Inorganic Chemistry> Largest in size out of Na+ Ne and F- is. ...

Largest in size out of Na+NeandF- is. Explain your answer.

Piyush Raj , 7 Years ago

Grade 11

9 Answers

IHate Ppts

Last Activity: 7 Years ago

F- is the most electronegative ion. Thus, it has the smallest size possible in atomic table. Coming to Ne, it has no charge on it. We know that when an atom has positive charge, effective nuclear charge decreases which causes it to have more size.Therefore, Na+ is the biggest of all of these three.

kanu parmar

Last Activity: 7 Years ago

F- is the biggest in size. As we know that when we go right in a period atoms become smaller in size because no of orbit is not changing ,no of proton is increasing ,no of electron is increasing so that attraction between protons and electrons will increase so the size will decrease. In F- ,no of electrons is more than no of protons so that protons can`t attract electrons very heavily. In Na+ no of electrons is less than no of protons so that protons will attract electrons more heavily thus its size will decrease. Ne has the same no of electrons and protons. F- Ne Na+ will be the order. And for knowledge these are isoelectronic species in these type of speceis the more negative charge on the atom it will have bigger size.

hari

Last Activity: 6 Years ago

F->Ne>Na+ is the order. Na+ has more nuclear charge. so it is small in size.Ne, it has no charge on it.

Pranav Prakasan

Last Activity: 6 Years ago

Well, F- has the largest size, and that is because of it`s ionic radius which has increased on accepting an extra electron than its stable octet. Na+ is the smallest as it has lost an electron from its outermost shell thereby pulled a bit more towards the nucleus.

S. Giri

Last Activity: 6 Years ago

1)As we go right side of the periodic table no. Of shell remain same while no. Of protons and electrons increases .. ,,,,as protons are positively charged and electrons are negatively charged force of attraction between them increases due to which size decreases. F belongs to 17 group of 2nd period, F- has an extra electron hence it is little big than F. 2) Na belongs to 1st group and third period 3) Na has a extra shell than F- hence obviously it is bigger than it. 4) also also Na present most right side of third period . 5) Now let`s compare between Na and Na+ .6) as Na+ has extra proton that means high positive charge in nucleus than negative charge on electrons. Positive charge will attract electrons with more electrostatic force than Na hence size decreases. 7) also Na + is bigger than F- bcoz has extra shell 8)overall conclusion that came is that F-

zia ullah

Last Activity: 6 Years ago

Na+ has largest size because it contain more no of shell less electroforce of attraction in other two F has more size than Ne because left to right size decrease due to nucear charge increase electrostatic force of attraction increase

Rohit

Last Activity: 6 Years ago

F^- has the largest size. As the atomic no. Increases nuclear charge increases. Hence Na will have more nuclear charge. Now it is a sodium cation so one electron from outer shell is reduced but the nuclear charge remains same. And acts in the present case on only 10 electrons hence the outer shell contracts and size decreases. Based on above explaination we conclude that anions have larger size than cations and neutral molecules.

Hope it helps you.

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Rohit

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Nidhi

Last Activity: 6 Years ago

Well, the correct order should be: Ne > F- > Na+

If you notice carefully all of these are isoelectronic species with a total of 10 electrons. Now, Ne is a noble gas therefore it’s radius is Van Der waal radius which is greater than ionic radius, so Ne has the greatest radius. Out of F- and Na+, since F gains one electron, the hold of F nucleus on outer shell electrons decreases per electron because the number of electrons in outer shell or the valence shell has increased, hence it expands whereas in case of Na, it loses 1 electron so the hold of nucleus on each outer shell electron increases (due to less outer shell electrons now) so it contracts which gives us this order.

Vibhav S Chaudhary

Last Activity: 3 Years ago

it is Na^+ and Fe^- not Na ir F so Na^+ will be smaller as compared to F^- because both have same no of electrons but Na has more no of protons and protons will attract electrons towards it and will make it smaller .