Question

The outer electronic configuration of the element Mo (Z=42) is:
A.$5{{s}^{2}}4{{d}^{4}}$
B.$5{{s}^{1}}4{{d}^{5}}$
C.$5{{s}^{2}}5{{p}^{4}}$
D.$4{{s}^{2}}3{{d}^{4}}$

Answer 1

Hint: Electronic configuration: The distribution of electrons of atoms and molecules in its atomic orbital is termed as Electronic configuration. There are four kinds of atomic orbitals present s, p, f, d. The electrons in atomic orbitals are filled according to order of their increasing energy.

Complete step by step answer:
 The increasing order of atomic orbital in context with energy is given by Aufbau principle. Example: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p…..
According to Aufbau principle, electronic configuration of $Mo(Z=42)$is:
Atomic number (Z) = 42
Electronic Configuration: $1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{10}},4{{p}^{6}},5{{s}^{2}},4{{d}^{4}}$
But, to attain stability the actual electronic configuration of $Mo(Z=42)$is $1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{10}},4{{p}^{6}},5{{s}^{1}},4{{d}^{5}}$ i.e. one electron of s-orbital jumps into d-orbital because half filled orbital is more stable. Now to avoid extra length we convert the electronic configuration to noble gas electronic configuration:$[1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{2}},3{{d}^{10}},4{{p}^{6}}],5{{s}^{1}},4{{d}^{5}}$
The outer E. C. is: $[Kr]5{{s}^{1}}4{{d}^{5}}$

So, the correct answer is “Option B”.

Note:
The total no. of protons present in an atom gives the atomic no. of an element. The mass no. is given by the sum of protons and neutrons so we can easily find no. of neutrons by subtracting atomic no. from mass no. of a particular element. One should be familiar with Aufbau principle i.e. electrons first occupy the lowest energy orbital available to them and enter higher energy orbitals only when the lower energy orbitals are filled. The name of the noble gas should be put into the square bracket i.e. [ ].