me the element (using periodic trends and not looking up information): 1. a 3rd

Question

me the element (using periodic trends and not looking up information): 1. a 3rd period element whose atoms have 2 unpaired p electrons 2. another 3rd period element whose atoms have 2 unpaired p electrons 3. an element that forms a 4t ion with the electron configuration, [Kr)4d10 4. an alkaline earth metal whose cation is isoelectronic with argon 5. the element with the smallest IE1 in period 4 6. the element with the highest (most favorable) EA in group 5A 7. a nonmetal element that has filled f orbitals 8. a period 5 element that forms a 3t ion with a pseudo-noble gas configuration

Explanation / Answer

From the electronic configuration, it is shown that the Silicon (Si) and sulfur (S) have two unpaired P electrons.

Silicon = 1s2, 2s2, 2p6, 3s2, 3p2

According Hund’s rule electrons always enter an empty orbital before they pair up.

So the P orbital of silicon is shown as below

In the case of Sulfur

Here 4 electrons in the P orbital. It will arrange according to the hunds rule is as follows

So here also have two unpaired electrons.

Electronic configuration of Tin = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2,4p6,5s2,4d10, 5p2

Electronic configuration of krypton = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s2,4p6.

4+ ion of tin contain 4 electron less. The four electrons will go from the outer shells. That is 5s2 and 5p2 so the electronic configuration of 4+ ion of Tin is

[Kr]4d10

4. the alkali element is Potassium.

electronic configuration of Argon = 1s2, 2s2, 2p6, 3s2, 3p6    

electronic configuration of Potassium = 1s2, 2s2, 2p6, 3s2, 3p6, 4s1

electronic configuration of Potassium ion (K+) = 1s2, 2s2, 2p6, 3s2, 3p6                                     

The 4s1 electron will remove to form the K+ ion so it is same as that of argon in number of electrons.

So K+ ion is isoelectronic with argon

5. Potassium have the least 1st ionisation energy. Because the outer 4s1 electron will have very less attraction from the nuclear charge and also it is not pared. So it can be easily removed. And also if we move from left to right of the period of the periodic table, the ionization energy will increase because of the effective nuclear charge.

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