Consider the following successive ionization energies in $k J / m o l$ :

$I E_{1} = 580, I E_{2} = 1815, I E_{3} = 2740, I E_{4} = 11, 600$ , and $I E_{5} = 14, 842$

Which element could have this successive ionization energy data?

N a

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

M g

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

A l

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

S i

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

Open in App

Solution

The correct option is D $A l$
The fourth IE is significantly higher than the third IE. Hence the element attains inert configuration on the removal of $3$ electrons i.e., it belongs to either IIIA or IIIB

As only such option is of $A l$ (IIIA).

Hence, the correct option is $C$

Suggest Corrections

Similar questions

I E_{1}

250 k J m o l^{- 1}

I E_{2}

820 k J m o l^{- 1}

I E_{3}

1100 k J m o l^{- 1}

I E_{4}

1400 k J m o l^{- 1}

Successive ionization enthalpies of an element 'X' are given below

IE1. IE2. IE3. IE4

165. 195. 556. 595

Electronic configuration of the element'X' is?

^{- 1}

N i

P t

N a, M g, A l

S i

Join BYJU'S Learning Program