The atomic radius of Ag is closest to

We first recall that the metallic (atomic) radius is the distance from the centre of an atom’s nucleus to the point where its outermost electrons spend most of their time in the metallic lattice. Numerical values for commonly-quoted metallic radii are:

$$r_{\text{Ni}} \approx 124\ \text{pm},\qquad r_{\text{Cu}} \approx 128\ \text{pm},\qquad r_{\text{Ag}} \approx 144\ \text{pm},\qquad r_{\text{Au}} \approx 144\ \text{pm},\qquad r_{\text{Hg}} \approx 150\ \text{pm}$$

Now we compare each option with the given radius of silver, $$r_{\text{Ag}} \approx 144\ \text{pm}.$$

For nickel, $$|r_{\text{Ag}}-r_{\text{Ni}}| = 144\ \text{pm}-124\ \text{pm}=20\ \text{pm},$$ a noticeable difference.

For copper, $$|r_{\text{Ag}}-r_{\text{Cu}}| = 144\ \text{pm}-128\ \text{pm}=16\ \text{pm},$$ still significant.

For gold, $$|r_{\text{Ag}}-r_{\text{Au}}| = 144\ \text{pm}-144\ \text{pm}=0\ \text{pm},$$ practically identical.

For mercury, $$|r_{\text{Ag}}-r_{\text{Hg}}| = 150\ \text{pm}-144\ \text{pm}=6\ \text{pm},$$ closer than Ni or Cu but not as close as Au.

Thus the element whose atomic radius is closest to that of silver is gold. This agreement is also expected from periodic trends because Ag and Au lie directly beneath each other in Group 11; the lanthanide contraction keeps Au’s radius from becoming much larger than Ag’s.

Hence, the correct answer is Option A.