You visited us 1 times! Enjoying our articles? Unlock Full Access!

Ionisation Energy

The ionizatio...

Question

The ionization potential values of an element are in the following order. The element is:
$I_{1} < I_{2} <<<< I_{3} < I_{4} < I_{5}$ .

an alkali metal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

chalcogen

No worries! We‘ve got your back. Try BYJU‘S free classes today!

halogen

No worries! We‘ve got your back. Try BYJU‘S free classes today!

alkaline earth metal

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D alkaline earth metal
The metal is an alkaline earth metal as the difference between second and third ionisation potential values is very large. Alkaline earth metal has 2 valence electrons. It loses these two electrons to form dipositive cation with stable electronic configuration of nearest noble gas in which octet is complete. Removal of third electron will break this octet and will require very large amount of energy.

Suggest Corrections

Similar questions

(X)

I_{1}, I_{2}, I_{3}, I_{4}

I_{5}

800, 2427, 3658, 25024, 32824 {kJ mol}^{- 1}

A B C D

O

1, 2, 3

4

I_{1}, I_{2}, I_{3} & I_{4}

I_{n} = \int_{0}^{π / 4} t a n^{n} x d x

I_{2} + I_{4}, I_{3} + I_{5}, I_{4} + I_{6}, I_{5} + I_{7}

I_{1} = \int_{0}^{π / 2} {sin}^{4} x d x

I_{2} = \int_{0}^{π / 2} {cos}^{6} x d x

I_{3} = \int_{0}^{π / 2} {sin}^{8} x d x

I_{4} = \int_{0}^{π / 2} {cos}^{2} x d x

I_{1}, I_{2}, I_{3}, I_{4}

I_{n} = \frac{π}{2} \int \frac{π}{4} {cot}^{n} x d x

Join BYJU'S Learning Program