If the Aufbau principle had not been followed, Ca $\left( {Z = 20} \right)$ would have been placed in the-
A.s-block
B.p-block
C.d-block
D.f-block
Answer
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Hint: Calcium is placed in s-block in group $2$ and period $4$because its electronic configuration is$n{s^2}$. The electronic configuration of all the elements in the periodic table is given according to the Aufbau principle which is based on the increasing energy of atomic orbitals.
Complete step by step answer:
Given, the atomic number of Calcium is $20$ and it is placed in s-block as it has electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}$ in which the last electron is filled in s-orbital. Here the electrons are filled in$4s$ before $3d$ because the energy of $4s$ orbital is less than the energy of $3d$. The s-block elements have the following properties-
1.They are metals.
2.They are white or silvery in colour and soft.
3.They have low melting and boiling point.
4.They are strongly electropositive in nature.
5.They impart characteristic colours to the flame.
6.They have low ionization enthalpy.
If Aufbau principle is not followed then the electrons will be filled in $3d$ before going to $4s$ so the electronic configuration will change to ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}3{{\text{d}}^{\text{2}}}$. Since in this electronic configuration the last electrons are placed in d- orbital so the calcium element would have been placed in d-block.
The correct answer is option C.
Note:
The properties of Calcium are-
1.It is a silvery-white and soft metal.
2.It imparts red brick colour to the flame.
3.It corrodes in the air and reacts with water.
4.It is used as a reducing agent.
5.It is a highly reactive metal.
6.It is used to make two important compounds- Gypsum and plaster of Paris.
Complete step by step answer:
Given, the atomic number of Calcium is $20$ and it is placed in s-block as it has electronic configuration ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{s}}^{\text{2}}}$ in which the last electron is filled in s-orbital. Here the electrons are filled in$4s$ before $3d$ because the energy of $4s$ orbital is less than the energy of $3d$. The s-block elements have the following properties-
1.They are metals.
2.They are white or silvery in colour and soft.
3.They have low melting and boiling point.
4.They are strongly electropositive in nature.
5.They impart characteristic colours to the flame.
6.They have low ionization enthalpy.
If Aufbau principle is not followed then the electrons will be filled in $3d$ before going to $4s$ so the electronic configuration will change to ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}3{{\text{d}}^{\text{2}}}$. Since in this electronic configuration the last electrons are placed in d- orbital so the calcium element would have been placed in d-block.
The correct answer is option C.
Note:
The properties of Calcium are-
1.It is a silvery-white and soft metal.
2.It imparts red brick colour to the flame.
3.It corrodes in the air and reacts with water.
4.It is used as a reducing agent.
5.It is a highly reactive metal.
6.It is used to make two important compounds- Gypsum and plaster of Paris.
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