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sp2606 [1]
3 years ago
12

How many unpaired electrons are in the following complex ions?

Chemistry
1 answer:
Oduvanchick [21]3 years ago
8 0

Solution :

A complex ion may be defined as a metal ion that is located at the center with other molecules or ions surrounding the center ion.

In the context, the number of the unpaired electrons for the following complex ions are :

1. $Ru(NH_3)62+$  (low spin)

  $Ru^{2+} = [Kr] \ 4d^6 \ 5s^0$

   Here all the d-electrons are paired. Therefore, there are no unpaired electrons.

2. $Ni(H_2O)62+$  (high spin)

   $Ni^{2+} = [Ar] \ 4s^0 \ 3d^8$

   Therefore, it has two unpaired electrons.

3. $V(en)33+$  (high spin)

    $V=[Ar] \ 3d^2 \ 4s^0$

   It has 2 unpaired electrons

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Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
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Answer:

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Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

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Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

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Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

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GIVING BRAINLIEST AND THE REST OF MY POINTS!!!!! :D
Alik [6]
It’s d I did this before
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