Question

The hot glowing gases around the Sun, the corona, can reach millions of degrees Celsius, high enough to remove many electrons from gaseous atoms. Iron ions with charges as high as 14+ have been observed in the corona. Which ions from Fe⁺ to Fe¹⁴⁺ are paramagnetic? Which would be most strongly attracted to a magnetic field?

Answer 1

Answer:

All ions except $Fe^{+4} \, \, , Fe^{+8}$ are paramagnetic as they have unpaired electrons while $Fe^{+} \, \, , Fe^{+3}$ will be strongly attracted to a magnetic field.

Explanation:

Here the ions of Fe which will be available are as follows

$Fe^{+} \, \, , Fe^{+2} \, \, , Fe^{+3} \, \, , Fe^{+4} \, \, , Fe^{+5} \, \, , Fe^{+6} \, \, , Fe^{+7} \, \, , Fe^{+8} \, \, , Fe^{+9} \, \, , Fe^{+10} \, \, , Fe^{+11} \, \, , Fe^{+12} \, \, \\Fe^{+13} \, \, , Fe^{+14}$Out of these ions the electrons  distribution is given as such that

$Fe^{+} \, \, , Fe^{+2} \, \, , Fe^{+3} \, \, ,Fe^{+5} \, \, , Fe^{+6} \, \, , Fe^{+7} \, \, , Fe^{+9} \, \, , Fe^{+10} \, \, , Fe^{+11} \, \, , Fe^{+12} \, \, \\Fe^{+13} \, \, , Fe^{+14}$

These ions will have one or more unpaired electrons. Since paramagnetism is dependent on the unpaired electron, therefore,these ions will be paramagnetic.

$Fe^{+} \, \, , Fe^{+2} \, \, , Fe^{+3} \, \, ,Fe^{+5} \, \, , Fe^{+6} \, \, , Fe^{+7} \, \, , Fe^{+9} \, \, , Fe^{+10} \, \, , Fe^{+11} \, \, , Fe^{+12} \, \, Fe^{+13} \, \, , Fe^{+14}$

So all ions except $Fe^{+4} \, \, , Fe^{+8}$ are paramagnetic as they have unpaired electrons

Out of these the most strongly attracted towards the magnetic field would be pairs having the highest number of unpaired electrons which is

$Fe^{+} \, \, , Fe^{+3}$

$Fe^{+} \, \, , Fe^{+3}$ will have the strongest effect of paramagnetism as they will have maximum unpaired electrons.