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raketka [301]
4 years ago
9

When an atom forms a positive ion (by losing electrons). Will the radius of an ion change from the radius of an atom? Explain wi

th the example of Sodium atom (Atomic no. 11) and Sodium ion.
Chemistry
1 answer:
Colt1911 [192]4 years ago
8 0

Radius of the positive ion will be smaller than the radius of its parent atom.

<h3>Explanation</h3>

Sodium Na has atomic number 11. A neutral Na atom has 11 electrons. It has

  • 2 electrons in the first main shell,
  • 8 electrons in the second main shell, and
  • 1 electron in the third main shell.

Na tends to lose one electron. It would form an Na⁺ ion. The ion is positive and has ten electrons. It has

  • 2 electrons in the first main shell, and
  • 8 electrons in the second main shell.

Three of the main shells of Na contain electrons. Only two main shells contain electrons in Na⁺. The energy of the second main shell is higher than the first. It is more distant from the nucleus than the first electron shell. Similarly, the third main shell is further away from the nucleus than the second.

The outermost shell in Na is the third main shell. The outermost shell in Na⁺ is the second main shell. As a result, the outermost electrons in Na are further away from the nucleus that electrons in Na⁺. As a result, the radius of Na appears to be larger than that of Na⁺.

When an atom loses electrons to form a positive cation, it tends to lose all electron in its outermost shell (the valence shell). That will give the ion an octet, which is more stable than having one or two electrons left in the valence shell. The rule on Na shall still apply. The positive ion will have one less main shell; its radius shall be smaller than that of the parent atom.

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3 years ago
During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).
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Answer:

The answers to the question are as follows

First part

The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g

Second part

The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F  

Explanation:

To solve this, we write the equation for the discharge of the lead acid battery as

H₂SO₄ → H⁺ + HSO₄⁻

Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻

at the cathode we have

PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O

Summing the two equation or the total equation for discharge is

Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

From the above one mole of lead and one mole of PbO₂  are consumed simultaneously hence

Number of moles of lead contained in 405 g of Pb with molar mass  = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb

Hence number of moles of  PbO₂ reduced at the cathode = 1.95 mole

mass of  PbO₂ reduced at the cathode = (number of moles)×(molar mass)

= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode

Part B

Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred

Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs

hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge

or transferred electrical charge = 377186.86 C or 3.909 Faraday

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