the correct answer is sodium as it’s located at the left of aluminum and closer to the left of the periodic table, albeit being in the same period

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Answer:

$\rm {Cr_2O_7}^{2-} + 6 \; Fe^{2+} + \underbrace{\rm 14\; H^{+}}_{\text{From}\atop \text{Acid}}\to 2\; Cr^{3+} + 6\; Fe^{3+} + 7\; H_2 O$ .

Explanation:

Consider the oxidation state on each of the element:

Left-hand side:

O: -2 (as in most compounds);
Cr: $\displaystyle \frac{1}{2}(\underbrace{-2}_{\text{ion}} - \underbrace{7\times (-2)}_{\text{Oxygen}}) = +6$ ;
Fe: +2 (from the charge of the ion);

Right-hand side:

Cr: +3;
Fe: +3.

Change in oxidation state:

Each Cr atom: decreases by 3 (reduction).
Each Fe atom: increases by 1 (oxidation).

Changes in oxidation states shall balance each other in redox reactions. Thus, for each Cr atom on the left-hand side, there need to be three Fe atoms.

Assume that the coefficient of the most complex species $\rm Cr_2O_7^{2-}$ is 1. There will be two Cr atoms and hence six Fe atoms on the left-hand side. Additionally, there are going to be seven O atoms.

Atoms are conserved in chemical reactions. As a result, the right-hand side of this equation will contain

two Cr atoms,
six Fe atoms, and
seven O atoms.

O atoms seldom appear among the products in acidic environments; they rapidly combine with $\rm H^{+}$ ions to produce water $\rm H_2O$ . Seven O atoms will make seven water molecules. That's fourteen H atoms and hence fourteen $\rm H^{+}$ ions on the product side of this equation. Hence the balanced equation. Double check to ensure that the charges on the ions also balance.