The first four ionisation energy values of an element are $191$,$578$,$872$ and $5962 \ kcal$. The number of valence electrons in the element is :-

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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The second ionization potential is

In the given graph,which point represents an alkali metal with the least atomic number (Period number $= 3$)?

From the following elements,which of them has the highest second ionisation potential?

If the successive ionisation energies of an element $A$ are $165$,$190$,$550$ and $595 \ kcal$,respectively,then the ground state electronic configuration of element $A$ is

Assertion $(A)$: The first ionisation energy of $Be$ is greater than that of $B$.
Reason $(R)$: $2p$ orbital has lower energy than $2s$ orbital.

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