Identify the elements X and Y using the ionization energy values given below:
Ionization energy (kJ/mol).
–
1^st
2^nd
X
495
4563
Y
731
1450

X = F; Y = Mg

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

X = Mg; Y = F

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

X = Na; Y = Mg

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

X = Mg; Y = Na

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

Open in App

Solution

The correct option is C
X = Na; Y = Mg

Explanation for the correct option:
Option (C ) is correct
Due to 2p⁶, noble gas electronic configuration, the second ionization enthalpy of Na is very high. That’s why has a large difference between IE₁, and IE₂
Mg⁺ is 2p⁶, 3s¹.
After the loss of one electron, Mg⁺ will be formed with noble gas electronic configuration. That’s why has less difference between IE₁ and IE₂, but their IE₃ is very high.
X= Na; Y=Mg
Na → [Ne] 3s¹ IE₁ is very low but IE₂ is very high due to the stable noble gas configuration of Na+.
Mg → [Ne] 3s² , $I E_{1} & I E_{2} \to L o w I E_{3}$ is very high.
Explanation for the incorrect option:
(a) X = F; Y = Mg
The outermost orbit configuration of Flourine is not fully stable but highly stable for Mg.
(b) X = Mg; Y = F
The outermost orbit configuration of Mg is highly stable but for Flourine it's not fully stable.
(d) X = Mg; Y = Na
For this option first element is not fully stable, so this is not a correct option.
Final Answer: So the correct answer is option C

Suggest Corrections

Similar questions

(kJ/mol)

\begin{matrix} 1^{s t} & 2^{n d} X & 495 & 4563 Y & 731 & 1450 \end{matrix}

k J m o l^{- 1}

k J m o l^{- 1}

X

Y

	1st Ionisation Enthalpy(KJ/mol)	2nd Ionisation Enthalpy(KJ/mol)
$X$	$495$	$4563$
$Y$	$731$	$1450$

Join BYJU'S Learning Program