The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders ?

CsH > RbH > KH > NaH > LiH
KH > NaH > LiH > CsH > RbH
NaH > LiH > KH > RbH >CsH
LiH > NaH > KH > RbH >CsH

Solution

The stability of alkali metal hydrides decreases from Lito Cs. It is due to the fact that M–H bonds become weaker with increase in size of alkali metals as we move down the group from Li to Cs. Thus the order of stability of hydrides is
LiH > NaH > KH > RbH > CsH
i.e. option(d) is correct answer.

In which of the following the hydration energy is higher than the lattice energy?

MgSO₄
RaSO₄
SrSO₄
BaSO₄

Solution

In alkaline earth metals ionic size increases down the group. The lattice energy remains constant because sulphate ion is so large, so that small change in cationic sizes do not make any difference. On moving down the group the degree of hydration of metal ions decreases very much leading to decrease in solubility
BeSO₄ > MgSO₄ > CaSO₄ > SrSO₄ > BaSO₄

Solution

On being placed in water, sodium peroxide not only produces an alkaline solution but also some bubbles. If we assume that the peroxide ion picks up two protons from water to produce a compound that can be seen as the dibasic conjugate acid of peroxide ion and then this compound undergoes a redox disproportion.Using the above information complete the following equation.
Na₂O₂(s) + H₂O(l) ⟶ (A) + (B)
(A) and (B) are

H₂O₂ and NaOH
H₂O and O₂
NaOH and O₂
Na₂O and NaOH

Solution

Which is not correctly matched
(1)Basic strength Cs₂O < Rb₂O < K₂O < Na₂O < Li₂O of oxides
(2)Stability of Na₂O₂ < K₂O₂ < Rb₂O₂ < Cs₂O₂ peroxides
(3)Stability of LiHCO₃ < NaHCO₃< KHCO₃ bicarbonates < RbHCO₃ < CsHCO₃
(4)Melting point NaF < NaCl< NaBr < NaI

1 and 4
1 and 3
1 and 2
2 and 3

Solution

Basic strength of the oxides increase in the order Li₂O < Na₂O < K₂O < Rb₂O < Cs₂O. The increase in basic strength is due to the decrease in I.E. and increase in electropositive character.The melting points of the halides decrease in the order NaF > NaCl > NaBr > NaI,as the size of the halide ion increases. The decrease in melting point is due to increase in the covalent character with increase in the size of an ion according to Fajan's rule.

A metal ‘M’reacts with N₂ to give a compound ‘A’ (M₃N).‘A’ on heating at high temperature gives back ‘M’ and ‘A’;on reacting with H₂O gives a gas B. ‘B’ turns CuSO₄ solution blue on passing through it. M and B can be

Al and NH₃
Li and NH₃
Na and NH₃
Mg and NH₃

Solution

Magnesium form Mg²⁺ and Mg⁺ because :

ionic radius of Mg(II) is smaller than of Mg(I).
hydration energy of divalent magnesium ion is higher.
magnesium (II) carbonate is insoluble in water.
generally higher oxidation states are preferred by metals.

Solution

Statement (b) is correct.

The melting point of lithium (181°C) is just double the melting point of sodium (98°C) because –

down the group, the hydration energy decreases
down the group, the ionization energy decreases
down the group the cohesive energy decreases
None of these

Solution

The atom becomes larger on descending the group, so the bonds becomes weaker (metallic bond), the cohesive force/energy decreases and accordingly melting point also decreases.

Select the correct statements :
I.Cs⁺ is more highly hydrated that the other alkali metal ions
II.Among the alkali metals Li, Na, K and Rb, lithium has the highest melting point
III.Among the alkali metals only lithium forms a stable nitride by direct combination with nitrogen

I, II and III
I and II
I and III
II and III

Solution

Amongst alkali metal Li ions are highly hydrated.

The correct order of radii is –

Li < Be < Mg
H⁺ < Li⁺ < H^-
Mn³⁺ < Mn²⁺ < Mn⁺⁷
K⁺ > Cl^- > S^2-

Solution

H⁺ cation is smaller than parent H^-atom. H^- and Li⁺ are isoelectronic species so Li+with higher nuclear charge(i.e.3) will be smaller than H–having smaller nuclear charge(i.e. 1).

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