Question

The element with Z=120 (not yet discovered) will be an/ a:

(A) Transition metal

(B) Non- transition metal

(C ) Alkaline earth metal

(D) Alkali metal

Answer 1

Hint: The alkaline earth metals have an outer s-orbital which is filled up with 2 electrons. They readily lose these 2 outer electrons. We need to keep in mind the electronic configuration of the metals to answer this question.

Complete step by step answer:

Given, Z=120 The electronic configuration of the element can be represented as  $[\text{Noble gas}]n{s}^2$ .

Transition metals are those elements which have a partially filled d subshell.

The general electronic configuration of transition metals is

 $[\text{Inert gas}](n-1)d^{1-10}n{s}^{0-2}$ 

Non-transition metals are those elements which are present in s and p blocks. They contain alkaline earth metal and alkali metal.

The alkaline earth metal belongs to the s block elements of the periodic group.

The general electronic configuration of alkaline earth metals is  $[\text{Noble gas}]n{s}^2$ 

The alkaline metals have 1 s electron in the outermost shell.

The general electronic configuration of alkaline earth metals is  $[\text{Noble gas}]n{s}^1$ 

Thus by using the electronic configuration of Z=120 we can see it is similar to the electronic configuration of alkaline earth metals.

Thus the given element is an alkaline earth metal.

Hence correct option is (C).

Note: Some examples of alkaline earth metals are, beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). In these types of questions first find out the general electronic configuration of the element. Then compare it with the electronic configuration of the elements given in the option.