Answer:
The elements mentioned in the series are Na, Mg, Al, Si, and P. It can be seen that all these elements are located in the same period. The atomic number of the mentioned elements are Na-11, Mg-12, Al-13, Si-14, and P-15.
a) There will be an increase in the ionization energy with the increase in the element's atomic number across a period. More energy is needed to withdraw an electron from a completely occupied shell in comparison to an incompletely occupied shell.
The atomic number of Na is 11. When one electron is withdrawn from Na, it gets converted into the inert gas configuration of Ne. Thus, it will require more energy to withdraw the second electron from Na. Hence, Na exhibits the lowest second ionization energy.
b) Across the period, with an increase in the element's atomic number, the atomic radii reduces from left to right. Thus, P exhibits the smallest atomic radius.
c) The metallic nature of the elements reduces from left to right across a period in the periodic table. Thus, P is the least metallic element.
d) Diamagnetic signifies towards the element that exhibits pair electrons in its sub-shells. The electronic configuration of Mg is,
1s2 2s2 2p6 3s2
In Mg, no unpaired electrons are present, while all the remaining elements mentioned exhibit unpaired electrons in their valence shell. Thus, Mg is the diamagnetic element.
Conduction is the act of touching 2 solids where one transfers heat
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It seems like I begin to forget whole organic Chemistry. Anyway, the answer is 2-methylheptane-3-yne.
To build a name for this molecule, you need to:
1. Determine where there is the lowest chain beginning with the double, triple bond, or there is other thing for other types of organic molecules.. In this case you have one triplet bond, and the lowest chain there is the one that is below the bond.
2. In this small chain you see one "forks": CH3 and another CH3 go from the one CH. Put number 1 where there are the most number of such compounds, in this case it doesn't matter because we have one CH3 in one way and one CH3 in the another.
3. Determine the largest chain which surely has the triple bond. Put the numbers from 2 to the final beginning with CH3 where you put number 1. Your largest chain consists of 7 particles CH3, CH2, and CH.
4. You see that you also have one unnamed particle, the one which is "the other way" when you are in the CH which is below the triple bond on the picture. Its name will be 2-methyl because it is connected with particle #2 and it consists only of one carbon. If there were 2, it would be ethyl, and so on.
5. Finally, let's write a full name: put 2-methyl first, "2-methyl." Then, write a full name for the chain of 7 carbons, which has to be heptane, if it had 5, you would name it pentane, ethane, and so on. So, we have "2-methylheptane." Finally, you see that you have a triple bond after the carbon #3. Put a dash and the number of this carbon. "2-methylheptane-3" you should have. Since it is alkyne (organic molecule with one triple bond), you have to write "yne" at the end.
the final name is "2-methylheptane-3-yne." You can answer the question "explain your naming process" by using the answer I provided. Hope it helps. Good luck!
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