1.) how many unpaired electrons are there in the fe3+ ion when it is in its ground state?
2.) which of these ions have noble gas electron configurations? (give the electron configuration of each ion.)
cr3+
al3+
se2−
3.) which of the following sets of species is/are isoelectronic?
s2−, cl−, k+
na+, mg2+, al3+
na+, k+, li+
for any that are isoelectronic, put them in order of increasing radius.
4.) using the periodic table classify each of these elements as a metal, nonmetal, or metalloid. does the element form an ion? if so, what will the charge of the ion be?
pb
k
xe
al
i (iodine)

1)5

2) Al3+ and Se2-

3)S2−, Cl−, K+    and  Na+, Mg2+, Al3+

Explanation:

1) Atomic number of Fe is 26. This means it has 26 protons and 26 electrons.

The electron configuration is as followed:

⇒Fe-1s2 2s2 2p6 3s2 3p6 3d6 4s2

⇒a neutral atom of iron would have two electrons in the 4s sublevel and six electrons in the 3d sublevel.

Fe3+ has 3 electrons less than Fe, so 23 electrons.

⇒The electron configuration for the Fe3+ ion is 1s2 2s2 2p6 3s2 3p6 3d5.  ⇒The positive charge of three in this ion shows that the atom lost three electrons.

⇒Electrons are taken from the outermost electron shell first, eliminating the two in the 4s sublevel. The other lost electron is then taken from the 3d sublevel.

⇒d subshell can occupy 10 electrons (in 5 orbitals). Pairing of electrons take place only after each orbital is singly occupied (Hund’s rule).

⇒ So, there are 5 electrons in the d subshell. So, each d orbital is singly occupied (unpaired).

⇒ This means there are 5 unpaired electrons in Fe3+

2) noble gas configuration means they have 8 electrons on the the outer shell.

⇒Cr3+ : Cr has 24 electrons of which 1 electron on the outer shell, so Cr3+ is not in the noble gass electron configuration

⇒1 s 2  2 s 2  2 p 6  3 s 2  3 p 6  4 s 0  3 d 3

⇒Note that it is  4 s 1  3 d 5  and not  4 s 2  3 d 4  because a half filled  d  orbital is more stable than a partially filled  d  orbital.

⇒Al3+: Al has 13 electrons, of which 3 are on the outer shell. Al3+ has 3electrons less, this means it loses the 3 electrons on the outer shell. So Al3+ has 10 electrons, of which 8 on the outer shell. This means it's in the noble gas electron configuration.

⇒The electron configuration of an aluminum ion, Al3+, is 1s2, 2s2, 2p6

⇒Se2-: Se has 34 electrons, of which 6 are on the outer shell. Se2- has 2 electrons more, this means it gains 2 electrons on the outer shell. So Se2- has 36 electrons, of which 8 on the outer shell. This means it's in the noble gas electron configuration.

⇒1 s 2  2 s 2 2 p 6  3 s 2  3 p 6  3 d 10  4 s 2  4 p 6

3) Isoelectronic species means those species which have same no. of total electrons :

S2-, Cl-,K+ :

⇒S has 16 electrons, S2- has 2 electrons more so 18 electrons

⇒Cl has 17 electrons,Cl- has 1 electron more so 18 electrons

⇒K has 19 electrons, K+ has 1 electron less so 18 electrons

⇔This is an isoelectronic set because they all have 18 electrons

⇔in order of increasing radius : S2- > Cl- > Al3+ : because in Sé- are the most electrons added.

Na+,Mg2+,Al3+

⇒Na has 11 electrons, Na+ has 1 electron less so 10 electrons

⇒Mg has 12 electrons, Mg2+ has 2 electrons less so 10 electrons

⇒Al has 13 electrons, Al3+ has 3 electrons less so 10 electrons

⇔This is an isoelectronic set because they all have 10 electrons

⇔in order of increasing radius : Na+ >Mg2+ > Al3+ : because in Sé- are the most electrons added.

Na+,K+,Li+

⇒Na has 11 electrons, Na+ has 1 electron less so 10 electrons

⇒K has 19 electrons, K+ has 1 electron less so 18 electrons

⇒Li has 3 electrons, Li+ has 1 electron less so 2 electrons

⇔This is no isoelectronic set

4) ⇒Pb is a metal. It's part of the p-block and the carbon group. It forms a Pb-ion ( Pb2+)

⇒ K is an alkalimetal ( so part of the metals). It forms K-ion (K+)

⇒ Xe is a nonmetal and part of the noble gasses. It doesn't form an ion, because it has 8 electrons on the outer shell, why it's stable.

⇒Al is a metal. It forms an Al3+ ion

⇒Iodine is a halogene., so a nonmetal. It forms I-, ion

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.