Question

How can I find valence electrons of transition metals?

Answer 1

Hint: There are various shells in the atoms which are denoted by K, L, M, N, etc. According to the number of electrons in the atoms are arranged in the shells and the last shell of the atom is the valence shell and the electrons in that shell are valence electrons this is true for representative elements but for transition elements, the last shell and (n-1)d sub-shell should also be considered.

Complete Solution :
We know that there are many types of elements in the periodic table i.e., representative elements, transition elements, and inner transition elements.
- The periodic table is the arrangement of elements according to their increasing number of electrons and these electrons are filled in the orbits according to their increasing energies. There are various shells in the atoms which are denoted by K, L, M, N, etc, which represents the specific energy of the shell. The shell K represents 1, the shell L represents 2, the shell M represents 3, the shell N represents 4. Each shell has specific sub-shells which are s, p, d, f, etc.
- According to the number of electrons in the atoms are arranged in the shells and the last shell of the atom is the valence shell and the electrons in that shell are valence electrons this is true for representative elements but in transition elements, the last shell and (n-1)d sub-shell should also be considered and these valence electrons are responsible for the bond formation.
- So, transition metals are those whose last electrons enters in the d-subshell, therefore, the number of valence electrons will be: $n{s}^{1-2}(n-1)d^{1-10}$. For example, the valence electron in iron will be 8 because the configuration of iron is $[Ar]3d^{6}4s^2$ so, the 6 electrons of d-subshell and 2 electrons of s-subshell is taken.

Note: It must be noted that the transition elements have a high number of valence electrons and due to this high number of valence electrons they can form coordination compounds and are very reactive.