Question

Arrange the following in order of increasing size: K, K+, Ar and Ca. Put charges in parentheses.

Answer 1

Answer:

Well, when we speak of atomic or ionic size, reasonably, we speak of the radius described by the valence electron(s)....

Explanation:

And for NEUTRAL atoms, reasonably, atomic size decreases across a Period, from left to right as we face the Table, and INCREASES down a Group.

Why? Well, atomic size is influenced by two factors: (i) nuclear charge,  

Z

; and (ii), shielding by other electrons. Incomplete electronic shells shield the nuclear charge very ineffectively. And thus addition of protons, positive charges, to the nucleus across the Period greatly contracts the electronic radius. However, once a shell is filled, shielding of the nuclear charge from the valence electrons becomes reasonably effective...and the valence electron lies in a shell that is farther removed from the nuclear core.

And so we gots  

C

l

−

,  

A

r

, and  

K

+

, as isolated particles... With respect to the parent atoms, potassium would be larger than argon, the which would be smaller than chlorine. But when we go to the ions, chloride,  

C

l

−

>>

A

r

...and potassium cation,  

K

+

, the which has a greater nucular charge, and whose valence electron has been REMOVED, the ionic radius should be smaller than both....

And now this is your job. AS a chemist, as a physical scientist, you interrogate data...find from somewhere estimates of the the atomic and ionic radii, and confirm that I have not been telling pork pies.

Answer 2

Explanation:

Atomic number of potassium (K) is 19 and its electronic configuration in both neutral and ionic form are as follows.

For K: $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}$

For $K^{+}$ :  $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$

Atomic number of argon is 18 and its electronic configuration is as follows.

                   $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$

Atomic number of calcium is 20 and its electronic configuration is as follows.

                $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}$

Hence, more is the number of electrons present in an atom more will be its atomic size.

Thus, we can conclude that order of increasing size is as follows.

                        $K^{+}$ = Ar < K < Ca