Consider the following electron configurations to answer the questions that follow: (i) 1s2 2s2 2p6 3s1 (ii) 1s2 2s2 2p6 3s2 (iii) 1s2 2s2 2p6 3s2 3p1 (iv) 1s2 2s2 2p6 3s2 3p4 (v) 1s2 2s2 2p6 3s2 3p5 The electron configuration belonging to the atom with the highest second ionization energy is .

Option (i) would have the highest 2nd Ionization Energy.

Option (i) is Sodium.

Can be Written as 2, 8 , 1

For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.

Now After Removing that Electron...

Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.

Neon has a very stable Octet.

It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.

So

Option (i) [Sodium] has the highest 2nd Ionization Energy

- calcium has greater affinity to bond with electron receptors than hydrogen does and hence, it has the ability to replace hydrogen in most of its compounds.

- lithium has 2 valence electrons in the product because it can make only 1 bond with hydrogen. beryllium has 4 valence electrons in the product because it can only make 2 bonds. it has fewer electrons than neon and can react with electron donors. each hydrogen has 2 valence electrons, the same as helium.

- the covalent bond formed between carbon and hydrogen is polar because carbon is more electronegative than hydrogen and it attracts the bonded electron pair closer to itself, thus giving hydrogen a partial positive charge.

- a hydrogen bond is a relatively weak bond that hydrogen atoms make with the electronegative atoms nitrogen, oxygen or fluorine.

so the best choice is a.) 2 calcium (ca) atoms