chemistry problem! please explain!!?

This isn't that easy of a problem, so don't feel bad about choosing the wrong answer.

Look at the d orbital patterns here:

http://en.wikipedia.org/wiki/File:D_orbitals.svg

And the p orbital patterns here:

http://www.chemtube3d.com/images/porbitals.png

O.K., to form a pi bond, the orbitals must overlap and be parallel to the x-axis (but not along the x-axis; IIRC, such electron density would interfere with the sigma bond, which is on that axis, and would thus have antibonding tendencies).

Let's examine the choices.

a. 3d xz and 3d xy

The nodes do not overlap very much at all.

b. 3d xz and 3s

The geometry involved would afford very poor overlap (the 3s electron has considerable density outside of the 3d xz orbital).

c. 2p y and 3d x2-y2

One of the 3d x2-y2 nodes falls on the x-axis, so a pi bond isn't likely for that reason;

d. 3d x2-y2 and 3d x2-y2

Again, one of the 3d x2-y2 nodes falls on the x-axis, so a pi bond isn't likely for that reason.

e. 3d xy and 2p y

These do appear the best, being that there will be a fair amount of overlap, plus there are no antibonding tendencies for these.

The mass in the periodic desk is an traditional of the mass of all isotopes however for the reason that their relative abundance. For example, assume you've gotten 100 atoms of an detail X that have 2 isotopes. Mass of isotope Nº1 is 50 uma and mass of isotope Nº2 is 60 uma. In case you have 40 atoms of isotope N1 and 60 of isotope N2, the traditional mass will be: [40x50 (mass of 40 atoms of N1) + 60x60 (mass of 40 atoms of N2)] / a hundred (total quantity of atoms) So, if this difficulty you must add the mass of each and every isotope elevated through they relative abundance (%) after which divide with the aid of one hundred.