Question

the first ionisation potential of Na,Mg,Al and Si follow the order

Answer 1

Mg > Si > Al > Na

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Answer 2

Answer:

The given options are :

(1) Na < Mg > Al < Si (The correct answer)

(2) Na < Mg < Al < Si

(3) Na > Mg > Al > Si

(4) Na > Mg > Al < Si

Explanation:

To answer the question fast, try to use elimination or verification method rather than going straight for a systematic answer.

We know that ionization potential is inversely proportional to atomic size; as the atomic size decreases, ionization potential increases and vice versa. Also, atomic size decreases across a period and increases down the group.

Since all of the elements are just in one period, we are more considered about the periods here rather than groups. Following the previous rule,

• Si is the smallest atom in  the period and therefore has greatest Ionization potential.
• Similarly, Na has the largest atomic size in the period and therefore has the least ionization potential.

By these two realizations alone, we can neglect the 4th and 3rd options.

Now after this, it might seem plausible to tick the 2nd option. But when we look at the electronic configurations of Na, Mg, and Al:

Na: 1s² 2s² 2p⁶ 3s¹

Mg: 1s² 2s² 2p⁶ 3s²

Al: 1s² 2s² 2p⁶ 3s² 3p¹

We see that in Na and Al, all the shells are not filled completely and hence  are more unstable and it is easier to take out an electron from them, that is, the ionization potential is smaller.

But in Mg, all the shells are completely filled (max capacity of s-subshell is 2 electrons and max. capacity of p-subshell is 6 electrons) and is more stable and therefore, it is harder to take out electron from it, and therefore has a higher ionization potential.

Thus we can say that the 1st option is the correct option.

(to understand fast, read only the bolded parts of the sentences)

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