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pickupchik [31]
3 years ago
10

Draw the electron configuration for a neutral atom of calcium

Chemistry
1 answer:
Nonamiya [84]3 years ago
6 0

In order to write the Calcium electron configuration we first need to know the number of electrons for the Ca atom (there are 20 electrons). When we write the configuration we'll put all 20 electrons in orbitals around the nucleus of the Calcium atom.

In writing the electron configuration for Calcium the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for Calcium go in the 2s orbital. The next six electrons will go in the 2p orbital. The p orbital can hold up to six electrons. We'll put six in the 2p orbital and then put the next two electrons in the 3s. Since the 3s if now full we'll move to the 3p where we'll place the next six electrons. We now shift to the 4s orbital where we place the remaining two electrons. Therefore the Calcium electron configuration will be 1s22s22p63s23p64s2.

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ΔH = +438 kJ  

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The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

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Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

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You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

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(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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