Question

Which of the respective orders are correct?
This question has multiple correct options.
(a) $N{{H}_{3}}>Sb{{H}_{3}}>As{{H}_{3}}>P{{H}_{3}}$: Thermal stability order
(b)Liquid hydrogen(c) $BeC{{O}_{3}}>SrC{{O}_{3}}>BaC{{O}_{3}}$: ${{k}_{p}}$value order during their thermal dissociation in closed container
(d) ${{O}_{2}}>K{{O}_{2}}>{{K}_{2}}{{O}_{2}}$: Magnetic moment order

Answer 1

Hint: Thermal stability depends on atomic size. Boiling point depends on surface area. Thermal stability of metal carbonate depends on the ionisation potential of the cations. Magnetic moment depends on the number of unpaired electrons.

Complete step by step solution:
Let us consider each option in the question
-In group 15 elements, the thermal stability of their hydrides decreases while moving down the group. The reason behind the decrease in stability is due to the increase in atomic size on going down the group. So, ammonia that is the hydride of nitrogen will be the most stable and the hydride of Sb will be the least stable. Hence, the order
is$N{{H}_{3}}>P{{H}_{3}}>As{{H}_{3}}>Sb{{H}_{3}}$. Option (a) is incorrect.

-Liquid hydrogen has a larger surface area than liquid helium. So, the boiling point of liquid hydrogen will be greater than liquid helium. So, option (b) is incorrect.

-Fajan’s rule states that, when the ion potential of metal cation increases then the thermal stability of its metal carbonate will decrease. So, there is increase in ${{k}_{p}}$ as the reaction moves forward, so, $MC{{O}_{3}}\rightleftharpoons MO+C{{O}_{2}}$

The ionisation potential for Beryllium is highest among strontium and barium. Ionisation potential for barium is the least. So, the correct decreasing order of ${{k}_{p}}$value during their thermal dissociation when kept in a closed container will be$BeC{{O}_{3}}>SrC{{O}_{3}}>BaC{{O}_{3}}$. The option (c) is correct.

-We know magnetic moment is given by the equation
Magnetic moment$=\sqrt{n(n+2)}$, where n=unpaired electrons. In the case of oxygen molecules, there are two unpaired electrons (determined from molecular orbital). So, n=2,
Magnetic moment $=\sqrt{2(2+2)}=\sqrt{8}$
In case of$K{{O}_{2}}$has one unpaired electron (one of the unpaired electrons of oxygen is involved in formation of bond with potassium), n=1. So
Magnetic moment$=\sqrt{1(1+2)}=\sqrt{3}$
For${{K}_{2}}{{O}_{2}}$, there are zero unpaired electrons (both the unpaired electrons of oxygen are involved in bond formation with potassium). So, n=0,
Magnetic moment=0

Thus, the decreasing order of magnetic moment is${{O}_{2}}>K{{O}_{2}}>{{K}_{2}}{{O}_{2}}$. (d) is the correct option.
Hence, correct orders are options (c) and (d).

Note: While moving down a group, the size of cations increases, so ionisation potential decreases. So, thermal dissociation of salts of the cations increases while going down the group and the stability decreases. But some salts of alkali and alkaline earth metals depend on their hydration energy and lattice energy.