Which one of the following relations is correct with respect to first $(I)$ and second $(I I)$ ionization potentials of sodium and magnesium?

I_{N a} > I_{M g}

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

I_{M g} = I I_{N a}

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

I I_{M g} > I I_{N a}

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

I I_{N a} > I I_{M g}

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

Open in App

Solution

The correct option is D $I I_{N a} > I I_{M g}$
After losing one electron sodium acquires stable noble gas configuration and thus second ionization potential of sodium is very large .

Therefore $I I_{N a} > I I_{M g}$

Option D is correct.

Suggest Corrections

Similar questions

I^{s t}

2^{n d}

(I) F^{-}

(I I) N a^{+}

(I I I) N^{3 -}

I^{s t}

2^{n d}

(I_{1})

(I_{2})

N a

M g

Join BYJU'S Learning Program