Math the following Groups with respect to its electron configuration:

$\begin{array} {ll} 1.\; 3B(3)& a.\;(n-1)d^3ns^2\\ 2.\; 4B(4)& b.\;(n-1)d^1ns^2\\ 3.\; 5B(5)& c.\;(n-1)d^5ns^1\\ 4.\; 6B(6)& d.\;(n-1)d^2ns^2\\ 5.\; 7B(7)& e.\;(n-1)d^5ns^2 \end{array}$

$\begin{array}{1 1}1 \to b \quad 2 \to d \quad 3 \to a \quad 4 \to c \quad 5 \to e \\1 \to c \quad 2 \to a \quad 3 \to b \quad 4 \to e \quad 5 \to d \\ 1 \to d \quad 2 \to a \quad 3 \to b \quad 4 \to c \quad 5 \to e \\1 \to b \quad 2 \to d \quad 3 \to c \quad 4 \to a \quad 5 \to e \end{array}$

Answer: $1 \to b \quad 2 \to d \quad 3 \to a \quad 4 \to c \quad 5 \to e$
In the transition metals, the five d orbitals are being filled in, and the elements in general have electron configurations of $(n-1)d^{1-10} ns^2$, although there are some exceptions when electrons are shuffled around to produce half-filled or filled d subshells.
Group 6B (6) elements have the electron configuration $(n-1)d^5 ns^1$, instead of the expected $(n-1)d^4 ns^2$ Since their d orbitals are one electron away from being half-filled (i.e., having one electron in each of the five d orbitals), an s electron can move into the d orbitals to make a more stable half-filled d-orbital configuration.
edited Aug 5, 2014