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Andrew [12]
3 years ago
5

Effective nuclear charge, Zeff, is defined as:

Chemistry
1 answer:
Alborosie3 years ago
6 0

Answer:

The effective nuclear charge for a valence electron in oxygen atom: Z_{eff} = 4.55

Explanation:

Effective nuclear charge  [Z_{eff}] is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.

<em>It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).  </em>

\Rightarrow Z_{eff} = Z - s

<u>For an oxygen atom</u>-

Electron configuration: (1s²) (2s² 2p⁴)

<em>The atomic number (actual nuclear charge): </em>Z = 8

The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:

⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45

<u><em>Therefore, the effective nuclear charge for a valence electron in oxygen atom is:</em></u>

Z_{eff} = Z - s = 8 - 3.45 = 4.55

<u>Therefore, the effective nuclear charge for a valence electron in oxygen atom:</u> Z_{eff} = 4.55

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Explanation:

The products of this reaction are given by:

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Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

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The following steps should be taken:

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Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)

  • balance oxygen atoms by adding 7 water molecules on the right:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the hydrogen atoms by adding 14 protons on the left:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

Similarly, tin(II) cation becomes tin(IV) cation:

Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-

Add them together:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)

Adding the ions spectators:

K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)

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