Na 2 B 4 O 7 $$\buildrel {heat} \over \longrightarrow $$ X + NaBO 2 in the above reaction the product X is :

Na 2 B 4 O 7 $$\buildrel {\Delta} \over \longrightarrow $$ B 2 O 3 + 2NaBO 2 Product X is B 2 O 3

p-Block Elements — Page 5 | NEET Chemistry

Q41

The correct order of the mobility of the alkali metal ions in aqueous solution is

A Rb + > K + > Na + > Li +

B Li + > Na + > K + > Rb +

C Na + > K + > Rb + > Li +

D K + > Rb + > Na + > Li +

Correct Answer

Option A

Solution

Ionic radii of alkali metals in water follows the order Li + > Na + > K + > Rb + > Cs + Thus in aqueous solution due to larger ionic radius Li + has lowest mobility and hence the correct order of ionic mobility is Rb + > K + > Na + > Li +

Q42

A solid compound X on heating gives CO 2 gas and a residue. The residue mixed with water forms Y. On passing an excess of CO 2 through Y in water, a clear solution Z is obtained. On boiling Z, compound X is reformed. The compound X is

A Ca(HCO 3 ) 2

B CaCO 3

C Na 2 CO 3

D K 2 CO 3

Correct Answer

Option B

Solution

The given compound X must be CaCO 3 .

Q43

In which of the following processes, fused sodium hydroxide is electrolysed at a 330 o C temperature for extraction of sodium?

A Castner's process

B Down's process

C Cyanide process

D Both 'b' and 'c'

Correct Answer

Option A

Solution

In Castner’s process, for production of sodium metal, sodium hydroxide (NaOH) is electrolysed at temperature 330ºC.

Q44

Identify the incorrect statement from the following:

A Nitrogen can form

p \pi-p \pi

multiple bonds with itself.

B

\quad \mathrm{P}\left(\mathrm{C}_2 \mathrm{H}_5\right)_3

and

\mathrm{As}\left(\mathrm{C}_6 \mathrm{H}_5\right)_3

form

\mathrm{d} \pi-\mathrm{d} \pi

bond with transition metals.

C Phosphorus, arsenic and antimony show catenation property.

D Nitrogen can form

d \pi-p \pi

bond with oxygen.

Correct Answer

Option D

Solution

Both nitrogen and oxygen do not contain d-orbitals so it cannot form $\mathrm{d} \pi-\mathrm{p} \pi$ bond.

Phosphorus and Arsenic can form $\mathrm{d} \pi-\mathrm{d} \pi$ bond with transition metals.

Since both have vacant d -orbitals by which it can interact with transition metals and can involve in $\mathrm{d} \pi-\mathrm{d} \pi$ interaction.

Phosphorous, Arsenic and Antimony show catenation property.

Q45

Identify the incorrect statement from the following:

A Carbon has the ability to form

p \pi-p \pi

multiple bond with itself.

B

\mathrm{ECl}_3(\mathrm{E}=\mathrm{B}

and Al

)

is a monomer when

\mathrm{E}=\mathrm{B}

and a dimer when

\mathrm{E}=\mathrm{Al}

.

C The order of catenation property of Group 14 elements is

\mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}

.

D Oxygen exhibits only -2 oxidation state.

Correct Answer

Option D

Solution

(1) $C$ can form $p \pi-p \pi$ multiple bond with itself.

It is observed when it forms $\mathrm{C}=\mathrm{C}$ and $\mathrm{C} \equiv \mathrm{C}$ (2) $\mathrm{ECl}_3(\mathrm{E}=\mathrm{B}$ and Al $)$ $\mathrm{BCl}_3$ does not form dimer. $\mathrm{AlCl}_3$ can form dimer in $\mathrm{Al}_2 \mathrm{Cl}_6$ .

(3) Catenation property of group $14 \mathrm{C} \gg \mathrm{Si}>\mathrm{Ge} \approx \mathrm{Sn}$ .

(4) Oxygen exhibits oxidation state of 0 in $\mathrm{O}_2,-2$ in oxides, -1 in peroxides.

So this statement is incorrect.

Q46

Identify the incorrect statement from the following :

A The acidic strength of

\mathrm{HX}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}

and

\mathrm{I})

follows the order:

\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}

.

B Fluorine exhibits

-

1 oxidation state whereas other halogens exhibit

+1,+3,+5

and

+7

oxidation states also.

C The enthalpy of dissociation of

\mathrm{F}_2

is smaller than that of

\mathrm{Cl}_2

.

D Fluorine is stronger oxidising agent than chlorine.

Correct Answer

Option A

Solution

Let's analyze each statement to identify the incorrect one: Option A: The acidic strength of

\mathrm{HX}(\mathrm{X}=\mathrm{F}, \mathrm{Cl}, \mathrm{Br}

and

\mathrm{I})

follows the order:

\mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI}

. This statement is incorrect. The correct order of acidic strength for hydrohalic acids is

\mathrm{HF} Option B: Fluorine exhibits

-1

oxidation state whereas other halogens exhibit

+1, +3, +5

and

+7

oxidation states also. This statement is correct. Fluorine is the most electronegative element and can only exhibit

-1

oxidation state. Other halogens can exhibit positive oxidation states due to the availability of d-orbitals. Option C: The enthalpy of dissociation of

\mathrm{F}_2

is smaller than that of

\mathrm{Cl}_2

. This statement is correct. The bond dissociation energy of

\mathrm{F}_2

is indeed smaller than that of

\mathrm{Cl}_2$$ because the F-F bond is weaker due to significant electron-electron repulsion between the non-bonding electrons in the small fluorine molecule.

Option D: Fluorine is a stronger oxidising agent than chlorine.

This statement is correct.

Fluorine is the most electronegative element, which makes it the strongest oxidizing agent among the halogens.

Therefore, the incorrect statement is Option A .

Q47

Among Group 16 elements, which one does NOT show -2 oxidation state?

A O

B

\mathrm{Se}

C

\mathrm{Te}

D Po

Correct Answer

Option D

Solution

The elements in Group 16 of the periodic table, also known as the chalcogens, include oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and polonium (Po).

Generally, these elements are known to commonly exhibit a -2 oxidation state because they have six electrons in their outermost shell and can gain two electrons to complete their octet, thereby achieving a stable electronic configuration similar to the nearest noble gas.

Oxygen, being the most electronegative element in this group, predominantly shows a -2 oxidation state in most of its compounds, such as water ( $\mathrm{H_2O}$ ) and oxides like $\mathrm{SO_2}$ .

Selenium and Tellurium also typically exhibit the -2 oxidation state.

For instance, in compounds like $\mathrm{H_2Se}$ (hydrogen selenide) and $\mathrm{H_2Te}$ (hydrogen telluride), these elements are in the -2 state.

Polonium (Po), on the other hand, is different from the lighter chalcogens.

Being a metalloid with more metallic character, it does not favor the -2 oxidation state nearly as strongly as the other elements in its group.

Polonium most commonly exhibits +2 and +4 oxidation states in its compounds, like $\mathrm{PoO_2}$ and $\mathrm{PoCl_2}$ .

Due to the relativistic effects and its position in the periodic table, the -2 oxidation state is unstable and rare in polonium compounds.

Given this, the correct answer is Option D: Po.

Polonium does not commonly exhibit the -2 oxidation state like the other Group 16 elements.

Q48

The element expected to form largest ion to achieve the nearest noble gas configuration is

A

\mathrm{F}

B

\mathrm{N}

C

\mathrm{Na}

D

\mathrm{O}

Correct Answer

Option B

Solution

\mathrm{F}^{-1}, \mathrm{~N}^{-3}, \mathrm{Na}^{+} ~\& ~\mathrm{O}^{-2}

all ions are isoelectronic containing

10 e^{-}

\mathrm{Z}_{\text{eff }} \rightarrow \mathrm{Na}^{+}>\mathrm{F}^{-}>\mathrm{O}^{-2}>\mathrm{N}^{-3}

order of radius

\rightarrow \mathrm{N}^{-3}>\mathrm{O}^{-2}>\mathrm{F}^{-}>\mathrm{Na}^{+}

$\rightarrow$ Nitrogen to achieve Noble gas configuration it gain

3 e^{-}

, & form

\mathrm{N}^{-3}

Q49

Which of the following reactions is a part of the large scale industrial preparation of nitric acid?

A Cu(NO 3 ) 2 + 2NO 2 + 2H 2 O

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}}

4HNO 3 + Cu

B NaNO 3 + H 2 SO 4

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}}

NaHSO 4 + HNO 3

C 4NH 3 + 5O 2 (from air)

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}}

4NO + 6H 2 O

D 4HPO 3 + 2N 2 O 5

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}}

4HNO 3 + P 4 O 10

Correct Answer

Option C

Solution

4NH 3 + 5O 2 (from air)

\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{\mathrm{500\,K,\,9\,bar}}^{\mathrm{Pt}}}

4NO + 6H 2 O 2NO(g) + O 2 (g) $\rightleftharpoons$ 2NO 2 (g) 3NO 2 (g) + H 2 O(l) $\to$ 2HNO 3 (aq) + NO(g) This is industrial method of preparation of nitric acid.

Q50

Na 2 B 4 O 7

\overset{{heat}}\longrightarrow

X + NaBO 2 in the above reaction the product X is :

A NaB 3 O 5

B H 3 BO 3

C B 2 O 3

D Na 2 B 2 O 5

Correct Answer

Option C

Solution

Na 2 B 4 O 7

\overset{{\Delta}}\longrightarrow

B 2 O 3 + 2NaBO 2 Product X is B 2 O 3