Tag: chemistry

Questions Related to chemistry

Hydorgen forms binary compounds called hybrides with several elements in the periodic table.Regarding these hydrides the wrong statement 

  1. Among lanthanoids europioum and ytterbium form ionic hydrides

  2. Density of ionic hydrides is less while the density of interstitial hydrides is more than the metals from which they are formed

  3. The rate of reaction of alkali metals towards hydrogen decreases from lithium to ceasium

  4. Hydrogen gas can be purified by the phenomenon of Occlusion of hydrogen by palladium


Correct Option: A

The hydride ion, $H^{-}$ is a stronger base than its hydroxide ion $OH^{-}$. Which of the following reactions will occur, if sodium hydride $(NaH)$ is dissolved in water?

  1. $H^{-} _{\left ( aq \right )}+H _{2}O\rightarrow H _{3}O^{-}$

  2. $H^{-} _{\left ( aq \right )}+H _{2}O _{\left ( l \right )}\rightarrow OH^{-}+H _{2}$

  3. $H^{-}+H _{2}O _{(l)}\rightarrow\ no\ reaction$

  4. $None\ of\ the\ above$


Correct Option: B
Explanation:

Sodium hydride $(NaH)$ react with water to form hydrogen gas and the sodium hydroxide $(NaOH)$.
$NaH+H _2O\rightarrow H _2+NaOH$
i.e., $Na^+, H^-$ and $H^+, OH^- $ give rise to $H _2$ and $NaOH$. So, it is similar to option B.

According to recent views, which is the current representation of hydrated proton in aqueous solution?

  1. $H^{+}$

  2. $H _{9}O _{5}^{+}$

  3. $H _{9}O _{4}^{+}$

  4. $H _{3}O^+$


Correct Option: D
Explanation:

The hydrated proton is represented as $H _3O^+$. Hence, the correct option is $D$.

Which of the following metals directly combine with hydrogen gas to give a hydride? 

  1. Au

  2. Ni

  3. Ca

  4. Cu


Correct Option: C
Explanation:

Amongst, $Au$, $Ni$, $Ca$ and $Cu$, $Ca$ is the most reactive metal. So it easily combines with $H _2$ to form hydride while the other metals do not combine directly.


The reaction can be given as:-
$Ca+H _2\longrightarrow CaH _2$

What is the trend of boiling points of hydrides of N, O, and F?

  1. Due to lower molecular masses $NH _{3}, H _{2}O$ and HF have lower boiling points than those of the subsequent group member hydrides

  2. Due to higher electronegativity of N, O and F; $NH _{3}, H _{2}O$ and HF show hydrogen bonding and hence higher boiling points than the hydrides of their subsequent group members.

  3. There is no regular trend in the boiling points of hydrides

  4. Due to higher oxidation states of N, O and F, the boiling points of $NH _{3}, H _{2}O$ and HF are higher than the hydrides of their subsequent group members.


Correct Option: B
Explanation:

The boiling point increases with increase in molecular masses. But in the case of $N, O$ and $F$ they show the higher boiling point in molecules. 

This is due to the higher electronegativity of $N, O$ and $F$ which leads to hydrogen bonding in $NH _3, H _2O$ and $HF$ due to which the boiling point of these increases & is higher than the hydrides of their subsequent group members.

Which of the following hydrides is electron-precise hydride?

  1. $B _2H _6$

  2. $NH _3$

  3. $H _2O$

  4. $CH _4$


Correct Option: D
Explanation:
Methane $CH _4$ is electron-precise hydride.
Diborane $B _2H _6$ is electron-deficient hydride
Ammonia $NH _3$ and water $H _2O$ are electron-rich hydrides.

Elements of group 14 form electron-precise (having required number of electrons to write the Lewis structure ) form precise hydrides.

Non-stoichiometric hydrides are produced by:

  1. palladium, vanadium

  2. manganese, lithium

  3. nitrogen, filorine

  4. carbon, nockel


Correct Option: A
Explanation:

The hydrogen deficient compounds formed by the reaction of $d-block$ and $f-block$ elements with dihydrogen are called Non-stoichiometric compounds. 

The d-block and f-block element form non-stoichiometric hydride because of the vacant d- and f-orbitals along with the small size.
Their elemental composition proportions cannot be represented in integers. They disobey the law of constant composition. Among the elements given, only vanadium and palladium form non-stoichiometric hydrides.


Answer: (A) palladium, vanadium

From group 6 only one metal forms hydride. This metal is:

  1. Mo

  2. W

  3. Cr

  4. Co


Correct Option: C
Explanation:

The low affinity of the elements of the group $6,7,8$ and $9$ towards hydrogen in their normal oxidation states prevent the formation of hydrides. This is called as hydride gap. 

In the group, $6$, Chromium $(Cr)$ is an exception as it is the only group $6$ element to form a hydride.

Answer: $(Cr)$

Only one element of ________ forms hydride. 

  1. group 6

  2. group 7

  3. group 8

  4. group 9


Correct Option: A
Explanation:

The only element of group $6$ i.e. Chromium $(Cr)$ forms hydride. The other elements do not form hydride due to low affinity towards hydrogen.


Answer: (A) group $6$

Elements of which of the following group (s) of periodic table do not form hydrides?

  1. Groups 7, 8, 9

  2. Group 13

  3. Groups 15, 16, 17

  4. Group 14


Correct Option: A
Explanation:

The elements of the groups $7,8,9$ show low affinity towards hydrogen due to which they lack the tendency to form hydrides.


Answer: (A) Groups $7,8,9$