Answer
See the explanation
Work Step by Step
a. $V,V^{2+},V^{3+},V^{5+}$:
The order of decreasing size for this group is:
$V > V^{2+} > V^{3+} > V^{5+}$
Explanation:
As the oxidation state of vanadium (V) increases, the number of electrons decreases, and the effective nuclear charge increases. This leads to a decrease in the atomic/ionic radius.
b. $Na^{+},K^{+},Rb^{+},Cs^{+}$:
The order of decreasing size for this group is:
$Cs^{+} > Rb^{+} > K^{+} > Na^{+}$
Explanation:
The alkali metal cations ($Na^{+}, K^{+}, Rb^{+}, Cs^{+}$) have the same electron configuration (1s^2 2s^2 2p^6) but different nuclear charges. As the atomic number increases, the atomic/ionic radius also increases due to the increased number of electron shells.
c. $Te^{2-},I^{-},Cs^{+},Ba^{2+}$:
The order of decreasing size for this group is:
$Te^{2-} >I^->Cs^{+} > Ba^{2+}$
Explanation:
The size of the ions depends on their charge and the number of electron shells. Larger anions ($Te^{2-}, I^{-}$) have a larger radius compared to smaller cations ($Cs^{+}, Ba^{2+}$).
d. $P,P^{-},P^{2-},P^{3-}$:
The order of decreasing size for this group is:
$P^{3-} > P^{2-} > P^{-} > P$
Explanation:
As the negative charge on the phosphorus ion increases, the effective nuclear charge decreases, leading to an increase in the atomic/ionic radius.
e. $O^{2-},S^{2-},Se^{2-},Te^{2-}$:
The order of decreasing size for this group is:
$Te^{2-} > Se^{2-} > S^{2-} > O^{2-}$
Explanation:
The size of the chalcogen anions ($O^{2-}, S^{2-}, Se^{2-}, Te^{2-}$) increases as the atomic number increases, as the number of electron shells also increases.
