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Which of the following pairs has the same size?
A
$Zn^{2+}$, $Hf^{4+}$
B
$Fe^{2+}$, $Ni^{2+}$
C
$Zr^{4+}$, $Ti^{4+}$
D
$Zr^{4+}$, $Hf^{4+}$
Detailed Solution
Zr (4d series) and Hf (5d series) belong to the same group; normally the size should increase from Zr to Hf.
But the 14 lanthanoids come between them. The 4f electrons shield the nuclear charge poorly, so the atomic and ionic radii decrease steadily across the lanthanoids (lanthanoid contraction).
This contraction cancels the expected increase in size from the 4d to the 5d series.
As a result $Zr^{4+}$ and $Hf^{4+}$ have almost identical radii (about 72 pm and 71 pm), and the two elements have very similar properties.
$Ti^{4+}$ (3d series) is clearly smaller than $Zr^{4+}$, and $Fe^{2+}$ and $Ni^{2+}$ differ because of increasing nuclear charge across the period.
Hence the pair with the same size is $Zr^{4+}$, $Hf^{4+}$.
But the 14 lanthanoids come between them. The 4f electrons shield the nuclear charge poorly, so the atomic and ionic radii decrease steadily across the lanthanoids (lanthanoid contraction).
This contraction cancels the expected increase in size from the 4d to the 5d series.
As a result $Zr^{4+}$ and $Hf^{4+}$ have almost identical radii (about 72 pm and 71 pm), and the two elements have very similar properties.
$Ti^{4+}$ (3d series) is clearly smaller than $Zr^{4+}$, and $Fe^{2+}$ and $Ni^{2+}$ differ because of increasing nuclear charge across the period.
Hence the pair with the same size is $Zr^{4+}$, $Hf^{4+}$.