Question

Which of the following pairs has the same size?

• A. ${\mathrm{Fe}}^{2+},{\mathrm{Ni}}^{2+}$
• B. ${\mathrm{Zr}}^{4+},{\mathrm{Ti}}^{4+}$
• C. ${\mathrm{Zr}}^{4+},{\mathrm{Hf}}^{4+}$
• D. ${\mathrm{Zn}}^{2+},{\mathrm{Hf}}^{4+}$

Answer 1

The correct option is C

${\mathrm{Zr}}^{4+},{\mathrm{Hf}}^{4+}$

The explanation for the correct answer

(C) ${\mathbf{Zr}}^{\mathbf{}\mathbf{4}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Hf}}^{\mathbf{}\mathbf{4}\mathbf{+}}$

Size and Periodic trend

• Down the group, the size of the atom increases.
• Across the period, the Size of the atom decreases.

Lanthanide contraction

• The steady decrease in the atomic size and ionic size of lanthanoids with an increase in atomic number is known as Lanthanide contraction.
• It is caused due to poor shielding effect of the $4$f electrons.
• ${\mathrm{Zr}}^{4+},{\mathrm{Hf}}^{4+}$ belongs to the same group and has the same size due to lanthanide contraction.
• The addition of $14$lanthanide elements before $\mathrm{Hf}$where electrons are added into the f-subshell results in poorly shielded outer electrons and thus contraction occurs.

Hence, Option (C) ${\mathbf{Zr}}^{\mathbf{}\mathbf{4}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Hf}}^{\mathbf{}\mathbf{4}\mathbf{+}}$is correct.

The explanation for the incorrect answer

(A) ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Ni}}^{\mathbf{2}\mathbf{+}}$

• $\mathrm{Fe}$ belongs to Group$8$ and $\mathrm{Ni}$belongs to Group$10$. Both are in the same period.
• Across the period, the size of an atom decreases.
• So ${\mathrm{Fe}}^{2+},{\mathrm{Ni}}^{2+}$don't have the same size.

(B) ${\mathbf{Zr}}^{\mathbf{}\mathbf{4}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Ti}}^{\mathbf{}\mathbf{4}\mathbf{+}}$

• $\mathrm{Zr}$ and $\mathrm{Ti}$belongs to the same Group$4$.
• Down the Group, the size of an atom increases.
• So ${\mathrm{Zr}}^{4+},{\mathrm{Ti}}^{4+}$don't have the same size.

(D) ${\mathbf{Zn}}^{\mathbf{2}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Hf}}^{\mathbf{}\mathbf{4}\mathbf{+}}$

• $\mathrm{Zn}$belongs to Group$12$ and $\mathrm{Hf}$belongs to Group$4$. Both are in different periods.
• So ${\mathrm{Zn}}^{2+},{\mathrm{Hf}}^{4+}$don't have the same size.

Therefore, Option (C) ${\mathbf{Zr}}^{\mathbf{}\mathbf{4}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{Hf}}^{\mathbf{}\mathbf{4}\mathbf{+}}$is correct.