Question

In the periodic table an element with atomic number 56 belongs to:
\[
  A.\;\;\;\;\;III{\text{ }}A{\text{ }}group,{\text{ }}{6^{th}}period \\
  B.\;\;\;\;\;IV{\text{ }}A{\text{ }}group,{\text{ }}{5^{th}}period \\
  C.\;\;\;\;\;II{\text{ }}A{\text{ }}group,{\text{ }}{6^{th}}period \\
  D.\;\;\;\;\;IV{\text{ }}A{\text{ }}group,{\text{ }}{6^{th}}period \\
 \]

Answer 1

Hint: We have to write the electronic configuration for the element \[\left( {Z = 56} \right)\] and analyse it with a periodic table for the answer.



Complete step by step solution:
In this question, we are having an element whose atomic number is 56. Atomic number 56 belongs to the barium atom. Barium is an alkali metal which means it is present in s-block and hence IIA group, 6th period.
But if you don’t know which element 56 belongs to then what to do? Well, you can go for electronic configuration of the atom, so let’s start with it.
Atomic number of the given atom is 56, so its electronic configuration will be
Electronic Configuration (atomic number 56) = \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^{10}}4{p^6}5{s^2}4{d^{10}}5{p^6}6{s^2}\]
So like this we have a total of 56 electrons in a total of 6 orbitals. Since 6 orbitals are being used this means the period number is 6. Along with that the 6s orbital is completely filled. And, hence the group number is IIA.
So, the answer to this question can be achieved in two different ways.
The answer is C. II A group, 6th period.



Note: We must know something about alkaline earth metals. When we burn, its gives various colored flames as follows: beryllium (white), magnesium (bright white), calcium (red), strontium (crimson), barium (green), and radium (red). Also the name "alkaline earths" comes from an old name for the oxides of the elements. Therefore, we can alkaline because they form solutions with a pH greater than 7, making those bases or "alkaline."