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3 years ago

27. Apply the Pauli exclusion principle, the aufbau princi-

1 answer:

5 0

The Pauli exclusion principle and Hund's rule are important principles to follow when writing electron configuration or drawing orbital diagrams.

Hund's rule is applicable when filling degenerate orbitals. Electrons are first filled singly before they are paired in degenerate orbitals.

Pauli exclusion principle states that two electrons can not have the same value for all the four quantum numbers.

As such; the electron configuration of the elements are;

silicon: 1s2 2s2 2p6 3s2 3p2

fluorine: 1s2 2s2 2p5

calcium: 1s2 2s2 2p6 3s2 3p6 4s2

krypton: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6

The orbital diagrams are shown in the images attached;

Image 1: Fluorine

Image 2: Silicon

Image 3: Calcium

Image 4: Krypton

Learn more: brainly.com/question/14283892

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Calculate the enthalpy of the reaction

harkovskaia [24]

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Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

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