JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The element with Z = 120 (not yet discovered) will be an/a -
- AAlkaline earth metal
- BAlkali metal
- CTransition metal
- DInner transition metal
View written solutionFree
Correct answer: A
-
Identify the electronic filling after known elements
The last known noble gas in the currently predicted sequence is element with configuration ending in:
-
Next elements after
After is completely filled at , the next electrons are expected to enter the subshell.
-
For : This would be an alkali metal.
-
For :
-
-
Determine the group from valence configuration
An outer electronic configuration of corresponds to Group 2 elements.
Group 2 elements are called alkaline earth metals.
-
Check options
- A: Alkaline earth metal Correct
- B: Alkali metal Incorrect, that would correspond to
- C: Transition metal Incorrect, transition metals involve filling of subshells
- D: Inner transition metal Incorrect, inner transition metals involve filling of subshells
-
Final answer
Therefore, the element with will be an alkaline earth metal.
More from Periodic Table and Periodicity
- For , , and ; the correct order of increasing ionic radii is :2018 · MCQ
- The correct order of electron affinity is :2018 · MCQ
- Consider the following ionization enthalpies of two elements 'A' and 'B' . Which of the following statements is correct ? Includes table2017 · MCQ
- Which one of the following is an oxide?2017 · MCQ
- The electronic configuration with the highest ionization enthalpy is :2017 · MCQ
- The following statements concern elements in the periodic table. Which of the following is true ?2016 · MCQ
- The ionic radii (in Å) of , and are respectively:2015 · MCQ
- The first ionization potential of is 5.1 eV. The value of electron gain enthalpy of will be:2013 · MCQ