Which one of the following is most likely to gain electrons when forming an ion, based on the natural tendency of the element? A Ni
B S
C Na
D Cr
E Be

Answer

B) Sulphur (S)

Explanation

Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.

Now, let us understand what is a metal and a non-metal.

So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.

Now let us also look at the electronic configuration of Sulphur to get the picture more clearly

so here the p-subshell is incomplete and is in need of 2 electrons.

Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.

To learn more about Ion Formation

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Option B. S

Explanation:

All of the options except sulphur, S is metal.

Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.

Sulphur, S has atomic number of 16 with electronic configuration as:

S (16) => 1s² 2s²2p⁶ 3s²3p⁴

From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.

Therefore, sulphur, S will gain two electrons to form ion.

As stated earlier, the rest option given are all metals which will form ion by losing electron(s).

c. 75.7

i think that is the right

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