[scifan]

For an octahedral complex of each of the following metal ions, determine the number of unpaired electrons for both high spin and low spin configurations. In some cases, both may be the same.

Zn2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:

Cr2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:

Co2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:

Rh3+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:

[cheese (MSW)]

Let me give you a worked example to give you a hand up.  How many unp e⁻ are in high and low spin Fe(III)?  (A classic case: F. A. Cotton, G. Wilkinson, C. A. Murillo, M. Bochmann, Advanced Inorganic Chemistry 6th ed (1999). p 784-786.)
CFT argument (LFT gives same overall picture).  In the six coordinate environment (the octahedral crystal field) the five d AOs do not have the same energy but are split into a lower energy set (t2g: dxy, dyz, dyz) of three d AOs  that point between the -ve atoms/ligands (less e⁻ repulsion) and a higher energy set of two AOs (eg set: dx^2-y^2,dz^2)) that point at the -ve ligands.  (The t2g→Δoct→eg energy gap is Δoct comes in the visible, hence the color of TM cmplxs).   
Go to the Wikipedia entry for iron and in the right-hand panel you will find the correct e⁻ configuration for Fe, namely, [Ar] 3d^6 4s^2 (note 3d < 4s).  Remove three e⁻s gives you [Ar] 3d^5.  There are two possibilities for the distribution of the five e⁻ over the d orbitals.  If Δoct < P (the pairing energy) the high spin case results:  t2g (↑)(↑)(↑)   (↑)(↑) eg with five unp e⁻ (μ = 5.9 BM).  This occurs in [Feox3]3- (the oxalate dianaion is a weak field ligand).  If however Δoct>P the low spin case results:  t2g (↑↓)(↑↓)(↑)            ()()eg that has one unp e⁻ (μ = 1.73 BM) and is found in [Fe(CN)6]^3- (CN^- is of course a strong field ligand).   Magnetic moments are readily determined even by undergrads!
 You should now be able to tackle your questions.  Good luck!