Question

list the following in order of increasing unpaired electrons: Fe,V,Sc,Mn

The answer is Sc<V<Fe<Mn, but can someone explain in detail how to fiqure out unpaired electrons to solve.

Answer 1

Aufbau principle : Subshells are arranged with increasing energy and electron filling occurs in same way. Generally the shell to be filled is always the one for which the sum of principle quantum number (n) and azimuthal quantum number (l) (i.e. n+l value). N = shell number like 1,2,3….. and l = azimuthal q.n. its 0 for S, 1 for P, 2 for d, 3 for f suborbitals.

e.g. for 4S ; n =4 and s = 0 hence n+l = 4

and for 3d, n = 3 and d = 2 hence n+l = 5

It means filling of 4S suborbital will take prior to 3d.

Hund’s rule of maximum multiplicity : While filling degenerate subshells pairing does not takes until and unless all degenerate subshells take 1 electron each. After that pairing starts. (maximum capacity of 1 subshell is 2 electrons)

e.g. in 3d suborbital there are 5 degenerate subshells if we have 6 electrons to fill we should not fill them only in 3 subshells (2x3 =6). According to Hund’s rule first give each orbital 1 e and 5 electrons filled. Then remaining 1 electron can be filled in any of these 5 degenerate subshell. It will pair up with electron there already.

Let us write the valence shell electronic configuration of all these metals using aufbau principle, Hund’s rule of maximum multiplicity,

1)Fe. Atomic number = 26

V.S.E.C. of Fe = [Ar] 4S2, 3d_xy², 3d_yz¹, 3d_xz¹, 3d_x2-y2¹, 3d_z2¹

Argon core take 18 electrons. We left with 26-18 = 8 e.

Now 4S has lower energy than 3d sub-orbital hence first filling of 4S sub-orbital takes place By aufbau principle). 4S will take 2 e and we left with 6 e.

Now 3d suborbital start filling. 3d suborbital have 5 degenerate (equal energy subshells) subshells and firstly each subshell get 1 e (Hund’s rule of maximum multiplicity) and i.e. 5 electrons filled. We still left with 1 e and that can be placed in any of these 5 degenerate subshells.

This lastly filled electron will pair up with the electron already in that subshell and finally there will be 4 unpaired electrons.

Number of unpaired electrons in Fe = 4

Following this way rest V.S.E.C. shown here,

2)V. Atomic number 23

V.S.E.C. of V = [Ar] 4S², 3d_xy¹, 3d_yz¹, 3d_xz¹, 3d_x2-y2⁰, 3d_z2⁰

Number of unpaired electrons in V = 3

3)Scandium (Sc) : Atomic number 21

V.S.E.C. of Sc = [Ar] 4S², 3d_xy¹, 3d_yz⁰, 3d_xz⁰, 3d_x2-y2, 3d_z2⁰

Number of unpaired electrons in Sc = 1

4)Manganese (Mn): Atomic number = 25

V.S.E.C. of Mn = [Ar] 4S², 3d_xy¹, 3d_yz¹, 3d_xz¹, 3d_x2-y2¹, 3d_z2¹

Hence number of unpaired electrons in Mn = 5.

Hence,

Order of given metal with increasing number of electrons,

Sc(1) < V(3) < Fe(4) < Mn(5).

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