Question

Shown below is a plot of the radial probability distribution, 415r2R2, as a function of radial distance,r, for an atomic orbital of hydrogen. This orbital has precisely two angular nodes, with one angular node perpindicular to the x-axis, and the second angular node perpendicular to the y-axis. Use this figure to answer the next 3 questions. Orbital Probability Distribution ( 47tr2R2(r)) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 radius / Å Question 2 1 pts The boundary surface for this orbital extends to a radius R such that the probability of finding the electron at a distance less than or equal to R is 90%. From the choices below, select the distance that is closest to R. All values are given in Angstroms. O 6.5 12.0 47.7 30.0 21.5 Question 3 1 pts How many radial nodes does this orbital have, if any? Question 4 1 pts What is the name of this orbital? (For example, if the orbital is 2pz write it down as 2pz) 30

I am a little confused if I have done this question right, would anyone be able to check if I answered my questions right and if I did anything wrong explain why. Thank you!!

Answer 1

Question 2:

Option 4: From the graph we observe that the probability distribution at r= 21.5 A^o is 90% or less than that.

Question 3:

From the graph we see that at r= 12 we observe the probability distribution to be zero other than the center (r=0) and at r= infinite. So the radial nodes present=1.

Question 4:

There are 1 radial node and two angular nodes present, so total no of nodes= 3.

As the total no of radial nodes= n-l-1 (n= principle quantum number, i= azimuthal quantum number)

And total no of angular nodes= l

so here, l=2 = d orbital.

and n-2-1= 1

or, n-3=1

or, n= 4.

So it is a 4dxy orbital ( as the angular nodes are perpendicular to x and y axis)