For alkali metals, which one of the following trends is INCORRECT?

Question

For alkali metals, which one of the following trends is INCORRECT? (A) Ionization energy: Li > Na > K > Rb (B) Hydration energy: Li > Na > K > Rb (C) Atomic size: Li < Na < K < Rb (D) Density: Li < Na < K < Rb

Tardigrade Admin · Accepted Answer

In general, density increases on moving downward in a group but density of potassium

(K) is lesser than that of the sodium (Na).

This is because of the abnormal increase in atomic size on moving from Na

(86 p m)

to

K (227 p m)

.

Thus, the correct order of density is

L i < K < N a < R b

Q. For alkali metals, which one of the following trends is INCORRECT?

Ionization energy : $L i > N a > K > R b$

Hydration energy : $L i > N a > K > R b$

Atomic size : $L i < N a < K < R b$

Density : $L i < N a < K < R b$

In general, density increases on moving downward in a group but density of potassium

(K) is lesser than that of the sodium (Na).

This is because of the abnormal increase in atomic size on moving from Na $(86 p m)$ to $K (227 p m)$ .

Thus, the correct order of density is

$L i < K < N a < R b$

Q. For alkali metals, which one of the following trends is INCORRECT?

Ionization energy : Li>Na>K>Rb

Hydration energy : Li>Na>K>Rb

Atomic size : Li<Na<K<Rb

Density : Li<Na<K<Rb

In general, density increases on moving downward in a group but density of potassium (K) is lesser than that of the sodium (Na). This is because of the abnormal increase in atomic size on moving from Na (86pm) to K(227pm). Thus, the correct order of density is Li<K<Na<Rb

Ionization energy : $L i > N a > K > R b$

Hydration energy : $L i > N a > K > R b$

Atomic size : $L i < N a < K < R b$

Density : $L i < N a < K < R b$

In general, density increases on moving downward in a group but density of potassium

(K) is lesser than that of the sodium (Na).

This is because of the abnormal increase in atomic size on moving from Na $(86 p m)$ to $K (227 p m)$ .

Thus, the correct order of density is

$L i < K < N a < R b$