Question

For each element:  

1. Write the complete electron configuration the neutral atom  

2. Draw out orbital box diagram showing the distribution of electrons in the atom  

3. Characterize the neutral atom as either paramagnetic or diamagnetic.  

4. write quantum numbers for the electrons in the highest energy level of the atom  

5. Write the four quantum number last electron added to the atom according to the Aufbau principle order discussed in class.  

A complete electron configuration does not abbreviate closed electronic shells below the valence level using the a noble gas's element symbol in square brackets.  

65.Scandium

Answer 1

Scandium Atomic Number = 21

1) Complete electronic configuration = 1s² 2s² 2p⁶ 3s²3p⁶ 3s² 4d¹ (If we consider aufbau principle). If we consider energy wise filling then we can write as = 1s² 2s² 2p⁶ 3s²3p⁶ 4s² 3d¹ because 3d has higher energy than 4s. This can be explained as per n+l (where “n” is the principle quantum number and “l” is the azimuthal quantum number). n+l= 3+2=5 for 3d orbital and n+l= 4+0= 4 for 4s orbital. Hence, 3d has higher energy than 4s.

2) Orbital box diagram of scandium:

3) Since the electron in the 3d is only one. Hence, it is paramagnetic. It would be diamagnetic if there were more paired electrons in the 3d orbital.

4) We know from our discussion above that the highest energy level for scandium is 3d. So, the four quantum numbers of 3d¹ electron:

(principal) n= 3

(azimuthal) l=2 (since we know d-orbital has azimuthal quantum number of 2. In fact, s=0, p=1, d=2, f=3, g=4, etc.)

(magnetic) m_l=+2 (Actually it can be +2 or +1 or 0 or -1 or -2. Here we are taking the highest possible one).

(spin) m_s = +1/2 or -1/2 (Since it contains only one electron, it can either have a positive spin or a negative spin).

5. According to the aufbau principle, the outermost electron in scandium will be filled in the 4s orbital (which is the outermost shell). Hence the quantum numbers are:

(principal) n= 4

(azimuthal) l=0 (since we know s-orbital has azimuthal quantum number of 0. In fact, s=0, p=1, d=2, f=3, g=4, etc.)

(magnetic) m_l=0

(spin) m_s = +1/2 or -1/2 (the last electron will have a spin value of -1/2 or +1/2 with respect to the other electron in the 4s orbital. One will have a positive spin while the other will have a negative spin. Both the electrons cannot have the same spin which is governed by Pauli's exclusion principle).