The ionic radii (in
) of
and
, respectively, are: Option B:
for
for
The ionic radii (in
) of
and
, respectively, are: Option B:
for
for
Statement - I is incorrect Be(OH)2 dissolve in alkali due to it's amphoteric nature.
Statement - II is correct Solubility of alkaline earth metal hydroxide in water increases down the group due to rapid decreases in lattice energy as compared to hydration energy.
.tg .tg 48 116 96 96 77
Among , the metal having highest is Ge and lowest is .
Most stable oxidation state of Ge is +4 and In is +3 .
The element (M) is potassium (K).
It reacts with O2 to form KO2, which is paramagnetic in nature.
All other elements form oxides or peroxides which are diamagnetic in nature.
Statement I is correct because the 4f and 5f series elements, commonly referred to as the lanthanides and actinides, respectively, have unique properties and configurations that distinguish them from other elements.
Placing them separately in the Periodic table helps to maintain the orderly classification based on atomic numbers and avoids disrupting the pattern established by the periodic law.Statement II is incorrect because s-block elements, which include Group 1 (alkali metals) and Group 2 (alkaline earth metals), are indeed highly reactive.
They typically do not exist in their pure form in nature due to their reactivity.
Instead, they are mostly found in compound forms, combined with other elements.
For instance, sodium (Na) and potassium (K) from Group 1 react with water and are usually found as salts, not in their metallic, pure state.
Melting point :
Ionic radius
Atomic radius (in
) :
(A) The process of adding an to a neutral gaseous atom is not always exothermic it may be exothermic or endothermic.
In Be 2s subshell in fully filled So, need high energy to remove as compared to B . due to partially positive charge So, EN of (E) Cs is most electropositive.
On heating with excess of air
and
forms following oxides
1st I.E. of Be > B In case of Be, electron is removed from 2s orbital which has more penetration power, while in case of B electron is removed from 2p orbital which has less penetration power. 2p electron of B is more shielded from nucleus by the inner electrons than 2s electrons of Be It is easier to remove 2p electron than 2s electron.